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Arbane
2014-12-20, 02:36 PM
Since it's taken over about half of the 'DM's who want you to Roleplay but can only say "No."' thread, I donate this thread as the place to argue about the Monty Hall Problem and similar brain-breaking weirdness in probability theory.

Domochevsky
2015-01-18, 02:39 AM
So I have to wonder... if you don't know the outcome behind the three doors, why does it matter that Monty knows the outcome? You are the one who picks and you don't know what's behind them.

So you have a 1/3 chance to get it right on the first choice and a 1/2 chance to get it right on the second one. (Even if you choose to stick with your initial pick, which is still a choice made.)

Knaight
2015-01-18, 02:54 AM
So I have to wonder... if you don't know the outcome behind the three doors, why does it matter that Monty knows the outcome? You are the one who picks and you don't know what's behind them.

So you have a 1/3 chance to get it right on the first choice and a 1/2 chance to get it right on the second one. (Even if you choose to stick with your initial pick, which is still a choice made.)

Monty needs to know the door with the right thing behind it so that they don't open it. The door revealed after the first choice is the one that doesn't have the correct answer, unless you picked it right originally (1/3 chance), in which case it's either. That detail there is what makes switching equivalent to picking either of the two doors, because the one that Monty didn't open has to have the good thing if either of the two did, which makes it a 2/3 chance instead of a 1/3 chance, and thus a better choice.

Brother Oni
2015-01-18, 06:01 AM
Further to Knaight's comment, it also depends on the circumstances in which Monty opens the other door. I've seen some variants of the problem where Monty only opens the other wrong door if you've picked correctly the first time and the contestant isn't aware of this.

Pinnacle
2015-01-18, 11:54 AM
You had a 2/3 chance to get it wrong in the first place. The host opening a door doesn't retroactively change that.

If you got in wrong the first time, which you probably did, the unopened door is the right one.
If the host doesn't know the right answer, opening a door does not demonstrate knowledge, it just eliminates an option and changes the number from 2 to 3--which means different odds.

prufock
2015-01-18, 01:53 PM
The Monty Haul problem is easy to prove with a simple matrix.

Initial Door Chosen
Switch
No Switch

Right
Goat
Car

Wrong
Car
Goat

Wrong
Car
Goat

There are six outcomes. Since initially there are 2 wrong doors and only 1 right door, you are twice as likely to win if you switch. Switching wins 2/3 times, not switching wins 1/3 times.

Beta Centauri
2015-01-18, 04:35 PM
Since it's taken over about half of the 'DM's who want you to Roleplay but can only say "No."' thread, I donate this thread as the place to argue about the Monty Hall Problem and similar brain-breaking weirdness in probability theory. I haven't seen that thread and couldn't find it. Can you point me to it? Thanks.

Lord Torath
2015-01-19, 04:42 PM
Here:DM's who want you to Roleplay but can only say "No." (http://www.giantitp.com/forums/showthread.php?385992-DM-s-who-want-you-to-Roleplay-but-can-only-say-quot-No-quot&highlight=DMs+who+can+only+say+No)

Beta Centauri
2015-01-19, 05:19 PM
Here:DM's who want you to Roleplay but can only say "No." (http://www.giantitp.com/forums/showthread.php?385992-DM-s-who-want-you-to-Roleplay-but-can-only-say-quot-No-quot&highlight=DMs+who+can+only+say+No) Looks like it has ended, but thanks anyway.

JusticeZero
2015-01-19, 06:03 PM
Easy to demonstrate if you skew the numbers further. There are 20 doors. After you choose a door, Monty opens eighteen doors with goats behind them, leaving only your original choice and the one he didn't open..

gom jabbarwocky
2015-01-19, 06:10 PM
Easy to demonstrate if you skew the numbers further. There are 20 doors. After you choose a door, Monty opens eighteen doors with goats behind them, leaving only your original choice and the one he didn't open..

I am stealing this charming bit of insanity for my next Paranoia game...

jaydubs
2015-01-19, 06:19 PM
Might as well throw in some other similar problems.

Problem 1: Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Problem 2: Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

Edit: I labeled problem 1 and problem 2 for easier reference.

Gavran
2015-01-19, 06:49 PM
Might as well throw in some other similar problems.

Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

... okay, I'll bite. Given the problem that started this thread I'm reasonably confident that by going with "what makes sense" I'll be choosing wrong, but hey, first step to learning.

In both cases the only unknown is the sex of one child, which can be either a match or not. Therefore, I want to say that there is a 50% chance that both children are the same sex.

Now, following your lead I shall present the other similar "problem" that I'm aware of:

Suppose you're at a gathering of roughly thirty people. One member of this gathering comes up to you and says "I bet you that there are at least two people here who share a birthday." Do you think it's more likely that he is right, or that none of the thirty people have the same birthday? How likely do you think it is?

jaydubs
2015-01-19, 07:53 PM
... okay, I'll bite. Given the problem that started this thread I'm reasonably confident that by going with "what makes sense" I'll be choosing wrong, but hey, first step to learning.

In both cases the only unknown is the sex of one child, which can be either a match or not. Therefore, I want to say that there is a 50% chance that both children are the same sex.

Your instinct is correct - the seemingly obvious answer is not the correct one. Well, at least for the second problem.

In the first problem, it is indeed 50/50. The fact that the older child is a girl doesn't effect the chance that the younger child is a girl.

In the second problem, the chance that both children are boys is 1/3. Since we know at least one of the children is a boy, the possibilities are:

1) The elder child is a boy, the younger child is a girl.
2) The elder child is a girl, the younger child is a boy.
3) They are both boys.

So, 1 out of 3.

Now, following your lead I shall present the other similar "problem" that I'm aware of:

Suppose you're at a gathering of roughly thirty people. One member of this gathering comes up to you and says "I bet you that there are at least two people here who share a birthday." Do you think it's more likely that he is right, or that none of the thirty people have the same birthday? How likely do you think it is?

Not accounting for leap years, and assuming there are no shenanigans (like he already knows one way or another), I'd say the odds of no one having the same birthday is:

364! / 345! / (365^19)

But I don't have a decent calculator on hand at the moment.

The Random NPC
2015-01-19, 08:13 PM
Suppose you're at a gathering of roughly thirty people. One member of this gathering comes up to you and says "I bet you that there are at least two people here who share a birthday." Do you think it's more likely that he is right, or that none of the thirty people have the same birthday? How likely do you think it is?

Not accounting for leap years, and assuming there are no shenanigans (like he already knows one way or another), I'd say the odds of no one having the same birthday is:

364! / 345! / (365^19)

But I don't have a decent calculator on hand at the moment.

Birthday paradox (http://en.wikipedia.org/wiki/Birthday_problem)! It's around 60-70% if I remember correctly.

Jeff the Green
2015-01-20, 07:07 AM
Might as well throw in some other similar problems.

Problem 1: Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Problem 2: Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

Edit: I labeled problem 1 and problem 2 for easier reference.

I wonder if people who have studied genetics are better at this, because that's how I knew the answer. Problem 1 is fairly simple. Problem 2 isn't, but it's equivalent to the genetics of munchies (http://en.wikipedia.org/wiki/Munchkin_cat). Basically, there's two versions of a gene M and m. Two Ms results in a normal cat, two ms result in a miscarriage, and one of each results in a munchy. So when you breed two munchies, you end up not with the 1:2:1 ratio (50% munchies) you normally see in Mendelian genetics, but just a 1:2 ratio (67% munchies).

Birthday paradox (http://en.wikipedia.org/wiki/Birthday_problem)! It's around 60-70% if I remember correctly.

70.6% (http://www.wolframalpha.com/input/?i=1-365%21%2F%28365%5E30%28365-30%29%21%29), actually. It hits near certainty around 50.

hymer
2015-01-20, 07:31 AM
1) The elder child is a boy, the younger child is a girl.
2) The elder child is a girl, the younger child is a boy.
3) They are both boys.

So, 1 out of 3.

Why not

1) The elder child is a guaranteed boy, the younger child is a girl.
2) The elder child is a girl, the younger child is a guaranteed boy.
3) The elder child is a guaranteed boy, the younger child is a boy.
4) The elder child is a boy, the younger child is a guaranteed boy.

So 2 out of 4.

?

SpectralDerp
2015-01-20, 08:05 AM
Conditional probabilities work like this:

P(Both children are boys | At least one child is a boy) = P(Both children are boys and at least one child is a boy) / P(At least one child is a boy)

Assuming that it's equally likely for a child to be a boy or a girl, that every child is either a boy or a girl and that the sexes of both children are independent, we see that

P(At least one child is a boy) = 1 - P(the elder child is a girl and the younger child is a girl) = 1 - P(the elder child is a girl) * P(the younger child is a girl) = 1 - (1/2) * (1/2) = 3/4

and

So

[B]P(Both children are boys | At least one child is a boy) = (1/4) / (3/4) = 1/3

What you are doing is trying to calculate

P(Both children are boys | At least one child is a boy) = P(Both children are boys | the elder child is a boy or the younger child is a boy)

and make use of

P(Both children are boys | the elder child is a boy) = P(Both children are boys and the elder child is a boy) / P(the elder child is a boy) = (1/4) / (1/2) = 1/2

and

P(Both children are boys | the younger child is a boy) = P(Both children are boys and the younger child is a boy) / P(the younger child is a boy) = (1/4) / (1/2) = 1/2

But these Probabilities don't "interact" the way you think they do because

P(Both children are boys | the elder child is a boy) * P(the elder child is a boy) + P(Both children are boys | the younger child is a boy) * P(the younger child is a boy) = P(Both children are boys)

The mistake in your calculation is the false equation

P(Both children are boys | the elder child is a boy) * P(the elder child is a boy) + P(Both children are boys | the younger child is a boy) * P(the younger child is a boy) = P(Both children are boys | the elder child is a boy or the younger child is a boy)

Jeff the Green
2015-01-20, 08:16 AM
Why not

1) The elder child is a guaranteed boy, the younger child is a girl.
2) The elder child is a girl, the younger child is a guaranteed boy.
3) The elder child is a guaranteed boy, the younger child is a boy.
4) The elder child is a boy, the younger child is a guaranteed boy.

So 2 out of 4.

?

You're treating "guaranteed boy" and "boy" as separate categories with a background probability of 33% each. Continuing your pattern to get all the possibilities, there's also guaranteed boy-guaranteed boy, boy-boy, boy-girl, girl-boy, and girl-girl. Note that two thirds of all children are boys.

The proper way is to lay out all the possible combinations of boy and girl, and then eliminate the ones that don't fit.

boy-boy
boy-girl

girl-boy
girl-girl

Lord Torath
2015-01-20, 08:25 AM
Why not

1) The elder child is a guaranteed boy, the younger child is a girl.
2) The elder child is a girl, the younger child is a guaranteed boy.
3) The elder child is a guaranteed boy, the younger child is a boy.
4) The elder child is a boy, the younger child is a guaranteed boy.

So 2 out of 4.

?I'd also go with: The gender and age of one child have no effect on the gender of the other child. One child's gender is set, and the other child has a (roughly) 50% chance of matching the set gender.

Edit: Effectively ninja's by Jeff the Green. :smallwink:

hymer
2015-01-20, 08:39 AM
@ SpectralDerp: Sorry, that's all Greek to me. :smallfrown:

@ Jeff the Green: Let me rephrase it. There are two children, at least one of whom is a boy. Turn that into, there are two children, one whose name is Tim (and presume that Tim must be a boy). Same thing mathematically speaking, right?

So we have:

1: Tim is the eldest, he has a younger sister.
2: Tim is the eldest, he has a younger brother.
3: Tim is the youngest, he has an older sister.
4: Tim is the youngest, he has an older brother.

That should cover all permutations, and all be equally likely, right?

Jacob.Tyr
2015-01-20, 08:40 AM
I'd also go with: The gender and age of one child have no effect on the gender of the other child. One child's gender is set, and the other child has a (roughly) 50% chance of matching the set gender.

Edit: Effectively ninja's by Jeff the Green. :smallwink:

Without problem one, this trick doesn't really work. They are independent, but the purpose seems to be to get us hung up on ordering and thinking everything is conditional.

The probability that the other is a boy is 50%.

SpectralDerp
2015-01-20, 10:39 AM
@ SpectralDerp: Sorry, that's all Greek to me. :smallfrown:

@ Jeff the Green: Let me rephrase it. There are two children, at least one of whom is a boy. Turn that into, there are two children, one whose name is Tim (and presume that Tim must be a boy). Same thing mathematically speaking, right?

So we have:

1: Tim is the eldest, he has a younger sister.
2: Tim is the eldest, he has a younger brother.
3: Tim is the youngest, he has an older sister.
4: Tim is the youngest, he has an older brother.

That should cover all permutations, and all be equally likely, right?

Call the children Terry and Robin. Then the options are

Terry is a boy, Robin is a boy
Terry is a boy, Robin is a girl
Terry is a girl, Robin is a boy
Terry is a girl, Robin is a girl

The information "One of them is a boy" is incompatible with one of these options, the information "Terry is a boy" is incompatible with two.

[I have decided to remove this part because it is confusing, below is a better explanation]

If you know that one of them is named Tim and therefore a boy, then you might know more than just "one of them is a boy". If you know that only one is named Tim, then you know there's a boy named Tim and another child. With this knowledge, the probability of both children being boys is 1/2. If you don't know if both are named Tim, you can't split them up as Tim and not Tim.

hymer
2015-01-20, 11:03 AM
@ SpectralDerp: You've been very patient with me, thank you! And I think I get it now. If there are 400 cases of two siblings, they are statistically 100 each of BB, BG, GB and GG. But one of those, GG, doesn't fit the description and must be discounted. In the 300 remaining cases, 200 are a combination of boy and girl, giving a 2/3 chance that the other sibling is a girl if a random couple is picked.

Lord Torath
2015-01-20, 12:22 PM
To reiterate
The gender and age of one child have no effect on the gender of the other child.How is this different from saying "I flip a perfectly fair coin twice. One time it came up heads. What are the chances it came up heads both times?" The coin doesn't care what any previous or future flips may be. It has a 50% chance of coming up heads. So the unknown flip has a 50% of matching the known flip. Have I got that wrong?

And if so, how is Problem 1 (older child's gender is known) different from Problem 2 (random child's gender is known) since in both cases, the unknown child's gender is not affected by either the age or gender of the known child?

SpectralDerp
2015-01-20, 12:38 PM
The probability that the second coin (and thus both coins) ends up showing heads, given that the first one did, is 1/2, because the two coin throws are independant of each other.

The probability that both coins end up showing heads given that the first or second coin did is not 1/2, because the result of both coins is not independant of the result of the second one.

Look at my Terry and Robin example. "Terry is a boy" is a stronger restriction than "Terry or Robin is a boy". The first one does not tell you anything about Robin, the second one does.

Talakeal
2015-01-20, 12:52 PM
It looks to me like the second chld has fifty fifty odds of being a girl or a boy.

Looking it up on the internet the answer is dependant upon selection method, as if you simply take a random sample of families with two children and then eliminate all of them with two girls you are left with only 1 in 3 being boy girl.

However, if you simply take the hypothetical family in a vacuum there is an equal chance of either gender.

That is if i am reading it correctly, which is not a garuntee.

Pinnacle
2015-01-20, 01:28 PM
To reiterateHow is this different from saying "I flip a perfectly fair coin twice. One time it came up heads. What are the chances it came up heads both times?" The coin doesn't care what any previous or future flips may be. It has a 50% chance of coming up heads. So the unknown flip has a 50% of matching the known flip. Have I got that wrong?

And if so, how is Problem 1 (older child's gender is known) different from Problem 2 (random child's gender is known) since in both cases, the unknown child's gender is not affected by either the age or gender of the known child?

Knowing that the eldest child is a girl is knowing the results of one coin flip. It has no impact on the second coin flip. 50/50 odds that the second is also a girl.

If I--knowing the results of both coin flips--tell you that at least one of them is a boy, you don't know which one I've told you about. There are three possibilities that fit with the information I've given you, and all are equally likely. In only one of the three possibilities are both children boys.
This is actually true with coin flips. You're just as likely to get a mismatch as a match, and twice as likely to get a mismatch as any particular match. People don't necessarily think of two coin flips (or two die rolls) as distinct events with two distinct sets of probability (there are 36 combinations possible with 2d6, but as we generally only care about the final results and not individual rolls we think of there being only 11 possibilities), but specifying which one we're talking about is a very different puzzle from saying "at least one".
Even just "This child is a boy; what are the odds the other is also a boy?" is different, because we're dealing with a specific. Not the same as "at least one".

Consider 100 pairs of children, equally distributed--25 older brother younger brother, 25 older brother younger sister, 25 older sister younger brother, and 25 older sister younger sister.
I tell you that the family I'm looking at has an older sister--50 of the pairs fit that description, and half of them have two girls.
I now tell you that the family I'm looking at has at least one brother--75 of the pairs fit that description, and only one-third of them have two boys.

Knaight
2015-01-20, 01:54 PM
To reiterateHow is this different from saying "I flip a perfectly fair coin twice. One time it came up heads. What are the chances it came up heads both times?" The coin doesn't care what any previous or future flips may be. It has a 50% chance of coming up heads. So the unknown flip has a 50% of matching the known flip. Have I got that wrong?

It isn't. Basically, you have these four scenarios:
HT
HH
TH
TT.

Question 1: You flip a coin and it comes up heads. What is the chance of it coming up heads again?
Question 2: You flip a coin twice, and at least one of the flips is heads. What is the chance that the other is heads?

Question 1 restricts the problem space to HT, HH. There's a 50% chance of each.
Question 2 restricts the problem space to HT, HH, TH. There's a 33.3% chance of each. The chance that the other is tails includes both HT and TH, which is a total of 66.7%, whereas the chance that the other is heads is just HH, and so is 33.3%

Lord Torath
2015-01-20, 02:03 PM
Okay, I think I can get my head around that. Thanks for the help, ladies and gents!

huttj509
2015-01-20, 02:04 PM
To reiterateHow is this different from saying "I flip a perfectly fair coin twice. One time it came up heads. What are the chances it came up heads both times?" The coin doesn't care what any previous or future flips may be. It has a 50% chance of coming up heads. So the unknown flip has a 50% of matching the known flip. Have I got that wrong?

And if so, how is Problem 1 (older child's gender is known) different from Problem 2 (random child's gender is known) since in both cases, the unknown child's gender is not affected by either the age or gender of the known child?

The issue is a small change in wording changes the question.

your phrasing: "I flip a perfectly fair coin twice. One time it came up heads. What are the chances it came up heads both times?"

Alternate: "I flip a perfectly fair coin twice. The first time it came up heads. What are the chances it came up heads both times?"

Yours it's 1/3, alternate is 1/2.

With how you interpreted yours, you assumed "one head one tail" had a same chance as "two head." However, due to the phrasing, there's 3 options:

HT
TH
HH

HT and TH appear as the same "one head one tail" option, but each has its own chance of coming up in the flipping, so you need to make sure to count them separately.

The method of listing out all the possibilities and then removing the ones that don't apply is just a way to ensure you don't miss anything when counting.

Edit: There's also "I flip a perfectly fair coin twice. Only one time it came up heads. What are the chances it came up heads both times?" Where the answer is 0.

LudicSavant
2015-01-20, 03:51 PM
The Monty Hall Problem is great, because it challenges common false intuitions about probability and understanding it changes the way you approach a lot of problems in general for the better.

Here's another fun one:

You run an office that employs 23 people. What is the probability that two of your employees have the same birthday? For the sake of simplifying the problem, pretend leap years don't exist (no February 29th).

Oh, and here's one of my favorites, based on a true story:

Abraham's job is to review damaged planes returning from sorties over Germany in World War II. He has to review the damage of the planes to see which areas must be protected more.

Abraham discovers that the fuselage and fuel system of returned planes are much more likely to be damaged by bullets or flak than the engines. What should he recommend to his superiors?

http://cna.org/sites/default/files/research/0204320000.pdf

Abraham Wald made the recommendation to his superiors that the places most damaged on returning planes were the places that they didn't need to focus more on protecting.

Instead, he suggested that the places that were returning undamaged were in most dire need of additional protection. If an essential part of the plane, such as the engines, comes back consistently undamaged, that suggests that planes with shot-up engines aren't coming back.

Pinnacle
2015-01-20, 04:15 PM

For the second one:
Obviously planes that have those parts damaged still survive; those parts do not need to be more protected.
Looks like planes that get their engines damaged don't survive, which makes sense. Protect them from damage.

LudicSavant
2015-01-20, 04:24 PM
For the second one:
Obviously planes that have those parts damaged still survive; those parts do not need to be more protected.
Looks like planes that get their engines damaged don't survive, which makes sense. Protect them from damage.

Correct.

The first one's in the thread already. Oh. Whoops. Guess I'll just have to offer something else then.

Allan is looking at Jane, but Jane is looking at George. Allan is married but George is not. Is a married person looking at an unmarried person?

A) Yes
B) No
C) Cannot be determined

Pinnacle
2015-01-20, 04:37 PM
We have not been told whether or not Jane is married... but we don't have to be.
The answer is Yes either way, because Jane is either the married person doing the looking or the unmarried person being looked at.

I feel that I must have some puzzles I could come up with. :smallconfused:

Talakeal
2015-01-20, 05:05 PM
The Monty Hall Problem is great, because it challenges common false intuitions about probability and understanding it changes the way you approach a lot of problems in general for the better.

Here's another fun one:

You run an office that employs 23 people. What is the probability that two of your employees have the same birthday? For the sake of simplifying the problem, pretend leap years don't exist (no February 29th).

Oh, and here's one of my favorites, based on a true story:

Abraham's job is to review damaged planes returning from sorties over Germany in World War II. He has to review the damage of the planes to see which areas must be protected more.

Abraham discovers that the fuselage and fuel system of returned planes are much more likely to be damaged by bullets or flak than the engines. What should he recommend to his superiors?

http://cna.org/sites/default/files/research/0204320000.pdf

Abraham Wald made the recommendation to his superiors that the places most damaged on returning planes were the places that they didn't need to focus more on protecting.

Instead, he suggested that the places that were returning undamaged were in most dire need of additional protection. If an essential part of the plane, such as the engines, comes back consistently undamaged, that suggests that planes with shot-up engines aren't coming back.

That reminds me of another one.

In World War I the British army began issuing their soldiers steel helmets and found that head injuries immediately spiked as a result. Why is this?

Avian Overlord
2015-01-20, 05:27 PM
That reminds me of another one.

In World War I the British army began issuing their soldiers steel helmets and found that head injuries immediately spiked as a result. Why is this?

Simple. The soldiers started surviving being shot in the head.

Talakeal
2015-01-20, 05:40 PM
Simple. The soldiers started surviving being shot in the head.

Correct!*

Apparently it is not so simple though, I have heard that when they got this data the higher ups almost stopped issuing helmets because they thought it meant their soldiers were feeling invincible and getting hurt because they took stupid risks while wearing helmets.

*Technically it was shrapnel, not being shot. The helmets couldn't stop a bullet unless it was traveling at an extreme enough angle that the wound wouldn't have been fatal anyway.

goto124
2015-01-20, 07:28 PM
Or, it was cheaper for the soldiers to die and replace them with new recruits than to treat the injured soldiers.

Until they started running out of human resources...

Jeff the Green
2015-01-20, 08:52 PM
That reminds me of another one.

In World War I the British army began issuing their soldiers steel helmets and found that head injuries immediately spiked as a result. Why is this?

This is the same as the enormous increase in traumatic brain injury in the wars in Afghanistan and Iraq compared to previous wars. We have advanced medicine enough to keep more soldiers with head wounds alive, which means there are more soldiers with TBI we have to care for.

veti
2015-01-20, 10:37 PM
This thread reminds me of a game I learned as a kid...

You're going to start flipping a coin, and write down the result of each flip, which may be H or T. (Assume the coin and the flipping process are fair.) You'll soon have a string of 'H's and 'T's, in unpredictable order.

Ask another person to write down a sequence of three tosses (HHH, HHT, HTH etc.). That's their 'bet'. Then write down your own sequence.

Then start flipping and writing, and whoever's sequence comes up first in the resulting string is the winner.

How can you set your bet, such that you have a guaranteed better-than-50% chance of winning?

NecroRebel
2015-01-20, 11:42 PM
How can you set your bet, such that you have a guaranteed better-than-50% chance of winning?

The obvious solution would be to take the first two flips of their sequence and make those the last two of yours. That is, if they said HHT, you'd make yours HHH or THH. Only one coin flip would matter for you, but three would matter for them, because if you guessed right on the first in your sequence you'd win before the third flip in theirs ever came up, while they'd need to have guessed all three correctly.

Gavran
2015-01-21, 01:50 AM
Ooh. I like your idea NecroRebel. Clever. I like this thread. It does occur to me it's probably in the wrong board though (albeit because the thread it spawned from was here.)

I'm still trying to see a difference between "the oldest child is" and "at least one child is" that matters, because I feel like they both mean "one child is", but it took me a while to get the Monty Hall problem happy in my brain too so I'm not going to call anyone wrong. To use the coin analogy, it's like saying "I flipped one coin, it was heads, there's a 50% chance the next coin will be heads" - but I don't see how that's different from "two coins were flipped, one was heads, what are the odds the other is heads." I do get that two heads in a row is a different probability than 50% (not that I'm good at that probability math, but it's 25% right?) - it just seems to me there's one given heads in both scenarios.

SiuiS
2015-01-21, 03:42 AM
Jumping back to the basic set up of the Monty haul door problem, using the basics of three doors, and one wrog door will always be open after the first choice is made;

He folks trying to explain why it's 2/3 instead of 1/2 are hitting resistance because nothing said so far explains why the second choice is a choice between three options in a system instead of a choice between two.

To reiterateHow is this different from saying "I flip a perfectly fair coin twice. One time it came up heads. What are the chances it came up heads both times?" The coin doesn't care what any previous or future flips may be. It has a 50% chance of coming up heads. So the unknown flip has a 50% of matching the known flip. Have I got that wrong?

And if so, how is Problem 1 (older child's gender is known) different from Problem 2 (random child's gender is known) since in both cases, the unknown child's gender is not affected by either the age or gender of the known child?

"I have two American coins. Their value is .30¢, and one of them is not a nickel. What are the two coins?"
A quarter and a nickel.

Why? Because one of them is not a nickel. But the other one is. It's a Riddle. Wordplay.

LudicSavant
2015-01-21, 04:34 AM
He folks trying to explain why it's 2/3 instead of 1/2 are hitting resistance because nothing said so far explains why the second choice is a choice between three options in a system instead of a choice between two.

Nor should anything be said to that effect, because it would be completely wrong. The second choice is not a choice between three options, it is a choice between two options (switch or stay). One choice (switch) has a 2/3rds probability of success, the other (stay) has a 1/3rd probability of success.

In the initial choice, you have three options, in the form of three doors. Two of these doors contain a goat. Therefore, there is a 2/3rds chance that your initial choice was a goat.

There are two situations of note. Either you initially chose a goat, or you initially chose a car.

If you initially chose a goat, and the host reveals a goat (as he always does), then by process of elimination the remaining door must be a car, so you should switch. What's the chance you originally chose a goat? 2/3

If you initially chose a car, and the host reveals a goat (as he always does), then by process of elimination the remaining door must be a goat, so you should stay. What's the chance you originally chose a car? 1/3

Cazero
2015-01-21, 06:16 AM
He folks trying to explain why it's 2/3 instead of 1/2 are hitting resistance because nothing said so far explains why the second choice is a choice between three options in a system instead of a choice between two.

Swordsage'd, but since you spoke about nickels.

edited for clarity/better parrallel

The monty haul problem is similar to the weirdest slot machine in the world : one wheel with three pictures : two goat and a car, a goat worth nothing, the car is a jackpot; a window to hide the wheel that can be opened after spinning the wheel to show you the result and closes by itself before you can play again; three different start buttons labelled one, two and three that each set the wheel on a different position but can't be pushed again until you played three times. The position associated with each button change randomly every time they are reactivated.

Your friend Monty, a crazy gambler, offers you a nickel to play. When you push a button, but before you can see your result, he offers you to trade two nickels with your result.

Considering the wheel setup, accepting the trade (equivalent to switching doors) allows you to get both other results, with a garanteed goat (tricky part : equivalent to the door opened by the host, since the worthless value makes who opened it irrelevant), and the button you just pushed has 1/3 chance to win.

You have a 1/3 chance of winning by sticking with your result, while accepting the trade will get you all other possible results, meaning 2/3 chance of winning.

Knaight
2015-01-21, 09:09 AM
I'm still trying to see a difference between "the oldest child is" and "at least one child is" that matters, because I feel like they both mean "one child is", but it took me a while to get the Monty Hall problem happy in my brain too so I'm not going to call anyone wrong. To use the coin analogy, it's like saying "I flipped one coin, it was heads, there's a 50% chance the next coin will be heads" - but I don't see how that's different from "two coins were flipped, one was heads, what are the odds the other is heads." I do get that two heads in a row is a different probability than 50% (not that I'm good at that probability math, but it's 25% right?) - it just seems to me there's one given heads in both scenarios.

Again, here's the 4 combinations for two coin flips:
HH
HT
TH
TT

Either "the first is heads" or "the second is heads" reduces the problem to 2 possibilities (HH, HT for the first; HH, TH for the second). In either of those cases, there's a 50% chance that HH is chosen, and a 50% chance that the one with tails is chosen (HT or TH, respectively). If either of the two can be heads, the problem is reduced to 3 possibilities (HH, HT, TH). There's only a 1/3 chance of HH there.

Now, say we're looking at siblings. Assume that there's a 50% chance of a boy and a 50% chance of a girl (this is heavily oversimplified in this case, as the neither option is vastly more likely than the side option on just about any coin, and it's not quite a 1:1 ratio between boys and girls either). You have 4 possibilities:

GG
GB
BG
BB

From there, the exact same thing applies.

Pinnacle
2015-01-21, 11:30 AM
I'm still trying to see a difference between "the oldest child is" and "at least one child is" that matters, because I feel like they both mean "one child is", but it took me a while to get the Monty Hall problem happy in my brain too so I'm not going to call anyone wrong. To use the coin analogy, it's like saying "I flipped one coin, it was heads, there's a 50% chance the next coin will be heads" - but I don't see how that's different from "two coins were flipped, one was heads, what are the odds the other is heads." I do get that two heads in a row is a different probability than 50% (not that I'm good at that probability math, but it's 25% right?) - it just seems to me there's one given heads in both scenarios.
You've changed the wording, so the problem changes.
In "what are the odds the other is heads" you've designated one coin. It's probably the first one, but we don't really know... regardless, you have removed one from the problem, so there's only the one coin toss to determine.
"At least one is" is different from "this one is" because you could be talking about either or both. There are three different possibilities where "at least one is" is true (HH, HT, and TH), and all were equally likely. The information you have given us only eliminated one possibility (TT), the others all remain.

Precision matters here.
"This one is..." This is the first problem where the eldest is definitely a girl: 50% chance of match.
"At least one is..." This is the second problem, and it's given the least information: 33.3...% chance of match.
"Exactly one is..." This is a trick question, and it's already told us the answer: 0% chance of match.

Jumping back to the basic set up of the Monty haul door problem, using the basics of three doors, and one wrog door will always be open after the first choice is made;

He folks trying to explain why it's 2/3 instead of 1/2 are hitting resistance because nothing said so far explains why the second choice is a choice between three options in a system instead of a choice between two.
Just because there are two choices does not mean the odds are 50%
Let's say I offer to either give you a pile of money or punch you in the face. There are only two options, but nobody betting on your choice is going to flip a coin.

Although you could phrase it as a choice between three... Monty is essentially saying that you can stick with your door or choose [i]both[i] of the other two. He's just kindly opening one of the two doors for you. The extra information he's given you is "If the prize is behind one of these two doors, it's that one."

We can prove it by breaking it down completely.
Let's say I always choose A initially, and the correct answer is randomly determined (you could break this down further with nine different scenarios, one for each choice and each correct answer, if you wanted).

A is the correct answer 1/3 of the time. In this case, Monty can choose to open either B or C. Assuming his choice is completely random, he chooses each 1/2 the time--1/6 each of the overall scenario. It doesn't matter which, because either way switching means I lose and staying means I win.
This 1/3 of the time, switching loses.

B is the correct answer 1/3 of the time. In this case, Monty must open C. If I switch to B, I win.
This 1/3 of the time, switching wins.

C is the correct answer 1/3 of the time. In this case, Monty must open B. If I switch to C, I win.
This 1/3 of the time, switching wins.

This problem is hard to intuitively understand, which is what makes it famous, but it's relatively easy to prove if you write all the possibilities out.
Those three possibilities were all equally likely, and I won in two of them.

There are two choices, but three possible scenarios. You could say that when Monty opens C, he has reduced the possible scenarios to the first and second, but that's ignoring part of the information he's given us.
He has told us that, if either the second or third scenario is correct, the second scenario is correct.
His behavior is predictable, and thus a valid clue. We know that if the third scenario was correct, he would not have done that. The first scenario could have happened, but in that case he was half as likely to open door C. He did open C, so it's twice as likely that he had to than that he did it by chance.

Now, this pure probability doesn't really apply to the actual game show itself too much.
My understanding is that he didn't always open a door (never seen that show myself), so what makes him choose to do so? Is he more likely to do so when I chose right?
Are the three scenarios really equally likely, or do they, say, disfavor door C when hiding the prize?
When Monty does have a choice between two doors, does he really randomly choose or does he maybe, for example, favor doors closer to where he was standing?
All of those factors, and probably some I've overlooked, change it from pure logic.

Why? Because one of them is not a nickel. But the other one is. It's a Riddle. Wordplay.
Not really.
If we're being that precise about "one of them is" than there is 0% chance that there's a match because the other one is not.
That's why the original puzzle did not use that wording at all.

Anonymouswizard
2015-01-21, 01:00 PM
The argument over the boy-girl problem (in my opinion the maths states that the chance is 1/3, but I don't have better proof than the rest of the thread) reminds me of when my teachers were teaching me probability, and I would get into arguments with them, because the examples would go like this:

Teacher: "I throw a fair dice where every number is as likely as every other number to come up. What's the chance I roll a six?"
Me: "Zero/One-Eighth/One-Tenth/One-Twentieth."
Teacher: "No, it is, as everyone else has said, one-sixth."
Me: "I own several 'fair' dice with a different number of sides."
Teacher: "Everyone knows that a normal dice has six sides."

Which after I was diagnosed with Autism and started to excel at English changed to:

Teacher: "I throw a fair dice where every number is as likely as every other number to come up. What's the chance I roll a six?"
Me: "We do not have enough data to give the actual answer."

In short, you can only calculate the probability if you have all the information, and the way the question is worded and interpreted changes the perceived amount of data (In the case of the children the question seems to differ between "there are two children, at least one of which is a boy, what is the likely hood of them both being boys", and "there are two children, one of which is a boy, what is the probability that the unspecified child is born as a boy", which are two different questions [the first has the 2/3 answer, and the second has the 1/2 answer, due to how dependent and independent variables work]. It's just in normal everyday life the difference doesn't come up enough to avoid paying attention to. Also, the phrasing of the original problem supports the first interpretation, and the other posters have done the reasoning for me).

Talakeal
2015-01-21, 01:15 PM
It actually is a very important lesson for real life though. A lot of statistics can be heavily influenced by how the question is asked or the data is collected, and it is good to be mindful of this when someone uses statistics in an argument.

Pinnacle
2015-01-21, 01:22 PM
Yeah, the problem expects you to assume there's a 50/50 split between genders. Given that it doesn't say otherwise, that's probably a valid assumption unless it's a trick question and the answer they're looking for is "insufficient data," but it is the most correct answer.

huttj509
2015-01-21, 01:24 PM
It actually is a very important lesson for real life though. A lot of statistics can be heavily influenced by how the question is asked or the data is collected, and it is good to be mindful of this when someone uses statistics in an argument.

When political pollsters call up, I play a personal game of "guess which candidate funded this" and "spot the push poll questions."

"Have you heard about so-and-so's rumored misconduct?" Guess what, you have now.

Amphetryon
2015-01-21, 01:46 PM
Yeah, the problem expects you to assume there's a 50/50 split between genders. Given that it doesn't say otherwise, that's probably a valid assumption unless it's a trick question and the answer they're looking for is "insufficient data," but it is the most correct answer.

I was under the impression that the number of extant, viable genders from which to choose one's answer made '50/50' an impossibility, personally.

Pinnacle
2015-01-21, 01:57 PM
Fair to assume that was the intent of the puzzle, mean.

Segev
2015-01-21, 02:22 PM
I'm still trying to see a difference between "the oldest child is" and "at least one child is" that matters, because I feel like they both mean "one child is", but it took me a while to get the Monty Hall problem happy in my brain too so I'm not going to call anyone wrong. To use the coin analogy, it's like saying "I flipped one coin, it was heads, there's a 50% chance the next coin will be heads" - but I don't see how that's different from "two coins were flipped, one was heads, what are the odds the other is heads." I do get that two heads in a row is a different probability than 50% (not that I'm good at that probability math, but it's 25% right?) - it just seems to me there's one given heads in both scenarios.

As others have said, the difference comes in because of the nature of independent versus dependent results.

When you specify that "the older child" or "the coin on the left" or "the blue die" had a result, you are rendering the probability of the other one's result totally independent.

That is, if I have two coins, one blue and one green, and I tell you, "The green one is Heads," you now know that the GREEN coin came up heads. There is no possibility that the green coin came up tails. The question, "What is the probabilitiy that both coins came up heads?" is equivalent, now, to, "What is the probability that the blue coin came up heads?"

If, on the other hand, I tell you, "At least one of the coins came up Heads," you don't know if it's the blue coin or the green coin. The question, "What is the probability that both coins came up heads," is no longer functionally equivalent to the question, "What is the probability that the blue coin came up heads?"

This is because you now don't know if the green coin came up heads or tails. So whether both came up heads depends on unknown information about both coins. All you know is that, if the green coin came up tails, the blue one MUST have come up heads, and vice-versa. Because you know "both came up tails" is false.

To put it another way, if I tell you, "The green coin came up heads," the question about both coming up is just asking about the blue coin, because the only possible results are "green heads, blue tails," or "green heads, blue heads." If I tell you "at least one coin came up heads," you now know that, even if the blue coin came up heads, there's some chance that the green coin did not. So you have more chances for "both coins came up heads" to fail. The exact probability is a matter of math; I'm just trying to help illustrate the dependent nature of the probability that changes with the change in the question.

Anonymouswizard
2015-01-21, 02:53 PM
It actually is a very important lesson for real life though. A lot of statistics can be heavily influenced by how the question is asked or the data is collected, and it is good to be mindful of this when someone uses statistics in an argument.

I agree with you there, but the fact is even though in normal life we have to be wary of statistics ("more people support the Green Party than the Liberal Democrats" is one of the bigger political statistics here in the UK, but as far as I know it's based on membership numbers, which do not indicate who the general public will vote for due to most people not being motivated enough to join the party), but my experience is that, unless your job or hobby requires it you aren't coming across statistics or probabilities that you find important every day (I personally come across a such a statistic once a week, and expect that to increase now that a general election is coming up).

However, I agree that the 'check the intent of the question/information' is an important lesson for everyday life, as there is a lot of difference that a single word or piece of punctuation makes ("let's eat grandma" versus "let's eat, grandma"), I was just trying to refer to the more specific example. For example, I could tell you that the game my group has played the most is Dark Heresy, or the game that we have played our longest campaign in is Mutants and Masterminds, both with a slant towards that being my favourite game.

A lack of such deconstruction is what lead me to have no fun playing a zombie survival character in a zombie action game (until the end session arrived and just made no sense), because the GM just said he wanted to run a zombie game, and the only indication that it was going to be an action game was that he essentially forced my character away from the planned support role (which meant a party with about 4 'heroes', and 'battle scientist' and a medic). But off topic is off topic.

With statistics we also have the "9/10 experts agree" problem, in that I check with 100 experts and pick 10 of them to get my desired outcome after asking my loaded questions (which means that adverts tell me that at least 8/10 dentists want me to use multiple brands of toothpaste).

Segev
2015-01-21, 03:40 PM
To illustrate the real difficulty we, as people, have wrapping our heads around these things, despite the fact that I understand the distinction, I still come up with weird thought experiments.

Let's say that Talekeal, Amphytreon, and I (to pick two others in this thread) are doing this experiment.

I have the blue and green coins discussed previously. I will flip them both until at least one comes up "Heads," concealing the results from Telakeal and Amphytreon. I will then pass a note to Amphytreon that says, "At least one of the coins came up Heads. Did they both?" and a note to Talekeal that says, "The [correct color] came up HEads. Did they both?"

That is, both know that at least one coin came up Heads. Talekeal additionally knows which one (green or blue) did so.

According to the probabilities we've been discussing, Amphytreon should be right 1/3 of the times he guesses "yes," and 2/3 of the times he guesses "no."

Talekeal, on the other hand, should be right half the time, either way.

This is because Talekeal is supposedly only predicting the result of one of the coins (the one I have not told him the result of), while Amphytreon is predicting the results of both, albeit knowing that BOTH cannot show Tails (so he knows enough to guess with better than 1/4 odds).

If Talekeal and Amphytreon always answer "yes," then the probability calculations from before tell us that Talekeal should be right half the time, while Amphytreon is right only 1/3 of the time. Yet, the same results will come up against which to compare their answers.

How do we resolve this paradox? Is knowing WHICH coin, blue or green, really going to impact the likelihood that guessing whether both came up heads is correct?

Pinnacle
2015-01-21, 04:08 PM
Hmm.

Let's break it down. We'll say both always say "No", since that's the better choice for one of our guessers and it makes no difference for the other.

You flip two heads. You give them both a note. They are both wrong.

You flip heads, then tails. You give them both a note. They are both correct.

You flip tails, then heads. You give them both a note. They are both correct.

You flip two tails. This scenario is discarded.

They're both right 2/3s of the time.
But the information you're giving Telekeal varies. If you only gave information about one of the coins--as in our puzzle--the odds fall to 1/2.
This time, we'll only give Telekeal a note if the first coin comes up heads.

You flip two heads. You give them both a note. They are both wrong.

You flip heads, then tails. You give them both a note. They are both correct.

You flip tails, then heads. This scenario is discarded only for Telekeal, but Ampytreon gets a note. Ampytreon is correct.

You flip two tails. This scenario is discarded.

This is more similar to our original problem--and while the one only being told "one of..." has the 1/3 vs 2/3 odds, the other has 50:50.

Saying that the first coin is heads--or that the first child is a girl--is extra information. It discards two of the four possible scenarios when saying that at least one is heads--or at least one is a boy--only discards one.
By deciding in advance you're going to give me a note as long as at least one has a heads... It's more information about what you're going to put on the notes, but that limits the usefulness of the notes themselves. You're going to give me basically the same note in three out of the four scenarios, and I can only eliminate one as a possibility.
For the note you gave me, there is one situation where you would have had to, two where you couldn't have, and one where you had to randomly pick between two. Your behavior in deciding when to give me a note is somewhat predictable--obviously it is more likely that this is the note you had to give me than it is that you had two options and chose this one.

We're actually back to something that resembles the Monty Haul problem.

(For the record, I didn't know what the answer here was. Your question did succeed in confusing me. I just figured I'd figure it out as I was breaking it down, and I did.)

LudicSavant
2015-01-21, 04:57 PM
It actually is a very important lesson for real life though. A lot of statistics can be heavily influenced by how the question is asked or the data is collected, and it is good to be mindful of this when someone uses statistics in an argument.

It's not that the statistics are wrong, it's that people interpret the statistics wrong.

Most people don't understand probability or statistics on even a basic level, but think they do. This makes it very, very easy to manipulate public opinion with statistics that don't mean what many people think they mean, while still being able to say that you weren't lying. For example, I remember someone pointing out that vaccinations were bad because X% (some large majority) of the cases of Polio after year Y were caused by polio vaccines. What this person failed to understand was that this was because there were almost zero cases of polio after the vaccines were introduced and well distributed (after year Y). In fact, the success of the vaccine was so great that simply having a nonzero number of misfired vaccines would account for a significant percentage of surviving cases of polio.

The error is similar to previously mentioned things like "Oh, the helmets must not be working, since we have more head wounds now" and failing to realize that the statistics imply the exact opposite: That the soldiers have more head wounds because more soldiers hit in the head are surviving. In both cases, the statistics were accurate. The problem is that the listeners had such a poor understanding of statistics that they interpreted statistics indicating overwhelming success as indicating overwhelming failure.

http://imgs.xkcd.com/comics/frequentists_vs_bayesians.png
(For those who don't get the joke, the Frequentist statistician is making an error that is very obvious with Bayesian reasoning)

Segev
2015-01-21, 04:59 PM
That still doesn't sit right with me, though.

Trying it differently.

I will not discard any scenarios.

I have two notes I can possibly give to Amphytreon:

A) At least one of the coins is Heads; are both coins Heads?
B) At least one of the coins is Tails; are both coins Tails?

I have 4 notes I can possibly give Telakeal:

3) The blue coin is Tails; are both coins Tails?
4) The green coin is Tails; are both coins Tails?

I will flip three coins: one blue, one green, and one whose color is irrelevant which I will call "mine."

Blue Heads, Green Heads, Mine Tails: I will give Amphytreon note (A), and Telakeal note (2).

Blue Heads, Green Tails, Mine Heads: I will give Amphytreon note (A), and Telakeal note (1).

Blue Heads, Green Tails, Mine Tails: I will give Amphytreon note (B), and Telakeal note (4).

Blue Tails, Green Heads, Mine Heads: I will give Amphytreon note (A), and Telakeal note (2).

Blue Tails, Green Heads, Mine Tails: I will give Amphytreon note (B), and Telakeal note (3).

Blue Tails, Green Tails, Mine Heads: I will give Amphytreon note (B), and Telakeal note (3).

Blue Tails, Green Tails, Mine Tails: I will give Amphytreon note (B), and Telakeal note (4).

Once again, always answering "no" is better for Amphytreon, in theory, and indifferent for Telakeal. There are equal odds that I will hand any given note to the person who receives it. Perhaps adding "my" coin to the mix makes some sort of difference, though I'm not convinced it should at this stage; all it's doing is randomizing whether I ask about heads or tails (when the two coins are different), or whether I give Telakeal information about the blue or green coin (when they're the same).

In all notes to Amphytreon, I am eliminating 1 of 2 possibilities from our hypothetical HH/HT/TH/TT set (either the HH or the TT). Theoretically, in each case, I am further eliminating one of TH or HT with my note to Telakeal.

If they always guess "no..."

Hrm.

Of the possible outcomes, both seem to guess correctly (by always answering "no") 2/3 of the time. This implies that the extra information about which coin is the given value is irrelevant.

The fact that I'm changing whether I ask about Heads or Tails at random shouldn't matter when the answer is "no." The fact that I'm changing it to make sure the answer is "yes" shouldn't matter when both are the same. Otherwise, that implies something about the individual, one-time results' probabilities still rendering the information given irrelevant, somehow.

I remain a bit perplexed.

Yaktan
2015-01-21, 05:09 PM
Adding in the "mine" coin is messing it up. We can see this from your list of which are correct or not, and note that you have two of the 8 listed as never happening, so it looks like you have 6 possibilities. But when you look at the actual set of events, each of the two that are paired with a never-happen possibility can happen twice. So you actually have him being wrong half the time, as we would expect.

Segev
2015-01-21, 05:29 PM
Adding in the "mine" coin is messing it up. We can see this from your list of which are correct or not, and note that you have two of the 8 listed as never happening, so it looks like you have 6 possibilities. But when you look at the actual set of events, each of the two that are paired with a never-happen possibility can happen twice. So you actually have him being wrong half the time, as we would expect.

Er, if you look at the numbers, I did not count the "never happens" in the count of being right 2/3 of the time.

The "never happens" are because I will never ask about Tails if it's two Heads, and vice-versa.

Pinnacle
2015-01-21, 05:30 PM
You're doing more than eliminating possibilities, though.

Knowing that you're randomly determining which information you give to me is itself a lot more information.

If I just told you "at least one is heads" or "the first one is heads", you don't have much to go on. If you know why I'm telling you, that's part of the question.

I know you'll give me an all-but identical note every single time. I know there's a pair half the time.
With this data, all your notes tell me is which pair there is if we're in one of the two cases where there is a pair.

If you tell me at least one is heads, without any additional information that means there are three equally likely possibilities (HH, HT, TH). However, I know you'll give me that note if the correct answer is HH, and that you'll give me that note half the time if the answer is either HT or TH. That's more information. It means that HH is twice as likely to be the correct answer as HT or TH, which means it's as likely as either one of the two happening.
You've eliminated only one, yes, but you've lowered the chances of two of the remaining possibilities.
My odds are 50/50--they must be since there are four different equally possible scenarios in this new situation and two of the four are pairs.

Once again, always answering "no" is better for Amphytreon, in theory, and indifferent for Telakeal.
Er, why? That theory was about an entirely different question.
This set has four different possibilities, for one.
The thing that gave us the 1/3 vs 2/3 odds was that there were only three possibilities, and we were treating two of them the same. Now there are four, and you're treating two the same and the other two the same.

EDIT:

Er, if you look at the numbers, I did not count the "never happens" in the count of being right 2/3 of the time.

The "never happens" are because I will never ask about Tails if it's two Heads, and vice-versa.
Precisely, and therein lies your error.

There are four different possibilities: HH, HT, TH, and TT.

But we know that HH, HT, TH, and TT are all equally likely... If you ask about heads half the time you get HT and tails half the time you get HT, those situations come up half as often as HH. By treating as six equally likely scenarios, you've stated that HT or TH are twice as likely as HH or TT; that is not correct.

Remember you initially split it into eight, not six.

HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT
You then condensed HHH and HHT to one and TTH and TTT to one, since the final flip didn't determine which coin you asked about, but they're still distinct scenarios.
So assuming they both always say "No"...
HHH, wrong
HHT, wrong
HTH, right
HTT, right
THH, right
THT, right
TTH, wrong
TTT, wrong

Both of them are right only half the time.

Anonymouswizard
2015-01-21, 05:45 PM
It's not that the statistics are wrong, it's that people interpret the statistics wrong.

Most people don't understand probability or statistics on even a basic level, but think they do. This makes it very, very easy to manipulate public opinion with statistics that don't mean what many people think they mean, while still being able to say that you weren't lying. For example, I remember someone pointing out that vaccinations were bad because X% (some large majority) of the cases of Polio after year Y were caused by polio vaccines. What this person failed to understand was that this was because there were almost zero cases of polio after the vaccines were introduced and well distributed (after year Y). In fact, the success of the vaccine was so great that even an extremely small number of misfired vaccines would account for almost all of the surviving cases of polio. The error is similar to previously mentioned things like "Oh, the helmets must not be working, since we have more head wounds now" and failing to realize that the statistics imply the exact opposite: That the soldiers have more head wounds because more soldiers hit in the head are surviving.

http://imgs.xkcd.com/comics/frequentists_vs_bayesians.png
(For those who don't get the joke, the Frequentist statistician is making an error that is very obvious with Bayesian reasoning)

I think you've hit the nail on the head, considering that I was at one point called a 'wrong bad feminist' for calling a figure of a few thousand statistically insignificant (this was to do with the number of UK teenagers of both genders entering university). There was an entire argument between two groups, the 'several thousand is a big number' and 'after doing some ballpark maths we have decided we don't have enough data to call this significant', of which the second group eventually 'won' after we gained support for having more reasoning. But I see this happening all the time, usually in the 'X% of people of ethnicity Y I've heard about are Z, X is bigger than 50, therefore all people of ethnicity Y are Z', where was most of the time there is a deeper layer of statistics to the obvious (Obvious=head wounds are going up, Deeper=fatalities are going down; Obvious=all seats currently being gained are UKIP, Deeper=this is based on a small number of seats that have retained the same candidate; Obvious=the chance of the detector lying is unlikely, Deeper=the chance of the sun exploding even more so; Obvious=80% of polled people at this university said they were voting conservative, Deeper=this poll was taking at the university conservative society), and there is also the correlation implying causation problem, including the possibility of getting the causation the wrong way round (e.g. 99% of roleplayers own dice, therefore owning dice means you're a roleplayer. In reality this are likely to be mainly unrelated, but you could make an argument that being a roleplayer is a cause of someone owning dice [but not the only possible one]).

Please bare in mind two things: 1)that any statistics I've used in this post were made up on the spot, and 2)that MY knowledge of statistics as displayed here is probably a bit off, as it's been years since I've taken a class in it.

veti
2015-01-21, 06:07 PM
The obvious solution would be to take the first two flips of their sequence and make those the last two of yours. That is, if they said HHT, you'd make yours HHH or THH. Only one coin flip would matter for you, but three would matter for them, because if you guessed right on the first in your sequence you'd win before the third flip in theirs ever came up, while they'd need to have guessed all three correctly.

You're on the right lines, but if they bet HHT and you bet HHH, your chances of winning are only 50/50. (Because everything hinges on the single coin toss that comes directly after 'HH' first appears in the sequence.) The correct answer to that bet would be THH (because then they'll only win if the first two tosses are both 'H' - if that condition isn't met, then 'THH' has to come before 'HHT' can).

LudicSavant
2015-01-21, 06:22 PM
I think you've hit the nail on the head, considering that I was at one point called a 'wrong bad feminist' for calling a figure of a few thousand statistically insignificant (this was to do with the number of UK teenagers of both genders entering university). There was an entire argument between two groups, the 'several thousand is a big number' and 'after doing some ballpark maths we have decided we don't have enough data to call this significant', of which the second group eventually 'won' after we gained support for having more reasoning. But I see this happening all the time, usually in the 'X% of people of ethnicity Y I've heard about are Z, X is bigger than 50, therefore all people of ethnicity Y are Z', where was most of the time there is a deeper layer of statistics to the obvious (Obvious=head wounds are going up, Deeper=fatalities are going down; Obvious=all seats currently being gained are UKIP, Deeper=this is based on a small number of seats that have retained the same candidate; Obvious=the chance of the detector lying is unlikely, Deeper=the chance of the sun exploding even more so; Obvious=80% of polled people at this university said they were voting conservative, Deeper=this poll was taking at the university conservative society), and there is also the correlation implying causation problem, including the possibility of getting the causation the wrong way round (e.g. 99% of roleplayers own dice, therefore owning dice means you're a roleplayer. In reality this are likely to be mainly unrelated, but you could make an argument that being a roleplayer is a cause of someone owning dice [but not the only possible one]).

Please bare in mind two things: 1)that any statistics I've used in this post were made up on the spot, and 2)that MY knowledge of statistics as displayed here is probably a bit off, as it's been years since I've taken a class in it.

Some very common, very dangerous misunderstandings people tend to have about probability:
- Conjunctions are more likely than disjunctions. (Can lead to massive casualties when it comes to disaster readiness investments, and high rates of false imprisonment)
- Correlations imply causation. (A source of endless useless or counterproductive projects directed at trying to solve problems)
- Seeing low percentages means something can be dismissed ("Oh, only 1% of cases have problems" can mean overwhelming disaster in certain contexts, for instance), and seeing high percentages means it's highly impactful. (Such as the polio vaccine example, or the sun explosion one in the XKCD comic)
- Statistics don't contain any real useful information at all, it's just about how they're interpreted. (Basically, they learned just enough about statistics to climb Mount Stupid http://blogs-images.forbes.com/chrisbarth/files/2011/12/Mount-Stupid.gif)

SiuiS
2015-01-21, 06:30 PM
Nor should anything be said to that effect, because it would be completely wrong. The second choice is not a choice between three options, it is a choice between two options (switch or stay). One choice (switch) has a 2/3rds probability of success, the other (stay) has a 1/3rd probability of success.

In the initial choice, you have three options, in the form of three doors. Two of these doors contain a goat. Therefore, there is a 2/3rds chance that your initial choice was a goat.

There are two situations of note. Either you initially chose a goat, or you initially chose a car.

If you initially chose a goat, and the host reveals a goat (as he always does), then by process of elimination the remaining door must be a car, so you should switch. What's the chance you originally chose a goat? 2/3

If you initially chose a car, and the host reveals a goat (as he always does), then by process of elimination the remaining door must be a goat, so you should stay. What's the chance you originally chose a car? 1/3

There you go. Simple language and clear explanation!
Knew it was related in some way. Couldn't phrase it though.

I was under the impression that the number of extant, viable genders from which to choose one's answer made '50/50' an impossibility, personally.

Folks tend to conflate gender and sex in these sorts of questions. We are usually talking about assigned gender, if that helps.

Segev
2015-01-21, 06:31 PM
You're doing more than eliminating possibilities, though.

Knowing that you're randomly determining which information you give to me is itself a lot more information.Not...really.

If I walk up to you and tell you, "The Joneses have two children; the oldest is a girl," that is sufficient to make whether or not their younger child is also a girl an independent probability.

It doesn't matter how I decided to give you that information.

If I just told you "at least one is heads" or "the first one is heads", you don't have much to go on. If you know why I'm telling you, that's part of the question.Not...really. Not in this context, anyway.

I know you'll give me an all-but identical note every single time. I know there's a pair half the time.
With this data, all your notes tell me is which pair there is if we're in one of the two cases where there is a pair.

If you tell me at least one is heads, without any additional information that means there are three equally likely possibilities (HH, HT, TH). However, I know you'll give me that note if the correct answer is HH, and that you'll give me that note half the time if the answer is either HT or TH. That's more information.No, it really isn't. If I give you that information, you know that there are three possibilities, as you said. However, you know that, if it is not HH, there is only a 50% possibility that I would have told you about the coin I told you about.

It means that HH is twice as likely to be the correct answer as HT or TH, which means it's as likely as either one of the two happening.Er, I don't follow. Break down this probability for me, please?

You've eliminated only one, yes, but you've lowered the chances of two of the remaining possibilities.That can also be said for if I don't TELL you how I'm choosing.

If I played this game with a different pair of people 1000 times, we'd still expect the same probabilities of correctness (based on which answer they gave). I don't have to tell them how I'm deciding which note to give them. It doesn't change the probability of the accuracy of their guess, based on what their guess is.

My odds are 50/50--they must be since there are four different equally possible scenarios in this new situation and two of the four are pairs.Except we've just established that the odds are actually 2/3 that you'll get it right if you guess "no." At least, by the breakdown I gave in that post.

Er, why? That theory was about an entirely different question.
This set has four different possibilities, for one.
The thing that gave us the 1/3 vs 2/3 odds was that there were only three possibilities, and we were treating two of them the same. Now there are four, and you're treating two the same and the other two the same.Again, our odds come out to 2/3 that "no" is right, by the analysis above. Not 1/2.

There are four different possibilities: HH, HT, TH, and TT.

But we know that HH, HT, TH, and TT are all equally likely... If you ask about heads half the time you get HT and tails half the time you get HT, those situations come up half as often as HH.

Remember you initially split it into eight, not six.Alright...I'm trying to follow, but I don't think you're drawing correct conclusions here.

HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT
You then condensed HHH and HHT to one and TTH and TTT to one, since the final flip didn't determine which coin you asked about, but they're still distinct scenarios.
But...

Okay. Let's take a single instance.

HTH I give you, based on what's in the spoiler, the note that says, "At least one of them was Heads. Were they both heads?" Based solely on that note, you have a 2/3 chance of being right if you say "no."

The possible outcomes of the two coins, as YOU know them (assuming you don't know about "my" coin), are:

HH
HT
TH

If you know about my coin, the possible outcomes are:

HHH
HHT
HTH
THH

...and here I see where you're coming up with your 50% chance that you'd be right, rather than 2/3, by saying "no." As 2/4 results are "right" and 2/4 are "wrong" for "no."

However, this implies that you knowing MORE information reduces your odds of guessing correctly. If this were a "man on the street" game, and I just walked up to random people and offered them \$10 if they got it right, but I didn't tell them how I was determining what question to ask them, why does the probability change, in theory, to 2/3?

This is about whether or not, upon a stranger walking up to you and saying, "If the older sibling of two is a boy, what are the odds that they're both boys?" versus "If one of two siblings is a boy, what are the odds that they're both boys?"

How he decided to ask you about whether or not they're both boys (and that one was a boy), or of they're both girls (and one of them was a girl), should not matter to your odds of guessing correctly based on the information you are given.

So assuming they both always say "No"...
HHH, wrong
HHT, wrong
HTH, right
HTT, right
THH, right
THT, right
TTH, wrong
TTT, wrong

Both of them are right only half the time.

Hrm.

Let me attempt something here to see if it makes any difference; I'll explain the notation afterwards if it's useful.

HHB wrong
HHG wrong
HTB right
HTG right
THB right
THG right
TTB wrong
TTG wrong

Odds don't change, obviously; they shouldn't. What I've done here is simplified my rules a bit. The third coin is blue on one side, green on the other, and we're deciding that we are going to report on and ask about the colored coin that matches the flip of the third one.

So if we get "blue," we'll tell both players what is on the blue coin, and ask if both coins match. We'll tell Amphytreon only that one of the coins has the symbol that happens to be on the blue coin. We'll tell Talekeal that it is the blue coin whose symbol we're telling him.

If we, again, only ever use any given player once, in theory the situation is the same as if a single experiment were done, at least for that one person. Right?

So... what about this scenario makes it so that, when I perform every step except handing the notes to people before hand, both players have the same odds of getting it right, when somehow the questions asked before give differing odds based on how they were asked?

If, instead of, "are they both heads?" we ask, "what is the other coin?" that definitely reduces it to 50%, by probability. Because we've specified that it's "the other" one, making it independent.

So, let's say you are approached on the street by somebody offering you the following information: "I have flipped two coins. At least one of them was heads. Do you think they both were?"

Would you still have a 2/3 chance of being right if you said "no?" You don't know how he decided to tell you that one of them was heads. You know one of them is, because otherwise he'd be lying.

If said stranger repeated his question slightly differently, and said, "What do you think the other coin was?" why does that change your odds to be lower?

It is truly arbitrary which coin he flipped first. It's arbitrary whether he tells you the result of the first or second coin.

While statistics claim this distinction is real, and I even understand the independence argument, I'm having trouble with the "same situation, different information" aspect of it changing the odds in the fashion it claims to.

LudicSavant
2015-01-21, 06:43 PM
There you go. Simple language and clear explanation!

Happy to help.

Anonymouswizard
2015-01-21, 06:54 PM
Some very common, very dangerous misunderstandings people tend to have about probability:
- Conjunctions are more likely than disjunctions. (Can lead to massive casualties when it comes to disaster readiness investments, and high rates of false imprisonment)
- Correlations imply causation. (A source of endless useless or counterproductive projects directed at trying to solve problems)
- Seeing low percentages means something can be dismissed ("Oh, only 1% of cases have problems" can mean overwhelming disaster in certain contexts, for instance), and seeing high percentages means it's highly impactful. (Such as the polio vaccine example, or the sun explosion one in the XKCD comic)
- Statistics don't contain any real useful information at all, it's just about how they're interpreted. (Basically, they learned just enough about statistics to climb Mount Stupid http://blogs-images.forbes.com/chrisbarth/files/2011/12/Mount-Stupid.gif)

Considering that with statistics I'm currently climbing down Mount Stupid, I've just had to go and look up conjunction and disjunction, because I couldn't remember the names. That sounds like a stupid thing to assume, but I can see why people would do so.

Misunderstanding 2 is of course the classic one :smalltongue: it's also the one I'm most likely to encounter when I go into work, studying engineering and all. See also the XKCD 'cancer causes mobile phones' comic.

Number 3 is a weird case, because it is both true and false (5% difference in university places by gender one year can be dismissed, but if this holds true for e.g. 5 years then it is important, if you'll except the Mount Stupid oversimplification). It's the difference between 70% of my class at university deciding to take up drug overdosing and me releasing a plague that'll kill 1% of all humans alive, I have met people who'll assume that the first situation is worse, despite it only risking the lives of less than 70 people, instead of killing 700,000,000 people. But I also see the opposite reasoning everyday, of that because a number is greater than 100/1000 it's important no matter if the percentage it represents is 90% or 1%. In fact, I'd say that the 'absolute numbers' misinterpretation is slightly more common, but that's just personal experience.

There is some truth to misunderstanding four, but only because a significant number of statistics have been creatively misinterpreted to project a certain image. At the end of the day though, I've discovered that statistics are the most reliable source of information available, simply because you can look at all of them and compare them to each other, even doing some statistics on statistics if it's really that important to you.

LudicSavant
2015-01-21, 07:17 PM
Considering that with statistics I'm currently climbing down Mount Stupid, I've just had to go and look up conjunction and disjunction, because I couldn't remember the names. That sounds like a stupid thing to assume, but I can see why people would do so.

Misunderstanding 2 is of course the classic one :smalltongue: it's also the one I'm most likely to encounter when I go into work, studying engineering and all. See also the XKCD 'cancer causes mobile phones' comic.

Number 3 is a weird case, because it is both true and false (5% difference in university places by gender one year can be dismissed, but if this holds true for e.g. 5 years then it is important, if you'll except the Mount Stupid oversimplification). It's the difference between 70% of my class at university deciding to take up drug overdosing and me releasing a plague that'll kill 1% of all humans alive, I have met people who'll assume that the first situation is worse, despite it only risking the lives of less than 70 people, instead of killing 700,000,000 people. But I also see the opposite reasoning everyday, of that because a number is greater than 100/1000 it's important no matter if the percentage it represents is 90% or 1%. In fact, I'd say that the 'absolute numbers' misinterpretation is slightly more common, but that's just personal experience.

There is some truth to misunderstanding four, but only because a significant number of statistics have been creatively misinterpreted to project a certain image. At the end of the day though, I've discovered that statistics are the most reliable source of information available, simply because you can look at all of them and compare them to each other, even doing some statistics on statistics if it's really that important to you.

Number one is far more common than you might think, and once you start looking for it you'll see it EVERYWHERE and it will drive you insane at the stupidity of man. In number 3, the difference that must be understood is context. 1% in one phrasing can be more meaningful than 95% in another... with respect to the same question. As for 4, the creative misinterpretation isn't the statistic's fault, but the reader's.

Amphetryon
2015-01-21, 07:19 PM
Folks tend to conflate gender and sex in these sorts of questions. We are usually talking about assigned gender, if that helps.
Given how the specific language in the rest of the example(s) changes the nature of the answer statistically, I find it curious that we're assuming the language about gender is accidentally ambiguous.

Pinnacle
2015-01-21, 08:17 PM
If I walk up to you and tell you, "The Joneses have two children; the oldest is a girl," that is sufficient to make whether or not their younger child is also a girl an independent probability.

It doesn't matter how I decided to give you that information.
It does, though.
"The eldest of two children is a girl" is true of half of all pairs, and half of those pairs have two girls. If you tell me that about a randomly selected pair, the odds are 50/50. I you tell me that only about pairs that have two girls, though, I'm never going to be right by guessing there's no match.
How you choose sets what the correct answers are. How could it not affect whether my answers are right?

No, it really isn't. If I give you that information, you know that there are three possibilities, as you said. However, you know that, if it is not HH, there is only a 50% possibility that I would have told you about the coin I told you about.
It's HT 1/4 of the time, and half the time it's HT you gave me that note. 1/4 x 1/2 = 1/8 of the total.
It's TH 1/4 of the time, and half the time it's TH you gave me that note. 1/4 x 1/2 = 1/8 of the total.
It's HH 1/4 of the time, and when it's HH you definitely gave me that note. 1/4 of total.
It's TT 1/4 of the time, and when it's TT you definitely did not give me that note. 0/4 of total.
1/8 +1/8 + 1/4 = 1/2.
That's correct. We know that you gave me that note half the time.
Half of the time that you gave me that note--1/4 of the total cases--there was a match. The other half there was no match.

Er, I don't follow. Break down this probability for me, please?
HH happens 1/4 of the time, and you give me the heads note every time. 1/4 x 1 = 1/4 of the time you've given me a heads note and there is a match.
HT happens 1/4 of the time, and you give me the heads note half the time. 1/4 x 1/2 = 1/8 of the time you've given me a heads note and there is no match.
TH happens 1/4 of the time, and you give me the heads note half the time. 1/4 x 1/2 = 1/8 of the time you've given me a heads note and there is no match.
There are two different ways this could happen, and each happens 1/8 of the time. 1/8 + 1/8 = 1/4
HT happens 1/4 of the time, and you give me the tails note half the time. 1/4 x 1/2 = 1/8 of the time you've given me a tails note and there is no match.
TH happens 1/4 of the time, and you give me the tails note half the time. 1/4 x 1/2 = 1/8 of the time you've given me a tails note and there is no match.
There are two different ways this could happen, and each happens 1/8 of the time. 1/8 + 1/8 = 1/4
TT happens 1/4 of the time, and you give me the tails note every time. 1/4 x 1 = 1/4 of the time you've given me a tails note and there is a match.

1/4 of the time you give me a heads note and there is a match, 1/4 of the time you give me a heads note and there is no match, 1/4 of the time you give me a tails note and there is not match, and 1/4 of the time you give me a tails note and there is no match.

That can also be said for if I don't TELL you how I'm choosing.
Probability only comes into play when there's randomness.
If I tell you there's a bag of marbles and only one out of 100 is blue, and ask you whether I've taken a blue one out of the bag, you might say the probability is 99% that I did not.
However, I chose the blue one intentionally knowing you wouldn't pick that.
"At least one is heads" is true in 75% of cases, but you're only telling me that in 50% of cases. Your selection method means that, in the cases you've chosen, half of them have matches.
It's not a random sample. Different odds apply to your selected sample, because there are different proportions in it.

If I played this game with a different pair of people 1000 times, we'd still expect the same probabilities of correctness (based on which answer they gave).

It doesn't change the probability of the accuracy of their guess, based on what their guess is.
And the percentages wouldn't be the same as with a random sample.
You've changed what the right answers are. If the right answers aren't the same as with a random sample, the percentage of people who get the answer right will change.

Except we've just established that the odds are actually 2/3 that you'll get it right if you guess "no." At least, by the breakdown I gave in that post.
And you made an error. Your breakdown shows that you're twice as likely to get HT or TH as HH or TT.

Again, our odds come out to 2/3 that "no" is right, by the analysis above. Not 1/2.
Note that your analysis is impossible.
Flip pairs of coins. How often do you get a match? 50% of the time.
If your analysis shows that people guessing you didn't get a match are right 2/3 of the time, you can see that you've made a mistake.
It has to be 50%.

This isn't true of the original scenario, because we disregard half of the matches.

Okay. Let's take a single instance.

HTH I give you, based on what's in the spoiler, the note that says, "At least one of them was Heads. Were they both heads?" Based solely on that note, you have a 2/3 chance of being right if you say "no."

The possible outcomes of the two coins, as YOU know them (assuming you don't know about "my" coin), are:

HH
HT
TH
Those are the possible outcomes, but they're not all equally likely to get me that note.
I don't know that, so my assumptions about probability will be wrong, but it's still the case.

If you know about my coin, the possible outcomes are:

HHH
HHT
HTH
THH

...and here I see where you're coming up with your 50% chance that you'd be right, rather than 2/3, by saying "no." As 2/4 results are "right" and 2/4 are "wrong" for "no."
Yup. I'd only be right half the time.

However, this implies that you knowing MORE information reduces your odds of guessing correctly. If this were a "man on the street" game, and I just walked up to random people and offered them \$10 if they got it right, but I didn't tell them how I was determining what question to ask them, why does the probability change, in theory, to 2/3?
It doesn't.
See, when I know there's at least one heads, I have only 1/3 odds of guessing the combination correctly. The question, though was the odds that a particular result was not correct. Since that's true of two of my three choices, there are 2/3 odds.
When I know that the first one was heads, though, I have 1/2 odds of guessing the combination correctly--that's more.

No, it's not. Not exclusively.
In random coin flips, there are 75% of cases where it's true that there is at least one heads. If you'd give me the same information for any of those cases, one-third of them (25% of all cases) have matches. 1/3 of the total sample have matches, so guessing that there is not match will be correct 2/3 of the time.
In your experiment, there are 75% of cases where it's true that there is at least one heads but you only give me that information in 50% of cases. It's still true that 25% of all cases have matches, but that's only 1/2 of the total sample. Guessing that there is not a match is only right half the time.
The samples are different, and contain different percentages of matches. Based on the information you've given me I might guess that there's no match, but information I do not know means there are totally different odds.

This is about whether or not, upon a stranger walking up to you and saying, "If the older sibling of two is a boy, what are the odds that they're both boys?" versus "If one of two siblings is a boy, what are the odds that they're both boys?"

How he decided to ask you about whether or not they're both boys (and that one was a boy), or of they're both girls (and one of them was a girl), should not matter to your odds of guessing correctly based on the information you are given.
Of course it does. If he only asks when there's a match, the odds of guessing right are 100% if I say there is and 0% if I say there isn't.

Your example isn't as extreme, but it's still true that you've altered the percentage of matches in the sample. That, of course, changes the odds of an answer being correct.

If we, again, only ever use any given player once, in theory the situation is the same as if a single experiment were done, at least for that one person. Right?

So... what about this scenario makes it so that, when I perform every step except handing the notes to people before hand, both players have the same odds of getting it right, when somehow the questions asked before give differing odds based on how they were asked?
It's not that you haven't told them. It's that your sample is different.

So, let's say you are approached on the street by somebody offering you the following information: "I have flipped two coins. At least one of them was heads. Do you think they both were?"

Would you still have a 2/3 chance of being right if you said "no?" You don't know how he decided to tell you that one of them was heads. You know one of them is, because otherwise he'd be lying.
For the sake of a probability puzzle, we assume each possibility is equally likely when we haven't been told otherwise.
The person approaching me is an unpredictable variable if he's a real person, though.
The odds vary based on what's in the sample. If he would have told me that for every case where it's true, the odds are 66.7% that there is no match. If he would have told me that for only cases where there's a match, the odds are 0% that there is no match. If he would have determined what information to give me the way you're doing here, the odds are 50%.
The odds of different correct answers change, so my odds of getting the answer correct also change.
Given no other information I may assume the first scenario as in the puzzle and guess there's no match with 66.7% confidence... but if he chose in one of the other ways I'll be wrong about that.

If said stranger repeated his question slightly differently, and said, "What do you think the other coin was?" why does that change your odds to be lower?
Because it's not just the wording that's different. "What are the odds both are heads?" is different from "What are the odds that one is heads?"
In the first scenario, telling me one of the coins is heads changes the chance from 1/4 to 1/3 because it's eliminated one possible result. In the second scenario, telling me that some other coin is heads tells me nothing relevant.

It is truly arbitrary which coin he flipped first.
Not in probability. HT and TH are different results when counting the total number of possibilities.

It's arbitrary whether he tells you the result of the first or second coin.
Ah, but it's not.
If the result was HH, he definitely tells me that.
His behavior is only arbitrary in certain situations.

While statistics claim this distinction is real, and I even understand the independence argument, I'm having trouble with the "same situation, different information" aspect of it changing the odds in the fashion it claims to.
The information changes the situation.

If there can be any of HH, HT, TH, and TT, there are four possibilities.
By telling me that at least one is heads, there were four possibilities, but you've told me that one didn't happen. Three remain.
By telling me that the first is heads, there were four possibilities, but you've told me that two didn't happen. Two remain.

In your experiment, deciding what to tell me is itself part of the situation. It has different possibilities. You're giving different information and asking different questions.

In the original puzzle, we're given a pair of children that was chosen randomly and given a piece of information about them. Our odds involve all pairs of children where the information we were given was true.
The way you're giving information is totally different from that. Our odds don't involve all pairs of coins where the information we were given is true, but all pairs of coins where you gave us the information.
Now, somebody who doesn't know about that might get it wrong, but the odds of your experiment are always 50%.

As for 4, the creative misinterpretation isn't the statistic's fault, but the reader's.
Sometimes it's the presenter's fault, intentionally trying to mislead.
There can also be intentionally misleading statistics.

Given how the specific language in the rest of the example(s) changes the nature of the answer statistically, I find it curious that we're assuming the language about gender is accidentally ambiguous.
It is quite an old question.
We also generally assume all coin tosses and die rolls are fair. Unless it's intentionally a trick question with a correct answer of "There is insufficient data to draw a conclusion" (which you do see), the puzzle probably expects you to assume anything unspecified is fair/simple/straightforward.

Anonymouswizard
2015-01-21, 08:18 PM
Number one is far more common than you might think, and once you start looking for it you'll see it EVERYWHERE and it will drive you insane at the stupidity of man. In number 3, the difference that must be understood is context. 1% in one phrasing can be more meaningful than 95% in another... with respect to the same question. As for 4, the creative misinterpretation isn't the statistic's fault, but the reader's.

Oh, I can see why people would believe number 1, and now thanks to you I'll be looking for it. I understand that the difference in number three is context, I just tried to illustrate it with examples instead of saying it outright, I'll remember to be more clear in future.

Number 4 is tricky, as it's easy to place the blame for creative misinterpretation on the reader, but there are times where the blame can be put on the presenter. I can't think of a good example off the top of my head (the one I have is simply twisting of how unstatistically back facts can be twisted), but you do see it all the time in the media, especially with the 'link between X and Y' stories uncovered (normally relying on misinterpretation number 2 to get readers to believe what they want you to believe). The problem with it though is that most people aren't interested in raw statistics, so you have to present it in an engaging way, which almost always slants towards one interpretation. In essence, if I decide to creatively misinterpret the statistic of '30% of people who go to university end up in entry-level jobs anyway' as 'it's not worth going to university as there's a decent chance you've paid £50000 for the same job you can get with A-levels', and then present it in a way that supports that interpretation then I can get other people to misinterpret it in the same way. But again, it isn't the statistics fault, but the interpreter's fault, who in most cases is the reader.

LudicSavant
2015-01-21, 08:38 PM
There are a lot of other issues that have the same root problem as issue 4, such as "Correlation means nothing." I feel like I should have worded it in an even broader way.

Basically, there is a tendency for people to dismiss or reject things entirely once they find that it doesn't mean a certain thing that they once expected it did. Just because correlation doesn't necessarily mean causation doesn't mean that it doesn't mean anything. Just because statistics can be misconstrued or deceptively presented doesn't mean that numbers lie.

Fiery Diamond
2015-01-21, 09:22 PM
Segev's problem (that is, the problem Segev is having, not the statistics thought problem) isn't super obvious, but it is actually one of the most basic reasons that very intelligent people trip over probabilities so much. In short, this is the issue:

Issue: Probability is used, commonly, to refer to two completely different concepts. It is extremely rare that people discussing it actually call this out, possibly because many don't realize that it's being used to talk about two different concepts.

Concept One: "If I pick this, then based on the information I have, I deduce that the chance of my choice aligning with present/past facts is [fill in the blank]." This is a concept about level of certainty.

Concept Two: "What is the chance that certain facts will come to pass in the future in a particular fashion?" This is a concept about likelihood of occurrence.

These two things, while both things that probability covers, ARE NOT THE SAME. Much of the confusion surrounding people putting lots of thought into particular scenarios and completely mangling the probabilities is a result of conflating the two concepts. This is also why the boy/girl "at least one boy is" vs. "the older boy is" confusion happens, by the way. People confusing the two concepts assume that the sex of the siblings MUST be independent of the other, so it doesn't matter how it is worded - because they don't realize that they are trying to apply concept two logic to a concept one scenario, and concept one scenarios are HEAVILY reliant on exactly what information is revealed.

For example, in Segev's thought problem, Segev doesn't understand why it is relevant how he chose to deliver information. To back up and use the Monty Haul problem...

Let's pretend we have never seen the problem and don't know the rules. We merely observe the following:

1: There are three doors. We are told one has a car and two have goats.
2: Abigail Gamble picks door number 1.
3: Host opens door number 2. There is a goat.
4: Host asks Ms. Gamble if she wants to switch.
Question: Should she?

Answer: Without further information, we don't know. There are a couple of possibilities, from our uninformed perspective.

1: Host picks a goat door and opens it in step 2. Host will do this no matter what door Gamble picked. (This is the actual scenario.)
2: Host picks a door at random/picks the next door in sequence/picks door 2 unless Gamble did in which case he would have picked door 1 or door 3. The door just happened to have a goat behind it because Host was unlucky/lucky.

The thing is, we have to pick an assumption or be given further information. We can go ahead and assume that scenario one is more plausible and work from there (in essence pretending we have more information than we do), we can ask and be told that it is scenario one (be given more information) or we can say since we don't have information we have to assume scenario two.

Here's where misunderstanding how probability concepts work comes in. When I was a kid, I argued about this problem (wrongly) while trying to understand it, and here's what I said... (you'll see how this lines up with Segev's issue, or at least I hope that will be clear)

"Well, as far as Gamble and I know, any door could have had anything at any point. We assume scenario two, since because we don't know how he decided, it doesn't matter [mistake number one]. So Gamble had a 1/3 chance, from her perspective, of being right the first time. But now that a goat has been revealed, there are two doors left and only one has a goat! Clearly, my probability can't have changed from 1/3 to 1/2 without me doing anything [mistake number two]. So if I keep the door, I have a 1/3 chance of being right. BUT there are two choices now, so if I switch, that's 1/2 chance. That's better, so I should switch [correct conclusion, wrong reasoning]. Hang on, those don't add up to 1. What gives?"

Confusing level of certainty with likelihood causes a ton of people to trip up when it comes to "why is this information relevant?"

LudicSavant
2015-01-21, 09:27 PM
Here's one that's much harder than the Monty Hall problem:

Have fun :)

Pinnacle
2015-01-21, 10:30 PM

Nothing happens.

I looked up solutions, but they make no sense.

I saw one that explained why it's a problem if there's only 1 dragon. That is true.
It then explained why there is a problem when there's only 2 dragons. That is also true.
It then explained why there is a problem when there's only 3 dragons. That is also also true.
But then it said you can extrapolate from there, and... no you can't.

It's harmless for groups of four dragons or more.
Dragon A is aware that dragons B, C, and D have green eyes. Dragon A is also aware that dragon B is aware that dragons C and D have green eyes. Dragon A is thus aware that dragon B is aware that dragon C is aware that dragon D has green eyes.

Oh wait, I figured it out. Everybody dies.
Gosh, breaking these things down really does work.

On the first night, dragon A doesn't expect anybody to die. After all, dragon B is aware that the one with green eyes could be either C or D, and C and D are each aware that the one with green eyes could be either B or the other one.
However, if A didn't have green eyes, and B doesn't know that she has green eyes, A expects B to expect that C and D both think only the other one has green eyes. Therefore, A expects B to expect that both C and D will notice that they did not turn into birds on the first night, and then realize that there must be at least one dragon other than the other one who also has green eyes.
A therefore expects B to expect C and D to have noticed that the other one did not turn into a bird, and therefore A expects B to expect C and D to turn into birds on the second night.
A can see that B can see that C and D did not turn into birds. A expects B to come to the conclusion that C and D must be able to see a third dragon with green eyes. Therefore, since A does not know she has green eyes, she must expect B (and C and D) to realize that they must have green eyes on the third night and turn into birds on the third night.
On the fourth day, A realizes that B and C and D must be able to see a fourth dragon with green eyes. That must be herself. B and C and D have all gone through the same reasoning at the same time.
All four of them turn into birds on the fourth night.

Basically, each dragon expects the others to conclude that they'll all turn into birds on day X - 1, which is three in my simplified case and 99 in the full problem.
When they don't, each dragon realizes that they must also have green eyes and all the others were also expecting the other 99 to turn into birds.
So it happens the next night.

Anonymouswizard
2015-01-21, 10:38 PM
Here's one that's much harder than the Monty Hall problem:

Have fun :)

This is probably going to be entirely wrong reasoning, but...

There are three possibilities: 1) nothing changes, 2) some but not all of them realise that they have green eyes, 3) they all realise they have green eyes. However, as all the dragons are essentially identical, we can assume they all have the same thought pattern (as the problem suggests that they are all perfectly logical), and so we only have scenarios 1 and 3 to worry about.

There are several possibilities for how 3 could occur, number 1 is that one dragon mentions that all dragons they have seen on the island have green eyes, and another dragon says that this is also the case for them (if this second part does not happen then we are left with one dragon on the island, but I suspect that with such interchangeable dragons multiple would mention it), and they all change. However, this introduces no new information.

2 is that they all knew that they shared an eye colour, but did not know what the colour green is. This means that you have introduced them to what the colour green looks like, and leads to either scenario three, or possibly scenario 4 (they change the rule).

I look forward to people telling me where I went wrong.

EDIT:

Nothing happens.

I looked up solutions, but they make no sense.

I saw one that explained why it's a problem if there's only 1 dragon. That is true.
It then explained why there is a problem when there's only 2 dragons. That is also true.
It then explained why there is a problem when there's only 3 dragons. That is also also true.
But then it said you can extrapolate from there, and... no you can't.

It's harmless for groups of four dragons or more.
Dragon A is aware that dragons B, C, and D have green eyes. Dragon A is also aware that dragon B is aware that dragons C and D have green eyes. Dragon A is thus aware that dragon B is aware that dragon C is aware that dragon D has green eyes.

Oh wait, I figured it out. Everybody dies.
Gosh, breaking these things down really does work.

On the first night, dragon A doesn't expect anybody to die. After all, dragon B is aware that the one with green eyes could be either C or D, and C and D are each aware that the one with green eyes could be either B or the other one.
However, if A didn't have green eyes, and B doesn't know that she has green eyes, A expects B to expect that C and D both think only the other one has green eyes. Therefore, A expects B to expect that both C and D will notice that they did not turn into birds on the first night, and then realize that there must be at least one dragon other than the other one who also has green eyes.
A therefore expects B to expect C and D to have noticed that the other one did not turn into a bird, and therefore A expects B to expect C and D to turn into birds on the second night.
A can see that B can see that C and D did not turn into birds. A expects B to come to the conclusion that C and D must be able to see a third dragon with green eyes. Therefore, since A does not know she has green eyes, she must expect B (and C and D) to realize that they must have green eyes on the third night and turn into birds on the third night.
On the fourth day, A realizes that B and C and D must be able to see a fourth dragon with green eyes. That must be herself. B and C and D have all gone through the same reasoning at the same time.
All four of them turn into birds on the fourth night.

Basically, each dragon expects the others to conclude that they'll all turn into birds on day X - 1, which is three in my simplified case and 99 in the full problem.
When they don't, each dragon realizes that they must also have green eyes and all the others were also expecting the other 99 to turn into birds.
So it happens the next night.

This is exactly like another puzzle I've seen, I just forgot the reasoning for it. All we need now is the new information.

LudicSavant
2015-01-21, 10:49 PM
Clarifications / hints:
- The dragons, in their wisdom, know what the color green is.
- Nobody is lying or guessing or using genetics or sign language or reflective surfaces of any kind. It's not a trick question.
- Just because 99 other dragons have green eyes does not necessarily mean that you do. Indeed, in some versions of this problem, not all of the dragons have green eyes. You can consider that a big hint: The fact that every dragon has green eyes is not relevant to the answer.

The correct answer to the 100 Green-Eyed Dragons problem (don't read it if you still want to figure it out yourself!):

All 100 dragons transform into sparrows on the 100th midnight. The new information you gave them is that every dragon has heard that at least one dragon has green eyes, and can be assumed to be acting based on that knowledge. Given the rules for dragon behavior, this makes it possible for the dragons to deduce their own eye color in 100 nights.

Talakeal
2015-01-21, 11:06 PM
I don't get it. What information are you providing?

They already KNEW that some of them had green eyes, they are looking at 99 green eyed dragons at that very moment, and have been for some time.

What does your statement of something they knew to be the absolute truth change their behavior in any way?

LudicSavant
2015-01-21, 11:16 PM
I don't get it. What information are you providing?

They already KNEW that some of them had green eyes, they are looking at 99 green eyed dragons at that very moment, and have been for some time.

What does your statement of something they knew to be the absolute truth change their behavior in any way?

Here is a complete explanation and proof of my answer:

Pinnacle
2015-01-21, 11:39 PM
I don't get it. What information are you providing?

They already KNEW that some of them had green eyes, they are looking at 99 green eyed dragons at that very moment, and have been for some time.

What does your statement of something they knew to be the absolute truth change their behavior in any way?

Each dragon is aware that there are at least 99 dragons with green eyes.

However, each dragon is not aware that he or she has green eyes, and also knows that each of the other 99 dragons is not aware that they have green eyes, and therefore thinks that each of the other dragons is only aware that there are 98 dragons with green eyes. Dragon 1 knows that dragons 2-99 have green eyes, but she knows that dragon number 2 is only aware that dragons 3-99 have green eyes--after all, if dragon number 2 is aware that all 99 of his associates also have green eyes, that must mean that dragon 1 does. And she doesn't, right?
As soon as it's proven that the other dragons also know there are 99, each one of them knows there's actually 100.

Talakeal
2015-01-22, 12:20 AM
Each dragon is aware that there are at least 99 dragons with green eyes.

However, each dragon is not aware that he or she has green eyes, and also knows that each of the other 99 dragons is not aware that they have green eyes, and therefore thinks that each of the other dragons is only aware that there are 98 dragons with green eyes. Dragon 1 knows that dragons 2-99 have green eyes, but she knows that dragon number 2 is only aware that dragons 3-99 have green eyes--after all, if dragon number 2 is aware that all 99 of his associates also have green eyes, that must mean that dragon 1 does. And she doesn't, right?
As soon as it's proven that the other dragons also know there are 99, each one of them knows there's actually 100.

Still not getting it. Does anyone have a step by step write out of the solution? Preferably one with only 3 or 4 dragons?

Pinnacle
2015-01-22, 12:43 AM
Still not getting it. Does anyone have a step by step write out of the solution? Preferably one with only 3 or 4 dragons?

If there's one dragon, he knows he's the one.
Done.

If there's two dragons, each dragon expects the other to be done at the end of the night. After all, unless I have green eyes, she knows she must be the one. She's not defeated on night one. She and I realize we both had the same reasoning, and are transformed on night two.

If there's three dragons, I look at the other two. Assuming I don't have green eyes, the two of them will look at each other and, well, use the reasoning for when there's only two. I'm irrelevant, unless I also have green eyes.
Each of them should be expecting the other to be transformed on the first night, and then both of them will realize the other also has green eyes and they'll both be transformed on the second night.
They're not, therefore, like myself, each of them must be seeing a second dragon with green eyes. That's me.
We're all transformed on the third night.

If there's four dragons, I expect each of the other three to, again, use the reasoning when there's only three. Again, unless I have green eyes they'll all be transformed on the third night. Dragon A doesn't know he has green eyes, and therefore expects B and C to each think the other is the only one with green eyes. When B and C aren't transformed on the first night, I now expect dragon A to think that dragons B and C will each realize that they both have green eyes and be transformed on the second night. When they're not, I expect dragon A (and B and C) to realize all three of them must have green eyes and be transformed on the third night.
When they're not, I realize they must all be aware of three dragons with green eyes and have gone through the same reasoning on the first three days as I did. We're all transformed on the fourth night.

If there's five and I'm dragon E, I expect dragon A to expect dragon B to expect dragons C and D to each think that the other one is the only one with green eyes.
After all, I'm assuming I don't. And I know that dragon A will assume he doesn't. So if A thinks I don't and he doesn't, he'll think that B and C and D are the only ones that do. So I expect A to expect that B will expect C and D to expect the other to be transformed on the first night. Then I expect A to expect B to expect C and D to realize it can't only be the other one, and both transform on the second night. Then I expect A to expect B and C and D to each realize that there must be a third, and then all transform on the third night. Then I expect A to realize that B and C and D must all be seeing a fourth green-eyed dragon--and I also expect B and C and D to have gone through the same reasoning as A--and I expect them to all transform on the fourth night.
They don't, so now I know they must each be seeing four, and that includes me. And all of us went through the same reasoning. We are all transformed on the fifth night.

Etc.

As long as you assume you don't have green eyes, you think all the others must have different reasoning than you do. When they don't, you must also have green eyes. And since they all have the same reasoning, they all figure it out at the same time.
If I don't have green eyes, I'm irrelevant to the others' reasoning. As soon as I'm clearly relevant, I must have green eyes.

It gets more complex at each stage, but as long as I assume I'm irrelevant, I can assume the others will go through the reasoning of the previous stage. And then they don't so obviously they know something that the hypothetical one-smaller group of dragons didn't. And that something must be that I also have green eyes.

LudicSavant
2015-01-22, 01:20 AM
If it is logically possible to deduce who had green eyes why haven't we already done so?

How would you do so before the human's announcement? If you can understand how you can deduce the answer AFTER the human's announcement, try and apply that solution before the human speaks, and see what happens.

Heck, after the humans announcement I would, if anything, be more confident that I didn't have green eyes, as I know for a fact every other dragon has green eyes and she implied that at least one of us didn't. No she did not. She did not say that, nor did she mean that.

Jeff the Green
2015-01-22, 03:38 AM
robability is used, commonly, to refer to two completely different concepts. It is extremely rare that people discussing it actually call this out, possibly because many don't realize that it's being used to talk about two different concepts.

Concept One: "If I pick this, then based on the information I have, I deduce that the chance of my choice aligning with present/past facts is [fill in the blank]." This is a concept about level of certainty.

Concept Two: "What is the chance that certain facts will come to pass in the future in a particular fashion?" This is a concept about likelihood of occurrence.

These two things, while both things that probability covers, ARE NOT THE SAME.

I'm not certain that they necessarily aren't the same. It's entirely plausible that, from a human perspective, concept two is a subset of concept one (at least for most events). If non-quantum events are purely mechanistic, the chance that certain facts will come to pass in the future in a particular fashion (e.g. this die will come up as a 2) are either 1 or 0 and never in between.

Segev
2015-01-22, 10:07 AM
Probability only comes into play when there's randomness.
If I tell you there's a bag of marbles and only one out of 100 is blue, and ask you whether I've taken a blue one out of the bag, you might say the probability is 99% that I did not.
However, I chose the blue one intentionally knowing you wouldn't pick that.
"At least one is heads" is true in 75% of cases, but you're only telling me that in 50% of cases. Your selection method means that, in the cases you've chosen, half of them have matches.
It's not a random sample. Different odds apply to your selected sample, because there are different proportions in it.Hm. This is building up to a different but related question for me, then. Bear with me; I need to trim a lot to try to synthesize this. My apologies if I miss anything.

And the percentages wouldn't be the same as with a random sample.
You've changed what the right answers are. If the right answers aren't the same as with a random sample, the percentage of people who get the answer right will change.The issue lies in the fact that, the moment I tell you, "at least one of them is heads," I have eliminated at least one possibility.

We're looking at the odds of you guessing correctly based on information I give you. If I have determined I am going to do this experiment, I obviously am going to ask you a question. If I simply decide I will always ask you if both were heads, and will tell you the result of at least one coin no matter what, we wind up with the following possibilities:

HH, "one of the coins is heads" "yes" correct
HH, "one of the coins is heads" "no" incorrect
HT, "one of the coins is heads" "yes" incorrect
HT, "one of the coins is heads" "no" correct
HT, "one of the coins is tails" "yes" incorrect*
HT, "one of the coins is tails" "no" correct
TH, "one of the coins is heads" "yes" incorrect
TH, "one of the coins is heads" "no" correct
TH, "one of the coins is tails" "yes" incorrect*
TH, "one of the coins is tails" "no" correct
TT, "one of the coins is tails" "yes" incorrect*
TT, "one of the coins is tails" "no" correct

If we assume that you are neither logically deficient nor deliberately answering wrong, the * options will never come up. You know from the information given that both were not heads.

A quick examination makes it look like, upon throwing out the * options, that you will guess right 2/3 of the time.

However...

The actual question is what are the odds that you'll guess correctly, given the information.

Given the information, "one of the coins is tails," there is not a 1/3 chance you'll guess "yes." There is a 100% chance you'll guess "no," and be correct, because "HH" is eliminated by the information "at least one is tails."

This still is dancing around it, and I can't quite put my finger on it. But the point here is that the very fact of trying to frame the question creates "added" information that you wouldn't be asking if both were tails. Semi-obviously; the hypothesis states that "TT" is the result eliminated by "at least one is heads."

I suppose the question here is, how do you even set up an experiment to actually test the hypothesis that telling somebody "at least one is heads" gives them a 2/3 chance of guessing correctly whether both were heads?

Or is that the wrong question? Maybe a better question is, "What does it mean that there is a 2/3 chance that both are not heads?"

Maybe the better "game" to test this is, "You win if both come up heads. You lose if one comes up heads and the other tails. You flip again if both come up tails." The alternative game would be, "You win if both come up heads. You may flip again if the green one comes up tails. You lose if the blue one is tails and the green one is heads." This gives the right odds, at least. 1/2 of winning the second, 1/3 of winning the first.

If, before you played the first game, you gave the player the option, "Do you want to win if both come up heads, or if one comes up heads and the other comes up tails? You will lose in the case you do not choose," he would be smart, in theory, to pick the HT/TH case. It gives him a 2/3 chance to win.

This brings us back to our original game, though: the test-giver can flip the coins until he has "heads" on at least one coin, and then tell the player that he has done so. The player is 2/3 likely to win if he guesses "one is heads, one is tails."

HOWEVER. The theory goes that the test-giver, by asking, "I have flipped two coins. One came up heads; what is the other?" has given the player enough information to make it only a 50/50 shot that he'll guess correctly.

We come back again to the test methodology. We know he specifically was looking for a situation where he could truthfully say, "at least one is heads" or "one is heads."

Perhaps the problem is that we're arguing in circles. On the one hand, we're saying that what the odds are depends on the information given. On the other, however, we're arguing that it's only the case because of how we decided what information to give. But that information can only be given if the experiment is set up to allow for it.

If somebody (truthfully) tells us, "I have flipped two fair coins; one came up heads," we do not know how he arrived at this set of flips.

Option 1: He flipped two coins and gave us information about an arbitrarily-selected one of them.

Option 2: He flipped two coins until one came up "heads" so that he could give us that specific information.

Option 3: He's attempting to play with our expectations, and actually flipped the coins until BOTH came up heads, and only then gave us the information.

Option 4: He actually deliberately flipped them until the green coin (or the one on the left, or whatever) came up heads.

It would seem, from this (very lengthy) examination, that it is which method he used to generate this instance, not the information we are given, which influences the result. This makes a bit more sense, since it's not really about information GIVEN so much as information about how the results are generated that SHOULD matter.

This makes me think the "The older child is a girl" information is actually badly worded as a problem, though. We don't know whether the question-asker looked at the first 2-child family he could find, and decided he would give us the information about the older child (option 1, above), or whether he looked for the first 2-child family with at least one girl, and told us about that child (choosing arbitrarily which to tell us about if both were girls) (option 2, above), or whether he looked for the first 2-child family where the older child was a girl so he could give us that specific information (option 4, above).

Because we do not know how he generated his result, "the older child" is irrelevant information. It has no bearing.

Without knowing the method by which the result is generated and the information given is to be determined, there actually isn't any difference between the questions, "At least one is heads; are they both heads?" and "One is heads; is the other one also heads?"

Both actually do have only a 50% chance of being guessed correctly. We know only the result of one of the coins. The extra information doesn't make any difference, because we don't know how it was generated.

We would have to know that "at least one is heads" means, "we chose only from samples where at least one was heads."

Again, I think this is best illustrated by choosing whether to phrase the question as, "At least one is heads; are they both?" versus, "One is heads; what is the other?"

These are the same functional question, with the same functional probability, because you don't know how it was decided what information would be given to you.

Therefore, I think the "older child is a girl" question is badly phrased, and that it needlessly creates confusion in the principle it is attempting to illustrate. It falsely attributes "information given" as the cause of the probability shift, where it should actually be "knowledge of how samples are chosen."

Pinnacle
2015-01-22, 12:01 PM
If we assume that you are neither logically deficient nor deliberately answering wrong, the * options will never come up. You know from the information given that both were not heads.

A quick examination makes it look like, upon throwing out the * options, that you will guess right 2/3 of the time.
You still asked me the questions those times, though. Saying I wouldn't have given those answers doesn't mean you should throw them out, it means you should assume I answered correctly instead.
This is because, in half of all situations, you already told me the answer. Thus, 50% of the time, you know I'm going to get it right.
I get it right half the other times. Total of 75%.

I suppose the question here is, how do you even set up an experiment to actually test the hypothesis that telling somebody "at least one is heads" gives them a 2/3 chance of guessing correctly whether both were heads?
Ask the same question every time. Only ask for situations where "at least one is heads" is correct.
See, my reasoning would be based on the knowledge that "In 1/3 of all cases where it is true that at least one is heads, there are two heads;" however, you're not presenting me with all situations where it's true. You're only presenting me with a subset of cases where it's true, and in fully half of those there is a match.

Or is that the wrong question? Maybe a better question is, "What does it mean that there is a 2/3 chance that both are not heads?"
Very simple. In 2/3 of all cases where it is true that "at least one is heads" is true, there is no match.
In this case 25% of the time there are two heads and you told me there is one, 12.5% of the time you flipped HT and told me there was a heads, 12.5% of the time you flipped HT and told me there was a tails, 12.5% of the time you flipped TH and told me there was a heads, 12.% of the time you flipped TH and told me there was a heads, and 25% of the time you flipped tails and told me there was a tails.
If we add that up, we see that 25% of the time you told me there was a heads and there was a match, 25% of the time you told me there was a heads and there was no match, 25% of the time you told me there was a tails and there was no match, and 25% of the time you told me there was a tails and there were two of 'em.
Half the time you gave me that information, there is a pair. Regardless of my answer, I will be right half the time I get that note. And 100% of the times I get the other note, of course.

Maybe the better "game" to test this is, "You win if both come up heads. You lose if one comes up heads and the other tails. You flip again if both come up tails." The alternative game would be, "You win if both come up heads. You may flip again if the green one comes up tails. You lose if the blue one is tails and the green one is heads." This gives the right odds, at least. 1/2 of winning the second, 1/3 of winning the first.
Exactly, the odds of getting any particular result are equal to the percentage of the result in the sample.
In this case, you're only presenting me with situations where information equivalent to the puzzle is true, and you are presenting me with all of those situations.

HOWEVER. The theory goes that the test-giver, by asking, "I have flipped two coins. One came up heads; what is the other?" has given the player enough information to make it only a 50/50 shot that he'll guess correctly.
Nah, 'cuz the "other" one could still be either one; they're still undesignated. It's just a weird way of phrasing our first problem.
If you do this every time at least one is heads, only in 1/3 of situations will the other be.

Perhaps the problem is that we're arguing in circles.
...Are we arguing? I thought you were trying to understand and I was trying to explain.

On the one hand, we're saying that what the odds are depends on the information given. On the other, however, we're arguing that it's only the case because of how we decided what information to give.
The more information you give me, the more likely I am to arrive at the correct conclusion. Only knowing that there is at least one heads (or girl), I know that there are three different possible combinations and I have 1/3 odds of guessing correctly. Knowing that the first is heads (or the eldest is a boy), I know that there are two different possible combinations and I have 1/2 odds of guessing correctly.
However, the percentage of cases I will actually guess right are quite dependent on the percentages of different possibilities in the sample. As an extreme case, if you only present the problem when there is a match, the odds of being correct if I say there is are 100% and the odds of being correct if I say there isn't is 0%--because in 100% of cases it is true. Now, your sample has both matches and non-matches so it's not that evidently extreme, but it still has different percentages than the set containing all cases does.
Both are true. Give me more information and I'll be more likely to get a correct answer, but alter the sample and you alter the odds.

If somebody (truthfully) tells us, "I have flipped two fair coins; one came up heads," we do not know how he arrived at this set of flips.

Option 1: He flipped two coins and gave us information about an arbitrarily-selected one of them.

Option 2: He flipped two coins until one came up "heads" so that he could give us that specific information.

Option 3: He's attempting to play with our expectations, and actually flipped the coins until BOTH came up heads, and only then gave us the information.

Option 4: He actually deliberately flipped them until the green coin (or the one on the left, or whatever) came up heads.

It would seem, from this (very lengthy) examination, that it is which method he used to generate this instance, not the information we are given, which influences the result. This makes a bit more sense, since it's not really about information GIVEN so much as information about how the results are generated that SHOULD matter.
Right. Based on the information we're given, we assume we're dealing with all cases where it's true for the sake of solving the puzzle.
However, each methodology has entirely different odds. If we don't know that, our assumption is incorrect.
Given that it's a puzzle, assuming all possible cases are in play unless told otherwise is fair--although the puzzle really should say so to be entirely complete.

This makes me think the "The older child is a girl" information is actually badly worded as a problem, though. We don't know whether the question-asker looked at the first 2-child family he could find, and decided he would give us the information about the older child (option 1, above), or whether he looked for the first 2-child family with at least one girl, and told us about that child (choosing arbitrarily which to tell us about if both were girls) (option 2, above), or whether he looked for the first 2-child family where the older child was a girl so he could give us that specific information (option 4, above).
All we have to go on is what we were told. We can probably assume all possible cases are in play and there's even odds of boy/girl since otherwise the only possible is "Insufficient data," but you're right that the puzzle really should include all this information.

Without knowing the method by which the result is generated and the information given is to be determined, there actually isn't any difference between the questions, "At least one is heads; are they both heads?" and "One is heads; is the other one also heads?"

Both actually do have only a 50% chance of being guessed correctly. We know only the result of one of the coins. The extra information doesn't make any difference, because we don't know how it was generated.
If two coins were flipped and we were given that information, there are three possibilities and only one of them is a pair. 1/3 odds.
That only changes if cases were discarded.
...although it's true that if the asker may have given us different information in the case of a mismatch, that does alter the odds as well.

Again, I think this is best illustrated by choosing whether to phrase the question as, "At least one is heads; are they both?" versus, "One is heads; what is the other?"
Those are the same, just one with bad wording.
Saying "This one is heads" and removing it from play is the equivalent of our girl problem. And it alters the odds, because we're eliminated the case where that one was tails and the other was heads. As long as I don't know what "one" is I also don't know what "the other" is.

Therefore, I think the "older child is a girl" question is badly phrased, and that it needlessly creates confusion in the principle it is attempting to illustrate. It falsely attributes "information given" as the cause of the probability shift, where it should actually be "knowledge of how samples are chosen."
Nah. In both cases the assumption is "all cases where this information is true". One of them just provides more information, and thus limits the sample because there are less cases where it is true.

Segev
2015-01-22, 01:12 PM
Nah, 'cuz the "other" one could still be either one; they're still undesignated. It's just a weird way of phrasing our first problem.
If you do this every time at least one is heads, only in 1/3 of situations will the other be.Eh... thing is, we're back to "how did I choose what extra information to present?"

If we assign equal probability to all three of options 1, 2, and 4 (discounting 3 because it's more mind-game than probability experiment), we could probably find the actual odds.

But that's a digression, sorry.

The problem is that we're arbitrarily calling something "sufficient" specification.

"One of these is heads; what is the other?" is not specifying which one. But "the one on the left is heads; what is the other?" is, somehow.

Again, it comes down to how we came to the decision of what information to present. In option 1, I had decided I would tell you about the left one, no matter what. Is, then, the fact that you know I'm telling you about the one on the left relevant? Or is it still a 1/3 chance that the other one (the one on the right) matches?

What makes the specification...specific? Rather than extraneous information no more designating than "one" and "the other?"

My conclusion in the last one is that, absent information about how you decided it was the older sister you'd tell me about, it is actually no more information than if you just picked one of the two children at random to tell me about. It's no different than if you flipped a coin to decide whether you'd tell me about the older or the younger child.

In fact, the ONLY way to generate the "at least one child is a girl; are they both girls?" 1/3 probability of "yes" is if we explicitly select for situations where that is the case.

That is, we can safely say that, if we only look at families of 2 children where at least 1 is a girl, the odds that both are girls in any one of those families chosen at random is 1/3.

However, actually presenting the question, "at least one child is a girl; are they both girls?" does not provide any more or less information than "the older child is a girl; are they both girls?" does.

This is illustrated by my earlier blue-coin/green-coin game, wherein I chose the information I gave such that it would always be true (and, based on the information, was not 100% known), but randomized it where it was possible to be truthful with two or more different pieces of information. It broke down to 50% odds, because of the nature of how I generated questions and information.

Similarly, "at least one child is a girl" does not tell us whether we would ONLY have been presented this situation if at least one was.

To put it another way: because we've reduced it to THIS specific sample of 2 children, there exist only two children. It really doesn't matter how or if you specify one versus the other; having learned "one," we know everything we need to know - it is not the other. There is no further specification that can be given that is meaningful for the purposes of our question.

Hence why the problem is so deceptive. In truth, the "at least one child" version is actually specifying one, because it's asking about one specific sample, not about a collection of samples wherein that condition is true. In any one specific sample of 2 children, if "at least" one is something, we know information about one of them. We can now specify them as "the one we know this about" and "the other, whom we do not."

Khedrac
2015-01-22, 01:26 PM
However, actually presenting the question, "at least one child is a girl; are they both girls?" does not provide any more or less information than "the older child is a girl; are they both girls?" does.
Now the statement I have quoted above is where you are going wrong as it is demonstrably false.

I think we all agree that for two children the four possibilities overall are:

1) Girl then Boy
2) Girl then Girl
3) Boy then Girl
4) Boy then Boy

Saying "The older is a girl" covers two of the options (1 and 2) whilst saying "at least one is a girl" covers three possibilities (1, 2 and 3).

If we ignore all the rest of the stats, since they provide for different list of possible outcomes the two statements provide very different information.

Once we have accepted that they provide different information, accepting that they may have different outcomes should be automatic.

Incidentally, once you see just how difficult to understand statistics can be it becomes really easy to see why people who know a little, often do not trust it.

Pinnacle
2015-01-22, 01:28 PM
"One of these is heads; what is the other?" is not specifying which one. But "the one on the left is heads; what is the other?" is, somehow.
There are three ways for the former to be true. There are only two ways for the latter to be true.

What makes the specification...specific? Rather than extraneous information no more designating than "one" and "the other?"
It's all about the proportions among the sample. The specification removes another one of the original four possibilities, and now there are only two.

My conclusion in the last one is that, absent information about how you decided it was the older sister you'd tell me about, it is actually no more information than if you just picked one of the two children at random to tell me about. It's no different than if you flipped a coin to decide whether you'd tell me about the older or the younger child.
That's true. As long as you tell me which one you're talking about, the odds are 50/50, because you've always eliminated two possibilities--the two where the thing you told me isn't true.

In fact, the ONLY way to generate the "at least one child is a girl; are they both girls?" 1/3 probability of "yes" is if we explicitly select for situations where that is the case.
I don't see how "all cases where this is true" is more specific than a subset of them. But sure, true 'nuff.

That is, we can safely say that, if we only look at families of 2 children where at least 1 is a girl, the odds that both are girls in any one of those families chosen at random is 1/3.
Correct.

However, actually presenting the question, "at least one child is a girl; are they both girls?" does not provide any more or less information than "the older child is a girl; are they both girls?" does.
Incorrect. The first clue tells us there are three possibilities; the second narrows it to two.

This is illustrated by my earlier blue-coin/green-coin game, wherein I chose the information I gave such that it would always be true (and, based on the information, was not 100% known), but randomized it where it was possible to be truthful with two or more different pieces of information. It broke down to 50% odds, because of the nature of how I generated questions and information.
Yes. You gave me the information that at least one is heads every time there was a match, but only [i]half the time it was a mismatch.
The odds of a match are 50/50 in any given pair of coin tosses (counting tails matches--1/4 otherwise).
The odds of a match are 1/3 in all cases that contain at least one heads.
The set you gave me didn't include all such cases, though; it only included a fraction of the ones where there was no match.

Similarly, "at least one child is a girl" does not tell us whether we would ONLY have been presented this situation if at least one was.
No. If we have only that one piece of information, we can only narrow the possibilities down to the three cases where that information is true.

Hence why the problem is so deceptive. In truth, the "at least one child" version is actually specifying one, because it's asking about [i]one specific sample, not about a collection of samples wherein that condition is true. In any one specific sample of 2 children, if "at least" one is something, we know information about one of them. We can now specify them as "the one we know this about" and "the other, whom we do not."
You don't know the right answer, though. We simply do not have the information to state it, since there is only one piece of information complete enough to determine it: The answer itself.
The question wasn't about the correct answer, though, it was about probability. The chances of each answer being correct are determined by the sample.
The sample where "at least one child is a boy" is a larger sample than "the eldest is a girl."
Which single case was the one chosen? No idea, other than it's one of the ones where the single clue is true.

Lord Torath
2015-01-22, 01:40 PM
Okay, let's try a slight change. Let's say I have twins Cameron and Kim. Fraternal, but nearly identical pre-pubescent twins. Cameron has a 50% chance of being a girl, and Kim also has a 50% chance of being a girl. There are three possible sex combinations: Boy-Girl (50%), Boy-Boy(25%), and Girl-Girl(25%). I put a sheet over each of them, and put them in my front room. I tell you that one of them, Cameron, is a girl. This eliminates the Boy-Boy possibility. So now the odds are Boy-Girl(66.7%) and Girl Girl(33%). Now you don't know which one is Cameron. But you know that Kim has a 50% of being a girl. So now we've got the paradox. There are two children in front of you (under sheets), and a 33% chance they are both girls. But there's a 50% chance that Kim is a girl, and thus that both are girls. How do you resolve this?

Pinnacle
2015-01-22, 01:49 PM
Permutations matter in probability.

It's not:
2 girls 25%
2 boys 25%
one of each 50%

It's:
Both are girls 25%
Cameron is a girl and Kim is a boy 25%
Cameron is a boy and Kim is a girl girl 25%
Both are boys 25%

By telling me that Cameron is a girl, you haven't just eliminated the possibility that both are boys, you have also eliminated the possibility that Cameron is a boy and Kim is a girl. You've eliminated half of the possibility that there's one of each along with the possibility that there's a pair of boys.
There's only two permutations left.

It's basically the same as the "eldest is a girl" premise, only with names instead of ages.

This stuff is really confusing, and I don't understand it too intuitively myself. Write out all possibilities and you'll always find the answer that way, though. Eventually.
Well, assuming you don't make a mistake of course. :smalltongue:

Our possibilities here are:
C-boy/K-boy
C-boy/K-girl
C-girl/K-boy
C-girl/K-girl
50% chance of match. 25% chance of match of girls, specifically.

If I know at least one is a girl, we're left with:
C-boy/K-boy
C-boy/K-girl
C-girl/K-boy
C-girl/K-girl
33.3..% chance of match of girls.

If I know Cameron is a girl, we're left with:
C-boy/K-boy
C-boy/K-girl
C-girl/K-boy
C-girl/K-girl
50% chance of match of girls.

Jay R
2015-01-22, 02:45 PM
If you want to understand conditional probability problems, take a probability course or read a probability book. Get at least through conditional probability.

If you will not do this, you will not understand conditional probability problems.

Nothing replaces learning the material.

Segev
2015-01-22, 03:16 PM
Now the statement I have quoted above is where you are going wrong as it is demonstrably false.

I think we all agree that for two children the four possibilities overall are:

1) Girl then Boy
2) Girl then Girl
3) Boy then Girl
4) Boy then Boy

Saying "The older is a girl" covers two of the options (1 and 2) whilst saying "at least one is a girl" covers three possibilities (1, 2 and 3).

If we ignore all the rest of the stats, since they provide for different list of possible outcomes the two statements provide very different information.

Once we have accepted that they provide different information, accepting that they may have different outcomes should be automatic.

Incidentally, once you see just how difficult to understand statistics can be it becomes really easy to see why people who know a little, often do not trust it.

This is actually not correct. We have changed the nature of the problem by reducing the sample size to one, and specifying nothing about the population from which the family was selected.

As a selection criterion, "at least one child must be a girl" does generate a population where only 1/3 of the 2-child families present have two girls.

Similarly, "the older child must be a girl" will generate a population where 1/2 are both girls.

However, as demonstrated by the blue-coin/green-coin game, where the content of the information and the question asked is determined by the nature of the result, a single sample presented to a given individual does not work that way.

You have to be told not, "at least one of them is a girl," but "chosen from a population of 2-child families where at least one was a girl." Because otherwise, it is equally possible that it was chosen from a population of all 2-child families, and the nature of the information given was based on the family chosen.

As demonstrated, that changes things.

Essentially, unless you specify that the information given is about the selection criteria for the sampled population, rather than information chosen based on the sample taken, you have rendered whether or not the older child is the girl irrelevant. By specifying, for a single sample, "at least one is a girl," you have created a situation where you know the answer to one of the two.

That is, you now can identify the children as "the one whose sex is known, and the one whose sex is unknown."

To be as clear as possible, this is because we do not know how you chose whether to ask us about both being girls, nor how you chose to tell us that at least one was a girl. Not knowing that, we do not actually have:

BB at 1/3
BG at 1/3
GB at 1/3

We have

BB "at least one is a boy" at 1/3
BG "at least one is a boy" at 1/6
BG "at least one is a girl" at 1/6
GB "at least one is a boy" at 1/6
GB "at least one is a girl" at 1/6

The stricken-thru ones are additional possibilities you've removed with the information you've given, that at least one is a boy. By telling us that, you tell us that it is not the half of the times BG is the result that you tell us there is a girl, nor is it the half of the times GB is a result that you tell us there is a girl.

That leaves us with normalized probabilities:

BB "at least one is a boy" 1/2
BG "at least one is a boy" 1/4
GB "at least one is a boy" 1/4

BB has twice the chance of the others because it doesn't matter which of them you tell me about; you'll give me the same information (that one is a boy).

Regardless, because it is a singular sample taken from an unspecified population with the criteria for how you chose whether to tell me about the older or younger child (without telling me which), you throw out enough additional possible results that it again becomes 1/2 chance.

Segev
2015-01-22, 03:23 PM
If you want to understand conditional probability problems, take a probability course or read a probability book. Get at least through conditional probability.

If you will not do this, you will not understand conditional probability problems.

Nothing replaces learning the material.

I have. I did understand it. This problem has bugged me anyway because of the paradox of the blue-coin/green-coin game as I tried to set it up.

I have come to the conclusion that, because setting up the coin game influences the probability such that the information given is no longer determining of the odds, the problem as presented is poorly set up.

To properly set it up, it would have to be phrased thusly:

"Of all families with 2 children wherein the older child is a girl, what are the odds that a single randomly-chosen family will have both children be girls?"

and

"Of all families with 2 children wherein at least one of the children is a girl, what are the odds that a single randomly-chosen family will have both children be girls?"

Those two questions properly frame it in a way that shapes the probabilities as specified in the "textbook answer."

As phrased earlier in this thread, the question arises about how you chose to give us the information about what "at least one" of the children's sexes was. Without being told that this was a random sample from a population where "at least one is a girl" is always true, we cannot know that the information and the question were not chosen AFTER the sample was taken, thus providing only information that lets us identify the children as "the one whose sex we know" and "the one whose sex we don't know." Or, if you prefer, introducing the now-excluded options of BG "at least one is a boy" and GB "at least one is a boy."

Heck, given how humans usually hand out information, it is reasonable to assume that the information came from observation of the chosen sample, rather than a criterion of the population.

Talakeal
2015-01-22, 03:41 PM
No she did not. She did not say that, nor did she mean that.

Hence the word "imply". In my experience you would only use the word "some" to mean "all" if you are deliberately trying to trick or mislead someone, as it is technically true, but not something you would say when there is a much more appropriate word.

Jay R
2015-01-22, 03:46 PM
I have come to the conclusion that, because setting up the coin game influences the probability such that the information given is no longer determining of the odds, the problem as presented is poorly set up.

You are absolutely right. It is a poorly designed math question. But when a puzzle has been set up to be hard to solve, it has been well set up, not poorly.

Segev
2015-01-22, 04:26 PM
You are absolutely right. It is a poorly designed math question. But when a puzzle has been set up to be hard to solve, it has been well set up, not poorly.

http://xkcd.com/169/

"Difficult to solve" is not the same as "incorrectly specified."

And in this case, as the effort to set up the experiment as a game demonstrated, the incorrect specification actually renders the sought-after answer wrong. Which means that it's as valid as: "What color is a tomato?" "Red!" "No, that's a common misconception. See, I was talking about this picture of a tomato in this coloring book, where my 5-year-old has colored it purple."

Yes, if you change the scenario not to be the one presented, with assumptions not in evidence, the sought-for answer is right. But it's wrong under not just reasonable assumptions, but the only assumptions the person being asked is even capable of making without being a mind-reader (or knowing this specific problem and how it's usually designed).

SpectralDerp
2015-01-23, 02:11 AM
{scrubbed}

Jenerix525
2015-01-25, 07:55 PM
Hence the word "imply". In my experience you would only use the word "some" to mean "all" if you are deliberately trying to trick or mislead someone, as it is technically true, but not something you would say when there is a much more appropriate word.

In my experience, you I would use the word "some" because there might be dragons I haven't met.
(Or because "all" would make it obvious to every dragon that they have green eyes, and I am too sentimental to guarantee their transformation knowingly. But that falls under deception, so you included that.)

Besides, as an 'infallibly logical' dragon, you are aware that such a word choice is unpopular but possible. Thus, you know it is possible that you have green eyes, and work it out as provided.
You can't choose to ignore the possibility, you're a hyper-intelligent dragon and the conclusion takes less than a milli-second for you. Plus, ignoring the truth is against your draconic nature.

Technically, the problem doesn't state that each dragon knows his peers to be logical. It does state that it is obvious, and leaves you to assume that the dragons are aware of said obvious fact.

The only certain new information is that "every other dragon knows that at least one has green eyes".

Dragon's noses are natural lie detectors, don'cha know.

2015-01-25, 09:26 PM
You are absolutely right. It is a poorly designed math question. But when a puzzle has been set up to be hard to solve, it has been well set up, not poorly.
I think the point of some statistics/logic problems being taught in the first place is to show how easy it is to be misled by intuition. It's a valuable thing to learn, for people dealing with mathematics. Because being that rigorously careful is annoying and difficult. And even if one has a lot of practice with logic, and is quite good at it, it's never a sure thing.

Plus, it's a good demonstration of how easy it is to accidentally make hidden assumptions. That is, if you can convince someone they've actually made that hidden assumption in the first place. (Sometimes it's quite hard to convince people they're absolutely wrong!)

LooseCannoneer
2015-01-26, 10:03 AM
People who think the probability is 1/2 are making a mistake. The question isn't worded, "Would you like to keep your door or switch to the UNOPENED door," its, "Would you like to keep your door or switch to ANY door. The not taking of the garbage prize is so obvious that people cut that step.

Segev
2015-01-26, 10:09 AM
I think the point of some statistics/logic problems being taught in the first place is to show how easy it is to be misled by intuition. It's a valuable thing to learn, for people dealing with mathematics. Because being that rigorously careful is annoying and difficult. And even if one has a lot of practice with logic, and is quite good at it, it's never a sure thing.

Plus, it's a good demonstration of how easy it is to accidentally make hidden assumptions. That is, if you can convince someone they've actually made that hidden assumption in the first place. (Sometimes it's quite hard to convince people they're absolutely wrong!)

The issue I'm having is that there are hidden assumptions in the asking of the "trick" question.

As evidenced by the attempt to construct a game from it with green and blue coins, you cannot actually reduce it to only the three options:

HH
HT
TH

without also reducing the odds of HT and TH to be half of what that presentation above makes them appear to be. That is:

HH (50%)
HT (25%)
TH (25%)

This is because, unless you specify that you deliberately first constructed the pool of possible families from those wherein at least one is a girl, it is impossible to tell if we don't actually only have a 50% chance of being told "at least one is a girl" vs. "at least one is a boy" in the HT/TH possibilities.

To word them in a way that illustrates, properly, the deceptive nature of intuition, one has to say:

The Smiths are one family in a sample group of 2-child families wherein the oldest child is a girl. What are the odds that both their children are girls?

and

The Joneses are one family in a sample group of 2-child families wherein at least one child is a girl. What are theo dds that both their children are girls?

Phrase it as originally phrased, and the blue-coin/green-coin game problem of having a 50/50 shot at being told about the boy or girl in the mixed cases reduces it again to a 50% chance that both are girls.

NecroRebel
2015-01-26, 01:04 PM

Look back at your post here. (http://www.giantitp.com/forums/showsinglepost.php?p=18693159&postcount=60) You'll note that before you started conflating the two questions, you found that always answering "no" was better for the person who was asked the "at least one" question. If, rather than randomizing the questions asked as you did there, you always asked the same question and simply re-did any flips that didn't fit the question (such as HH on an "at least one tails" scenario), you'd find that answering "no" is correct 2/3 of the time and answering "yes" is correct 1/3 of the time.

And when you get to the part of the post starting "If they always guess 'no...'" you should split the list into two parts, one where you always ask about heads and one where you always ask about tails. As is, in that section, you're adding the 3 possibilities for heads (along with the impossible TT) to the 3 possibilities for tails (along with the impossible HH) and concluding that since 4/8 are correct that it's a 50% chance, when it's really one 2/3 that is correct and another 2/3 that is correct, or total of 4/6 that are correct, for an ~66% chance.

Combining "I flipped these two coins and at least one was heads; what are the chances that both are heads?" and "I flipped these two coins and at least one was tails; what are the chances that both are tails?" is just exactly as valid as "There are twenty mountains in this mountain range; what are the chances one is taller than 10,000 feet?" and "I caught a thousand fish and at least one of them is a tuna; what are the chances all of them are tuna?" You wouldn't add the fish to the list of mountains and say of all of them "this isn't taller than 10,000 feet," because that's just crazy, but from a math standpoint, it's just exactly as crazy to add the HH case on the "at least one was tails" list. They're different questions. Don't conflate them.

Segev
2015-01-26, 02:00 PM

Look back at your post here. (http://www.giantitp.com/forums/showsinglepost.php?p=18693159&postcount=60) You'll note that before you started conflating the two questions, you found that always answering "no" was better for the person who was asked the "at least one" question.True! But it was also always better for the person I asked "the blue one" about.

If, rather than randomizing the questions asked as you did there, you always asked the same question and simply re-did any flips that didn't fit the question (such as HH on an "at least one tails" scenario), you'd find that answering "no" is correct 2/3 of the time and answering "yes" is correct 1/3 of the time.Also true. But, again, if I do that, the person I tell "the blue coin is heads" to has the same odds of getting it right as the person I tell "at least one is heads."

The information being given has to be about the population from which the sample is taken, not about the sample itself. Because if the information given is about the sample itself, it truly does bring us back to the problem where the question is semi-randomized.

It's the difference between rejecting samples that do not fit the question, and shaping the question to fit the sample.

The problems as presented initially are given in a form such that the one being asked does not know whether this is a truth about the sample population (and thus he would be asked the same question no matter what), or it is a question with information given after the results of THIS sample were known to the question-asker.

Not knowing, he has to assign probabilities to the question-asker's likelihood of asking a particular question and giving particular information. This brings us, as far as the information provided to the one being asked the question goes, back to the "50% chance he'll ask about this and tell me about that" consideration that alters probabilities to 50%.

The question has to be phrased to specify that the information is about the population from which the sample is taken in order for it to have the "textbook" answer as correct.

Combining "I flipped these two coins and at least one was heads; what are the chances that both are heads?" and "I flipped these two coins and at least one was tails; what are the chances that both are tails?" is just exactly as valid as "There are twenty mountains in this mountain range; what are the chances one is taller than 10,000 feet?" and "I caught a thousand fish and at least one of them is a tuna; what are the chances all of them are tuna?" You wouldn't add the fish to the list of mountains and say of all of them "this isn't taller than 10,000 feet," because that's just crazy, but from a math standpoint, it's just exactly as crazy to add the HH case on the "at least one was tails" list. They're different questions. Don't conflate them.

Except that it again goes to methodology of setting up the problem.

And your analogy is highly flawed, because "There are 20 mountains in this range; what are the chances all of them are tuna?" is not something that even makes sense in the context of possible results. Fish are not mountains.

Coins, on the other hand, can be heads or tails.

The two questions you contrast regarding chances of heads or tails are both questions you can arrive at if your methodology is, "I flipped two coins, and provide information about one of them, and then ask a question about something that has a non-zero chance."

You can see this clearly if you try the experiment your way:

Let's say that you reject all results where both coins are tails. You then take one of your results that has at least one heads and ask me the question, "I flipped two coins, one of which came up 'heads;' did both come up 'heads?'" I will always answer "no."

You then ask Pinnacle the question phrased thusly: "I flipped two coins, and the green one came up 'heads;' did both come up 'heads?'" She will also always answer "no," just to give us a solid basis of comparison.

By the postulate of the two questions originally posed about the Smiths and the Joneses, I should have a 2/3 chance of being right, and Pinnacle should have a 1/2 chance of being right. However, since both of us are choosing the same answer, we will be right (and wrong) exactly the same percentage of the time (which, statistically, should be 2/3).

This is because the information you are giving Pinnacle is extraneous and changeable based on the result. You will tell her about the blue coin if the blue coin came up "heads," but the green one didn't. And presuably will pick one arbitrarily (50% chance) if both are Heads.

The result is determined by your methodology, which you did not tell either of us in the question you asked.

Similarly, if you eliminate the possibility of changing any information given to Pinnacle, you will reject, say, all results where the green coin came up "tails."

Then, you will still phrase your question to me thusly: "I have flipped two coins, at least one of which came up 'heads;' did they both come up 'heads?'" and your question to Pinnacle thusly: "I have flipped two coins, and the green one came up 'heads;' did they both come up 'heads?'"

If we again both answer "no," we again will have statistically the same results on repeated experiments. This time, it will be 1/2 chance of being right.

Again, it's because of your methodology, of which you told neither of us anything.

Therefore, the two phrasings, "at least one of which" vs. "the green one," are functionally identical until they are used as selection criteria to form your population from which valid samples may be drawn.

In the questions as initially phrased about the Smiths and Joneses, there is no indication that the sample populations were restricted such that the questions could be asked as they were. To assume that they were is an unstated assumption which is no more correct than to assume that they were not, but rather than the information to be given about the Smiths and Joneses was chosen after they were selected, and thus is information specifically derived from their conditions.

In fact, because of how the questions are phrased, it is more reasonable to assume that the information to be given and the question asked was chosen based on the sample taken. No indication was given that the Smiths and Joneses were selected nor screened for anything. The information given therefore does not actually contain selection criteria about the population.

So far as the person being asked knows, you flipped two coins, looked at the results, then decided what to tell him and what to ask. That is, in fact, what the scenario sounds like, since specific selection criteria being applied is generally stated when problems are formulated.

That's why the problems are badly worded; they have their own unstated assumptions (which are less obvious, colloquial, or common than the unstated assumptions for which those who give the "wrong" answer are chided), without which the "textbook correct" answer isn't.

NecroRebel
2015-01-26, 03:34 PM
it was also always better for the person I asked "the blue one" about.[/i]

What? No it wasn't; it was a 50/50 chance for the person you asked "the blue one" about.

Also true. But, again, if I do that, the person I tell "the blue coin is heads" to has the same odds of getting it right as the person I tell "at least one is heads."

The information being given has to be about the population from which the sample is taken, not about the sample itself.

The only relevant thing is the population from which the sample was taken. The sample is the whole population. You're including things that aren't part of the sample - on a list of coin flips wherein at least one was heads, TT never appears, because there is not at least one heads in it.

It's the difference between rejecting samples that do not fit the question, and shaping the question to fit the sample.

Yes, and meaningful probabilities only come when the samples all fit the question. Mountains are not fish and asking questions about what sort of fish a mountain is is nonsensical. A TT flip is not a flip with at least one heads, and asking questions about what sort of flip with at least one heads a TT flip is is nonsensical.

The question has to be phrased to specify that the information is about the population from which the sample is taken in order for it to have the "textbook" answer as correct.

And it is. At least one of the coins was heads - that tells you what population the sample was taken from: flips wherein at least one of the coins was heads.

And your analogy is highly flawed, because "There are 20 mountains in this range; what are the chances all of them are tuna?" is not something that even makes sense in the context of possible results. Fish are not mountains.

Coins, on the other hand, can be heads or tails.

Yes, that's exactly my point - fish are not mountains, mountains are not fish. Flips wherein at least one flip is heads are not flips wherein at least one flip is tails.

We're not discussing coins. We're discussing pairs of flipped coins with certain properties. There's a difference. The fact that HT and TH flips are both pairs of coins wherein at least one is heads and pairs of coins wherein at least one is tails is irrelevant; we're discussing only one or the other.

If I said that I had a thousand fish, at least one of which was a tuna, and asked about what chance there was 200 of them were tuna, it would not be relevant to say that at least one of those same thousand fish was a shark.

By the postulate of the two questions originally posed about the Smiths and the Joneses, I should have a 2/3 chance of being right, and Pinnacle should have a 1/2 chance of being right. However, since both of us are choosing the same answer, we will be right (and wrong) exactly the same percentage of the time (which, statistically, should be 2/3).

Yes, because both answers are the same. You're running into the trap of the Monty Hall problem - the probabilities of a given answer for a given event are fixed when that event happens. In the classic Monty Hall problem, when you choose the first door, it's probability of being correct is set to 1/3, so when another door is opened and shown to have a 0/3 chance of being correct, your door is at 1/3, the now-open door is at 0/3, and the third door is at 2/3.

You're adding information to the scenario when you guess the first time. Thereafter, the two possibilities are no longer equally likely. This is to be expected - how would it make sense if you, answering "no," had a 2/3 chance of being correct, while someone else giving the same answer had a 1/2 chance of being correct?

If, on the other hand, you did the test and one person guessed with some information and then you did the test again and another person guessed with different information, the probabilities would be as expected - 2/3 and 1/2 of "one heads" being correct. Otherwise, you're no longer considering mathematical probabilities, but rather real-world scenarios which are much messier.

SpectralDerp
2015-01-26, 04:02 PM
Let's see how badly worded these problems actually are.

Problem 1: Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Problem 2: Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

Let's make the following assumptions:
- Any child is either a boy or a girl.
- For any child, the probability of that child being a boy is 1/2.
- The sexes of any two children are independent.

Let's make one more:
- The question "What is the probability that X?" is to be interpreted as "What is the probability that X, given the information you were just given?".

Given these assumptions, the answers to these problems are 1/2 and 1/3 (and their calculations have no need for notions like population and samples).

All of them are unstated, yet everyone figured them out. That's because they are typical conventions people use when discussing probability.

Your examples of coin games involve scenarios where "What is the probability that X?" means "What's the probability that X, given a probability space I am misleading you about by straying away from these basic conventions?". That's not evidence that the original problems are badly worded, the ones in your scenarios are.

Pinnacle
2015-01-26, 05:25 PM
Segev, I'm not really sure what you're trying to ask/argue/prove.

If you change the scenario, sample, assumptions, and question, you get different odds.
Well, yeah.

Segev
2015-01-27, 12:28 PM
What? No it wasn't; it was a 50/50 chance for the person you asked "the blue one" about.

That's not what your writeup in the linked post says.Er... the linked post was shown to have an error in it later, which I acknowledged. The end result, however, is this:

You can construct the scenario such that it has either a 1/2 or 1/3 chance of somebody always answering "no" to the question asked being correct. Once you have established the scenario, however, you do not get different odds of correctness based on the information you give the people answering the question.

If I decide what to tell and what to ask after seeing the results of the coin flips, the chances are 1/2.

If I decide on my question but not which coin I will give information about, and reject results which would make the question always "no" given the information, then choose the coin about which I will give information, the chances are 1/3.

If I decide on my question and which coin I will give information about, then reject any results which would always be "no" given the information and coin chosen, the chances are 1/2.

In all three cases, I can tell Alice "at least one coin is [result]; are they both [result]?" and Bob, "the [color] coin is [result]; are they both [result]?"

It will not make Alice be right a different fraction of the time than Bob.

The only relevant thing is the population from which the sample was taken.Correct.

The sample is the whole population.That is the most obvious assumption, and it actually comes out to the 1/2 result. This is because, if the sample is the entire population, then it must be the case that you were given this sample before you decided on what information to give and what question to ask.

It, in fact, makes the question part of the sample, which is the point I think I was missing way back when. That is, the possible scenarios, given that you have only this one sample and had it before you decided what information to give and what question to ask, are:

BB "the older is a boy, are both boys?"
BB "the younger is a boy, are both boys?"
BB "at least one is a boy, are both boys?" (You are telling me about the older one)
BB "at least one is a boy, are both boys?" (You are telling me about the younger one)
BG "the older is a boy, are both boys?"
BG "the younger is a girl, are both girls?"
BG "at least one is a boy, are both boys?" (You are telling me about the older one)
BG "at least one is a girl, are both girls?" (You are telling me about the younger one)
GB "the older is a girl, are both girls?"
GB "the younger is a boy, are both boys?"
GB "at least one is a boy, are both boys?" (You are telling me about the older one)
GB "at least one is a girl, are both girls?" (You are telling me about the younger one)
GG "the older is a girl, are both girls?"
GG "the younger is a girl, are both girls?"
GG "at least one is a girl, are both girls?" (You are telling me about the older one)
GG "at least one is a girl, are both girls?" (You are telling me about the younger one)

Now, I do have one more piece of information than the list of possible scenarios: I know which question and which information you gave me.

Let us now examine the two forms the examples took:

"The older child is a girl; are both girls?"

This gives us:

GB "the older is a girl, are both girls?"
GG "the older is a girl, are both girls?"

as the only two possibilities left. They had even odds of being the case before I knew which question and info you would give me. It is, as the textbook answer expects, a 50% chance that I will guess correctly.

"At least one child is a girl; are both girls?"

This gives us:

BG "at least one is a girl, are both girls?" (you are telling me about the younger one)
GB "at least one is a girl, are both girls?" (you are telling me about the older one)
GG "at least one is a girl, are both girls?" (you are telling me about the older one)
GG "at least one is a girl, are both girls?" (you are telling me about the younger one)

Again, each of those possibilities has an even chance. This is because it was only a 50% chance, if it was BG or GB, that you would have even told me about the girl and asked the associated question.

This again gives me a 50% chance of guessing correctly.

To reitterate: this is true because your "one sample" situation means you had zero controls on how you generated your sample. Thus, the true population is about the way you presented me the information and question, since this was generated based on the single sample.

To test this, consider the situation where you flip a blue coin and a green coin. Decide on your question and the information you will give. Give half the people you ask the information about WHICH coin you're telling them about, and don't give it to the other half. You will find 50% of them get it right.

You could also do this by flipping new pairs of coins each time, and generating your question as described in the post you linked; considering again that I'd made a mistake and not considered all cases then (I have, now, in this post), you'll find that you still get 50% right answers.

By not rejecting results - which you cannot do when you have literally one sample that is your population - you have made it a 50/50 proposition

If I said that I had a thousand fish, at least one of which was a tuna, and asked about what chance there was 200 of them were tuna, it would not be relevant to say that at least one of those same thousand fish was a shark.No, but it would be relevant to know what the odds of a given fish out of all possible fish were tuna, and whether you had looked at your sample of 1000 fish before deciding to a) tell me that at least one is tuna and b) to ask me if all of them were tuna.

If you did, in fact, wait to see what was in there, then my chances are really the chances of any collection of 999 fish having 199 tuna in it.

Just as, if you did, in fact, wait to see what the Jones's two children were before deciding to a) tell me that at least one was a girl, and b) to ask me if both were girls, my chances are really the chances of any collection of 1 children having 1 girl in it.

If, on the other hand, you tell me that you sought out a collection of 1000 fish that had at least 1 tuna in it, you have the case you're looking for.

That is, you'd have to tell me that the Joneses were selected because they have at least one daughter.

This tells me that you would have rejected any samples that had two boys. You already knew what you were looking for, and THAT criterion gives a 2/3 chance of one boy and one girl, with only a 1/3 chance of two girls (and no chance of two boys).

Whereas, without knowing you actively sought "at least one girl" in the pair, and therefore rejected all BB samples, I only know this was a random sample taken from all possible samples (BB, BG, GB, GG), and that you generated the question based on its results, leading to my lengthy discussion earlier in this post.

If, on the other hand, you did the test and one person guessed with some information and then you did the test again and another person guessed with different information, the probabilities would be as expected - 2/3 and 1/2 of "one heads" being correct. Otherwise, you're no longer considering mathematical probabilities, but rather real-world scenarios which are much messier.That again depends not on the information you give, but on the criteria you have for choosing what information and question to give and for accepting or rejecting your sample.

If you choose your information and question before accepting the sample, and reject samples that don't fit, you will get the results expected for the acceptence criteria you chose. If your acceptence criterion was "at least one heads," then it's a 2/3 chance soembody guessing "no" will be right, even if you tell them WHICH is heads (because there's a 50/50 shot of which coin you're telling him about - green or blue). If your criterion was "the blue coin is heads," then it's a 1/2 chance they'll get it right either way. Even if all you do is tell them "at least one was heads," they still have a 50/50 shot at a right answer because the only coin that could possibly not be heads is the green one.

IF you instead accept any sample, as my lengthy discussion shows, and then choose your information and question, they again have a 50/50 shot. That's the situation of "only one sample," as the question we're discussing asks it about the Joneses' two children. Absent information that the Joneses were specifically accepted because at least one was a girl (and thus would have been rejected and the Johnsons or the Jacksons or the first family that had at least one of their two children being a girl was accepted), it is more likely that you've flipped two coins/found the first two-child family you could, and asked a question and gave information based on what you found the sample to be.

And this post has gotten enormous, so I'll answer the otehrs in at least one separate post.

Segev
2015-01-27, 12:48 PM
Let's see how badly worded these problems actually are.

Let's make the following assumptions:
- Any child is either a boy or a girl.
- For any child, the probability of that child being a boy is 1/2.
- The sexes of any two children are independent.

Let's make one more:
- The question "What is the probability that X?" is to be interpreted as "What is the probability that X, given the information you were just given?".

Given these assumptions, the answers to these problems are 1/2 and 1/3 (and their calculations have no need for notions like population and samples).No, they're not.

Because again, the information I've been given does not include how the Smiths and Joneses were selected as samples. I do not know if you chose your question and information first, and rejected samples that would render the information false, or if you chose the samples and then picked whati nformation about the samples you would give me, and a question that was associated with that information.

(i.e., if it's BG or GB, 50/50 shot that you'll tell me about teh boy or the girl, and the question will ask if they're both the sex of the child you told me about.)

In fact, the base assumption that should be made, given how the questions were asked and the information presented, is that you did, in fact, just pick a 2 child family and formulate the information and associated question based on the facts of that sample.

To get the 1/3 result, you have to tell me not that at least one of the Joneses' two children is a girl, but that you selected the Joneses because at least one of their two children is a girl. i.e. that you have a population of families that consists solely of 2-child families with at least one daughter.

All of them are unstated, yet everyone figured them out. That's because they are typical conventions people use when discussing probability.Er...yes and no. This is a classic "trick question to show how normal people don't understand probability," so the formulation is such that anybody who knows the textbook answer understands the unstated assumption. The trouble is, it is probable that many don't actually realize the unstated assumption is being made. They've been given the textbook explanation, which leaves the assumption unstated even as it relies on it.

Thus, they will not appreciate the difference between "look at sample; formulate question" and "formulate question; choose sample about which it is a valid thing to ask."

This is exacerbated by the fact that the actual way people generally expect information to be given about a sample is as informative, rather than restrictive, details.

Your examples of coin games involve scenarios where "What is the probability that X?" means "What's the probability that X, given a probability space I am misleading you about by straying away from these basic conventions?". That's not evidence that the original problems are badly worded, the ones in your scenarios are.Except that the original problems are straying away from the conventiosn of standard information-giving. That is precisely why they are badly worded.

The coin-game uses exactly the base assumptions that are expected by standards of giving information.

To formulate the Joneses' question in a way that is not misleading by straying away from the assumptions, it would have to actually be, "The Joneses are participating in a program for 2-child families with at least one daughter. What is the probability that both of their children are girls?"

Or, alternatively, "I have sought out the Joneses because at least one of their two children is a girl. What is the probability that both of their children are girls?"

Phrased as it was, "The Joneses have 2 children. At least one is a girl. What is the probability that both are girls?" implies that you have first selected the Joneses and THEN started to give information about them. Not that there was a selection criterion in place that required them to match this information.

Segev, I'm not really sure what you're trying to ask/argue/prove.

If you change the scenario, sample, assumptions, and question, you get different odds.
Well, yeah.

What I'm trying to show is that the original question is badly formulated. It implies one set of assumptions based on how people usually report information, and then uses - without stating - a different set of assumptions.

Implied: "The Joneses were selected at random. There were no rejection criteria. We are giving you the following additional pieces of information: They have two children; at least one is a girl. You know, based on the fact that there were no rejection criteria, that if one is a girl and one is a boy, there is only a 50% chance that we would be asking you this question, and a 50% chance that we could be asking you instead if both were boys (and telling you at least one was a boy). What is the probability that both are girls?"

With Actually-Used Assumptions Stated: "The Joneses were selected at random, but would have been rejected if we could not truthfully give you the following information about them: They have two children; at least one is a girl. You know, because we would have rejected any which did not make that statement true, that this information restricted the possible results, rather than being determined by the results. What is the probability that both are girls?"

The answer to the implied question is the same as with the two-coin game wherein I flipped the coins THEN decided what information (and associated question) to give. The answer to the question with the actually-used assumptions stated is the textbook "1/3."

SpectralDerp
2015-01-27, 04:57 PM
No, they're not.

Yes they are and every step in the calculation is trivially correct.

Given the fourth assumption, the problem wants us to calculate the value of

P(Both children are boys | At least one child is a boy)

This is a conditional probability. Conditional probabilities are calculated via

P(A|B) = P(A and B) / P(B)

This mean the solution to the problem can be calculated via

P(Both children are boys | At least one child is a boy)
= P(Both children are boys and at least one child is a boy) / P(At least one child is a boy)

First, we calculate P(At least one child is a boy). From the first assumption, we can deduce that

P(At least one child is a boy)
= 1 - P(the first child is a girl and the second child is a girl)

because the two events here are complementary. Given the third assumption, it follows that

P(the first child is a girl and the second child is a girl)
= P(the first child is a girl) * P(the second child is a girl)

and given the second assumption, it follows that

P(the first child is a girl)
= 1/2

as well as

P(the second child is a girl)
= 1/2

Putting all of this together, we get

P(At least one child is a boy)
= 1 - (1/2) * (1/2)
= 3/4

Next, we will calculate P(Both children are boys and at least one child is a boy). First, we will simply rewrite it as

P(Both children are boys and at least one child is a boy)
= P(Both children are boys)
= P(the first child is a boy and the second child is a boy)

From the third assumption, it follows that

P(the first child is a boy and the second child is a boy)
= P(the elder child is a boy) * P(the younger child is a boy)

From the second assumption, it follows that

P(the first child is a boy)
= 1/2

and

P(the second child is a boy)
= 1/2

Putting this together, it follows that

P(the first child is a boy and the second child is a boy)
= (1/2) * (1/2)
= 1/4

Putting in these values gives us

P(Both children are boys | At least one child is a boy)
= (1/4) / (3/4)
= 1/3

Q.E.D.

The first problem is solved in an analogous way.

You need to drop the notions of populations, samples and selection. Probability theory does not need them. These concepts primarily play a role in statistics and are used to make educated guesses about probabilities.

Segev
2015-01-28, 09:49 AM
Yes they are and every step in the calculation is trivially correct.

Given the fourth assumption, the problem wants us to calculate the value of

P(Both children are boys | At least one child is a boy)

This is a conditional probability. Conditional probabilities are calculated via

P(A|B) = P(A and B) / P(B)

This mean the solution to the problem can be calculated via

P(Both children are boys | At least one child is a boy)
= P(Both children are boys and at least one child is a boy) / P(At least one child is a boy)

First, we calculate P(At least one child is a boy). From the first assumption, we can deduce that

P(At least one child is a boy)
= 1 - P(the first child is a girl and the second child is a girl)

because the two events here are complementary. Given the third assumption, it follows that

P(the first child is a girl and the second child is a girl)
= P(the first child is a girl) * P(the second child is a girl)

and given the second assumption, it follows that

P(the first child is a girl)
= 1/2

as well as

P(the second child is a girl)
= 1/2

Putting all of this together, we get

P(At least one child is a boy)
= 1 - (1/2) * (1/2)
= 3/4

Next, we will calculate P(Both children are boys and at least one child is a boy). First, we will simply rewrite it as

P(Both children are boys and at least one child is a boy)
= P(Both children are boys)
= P(the first child is a boy and the second child is a boy)

From the third assumption, it follows that

P(the first child is a boy and the second child is a boy)
= P(the elder child is a boy) * P(the younger child is a boy)

From the second assumption, it follows that

P(the first child is a boy)
= 1/2

and

P(the second child is a boy)
= 1/2

Putting this together, it follows that

P(the first child is a boy and the second child is a boy)
= (1/2) * (1/2)
= 1/4

Putting in these values gives us

P(Both children are boys | At least one child is a boy)
= (1/4) / (3/4)
= 1/3

Q.E.D.

The first problem is solved in an analogous way.

You need to drop the notions of populations, samples and selection. Probability theory does not need them. These concepts primarily play a role in statistics and are used to make educated guesses about probabilities.

Alright. Perform the experiment. Since you said we have no populations to worry about, we'll just do it with one pair of coins, one green and one blue. I will flip them and hand you the results.

You decide whether you're going to ask about heads or tails, and pick one of the coins which matches that result to give information about. The question needs to be in yes/no form, not "what is the probability" form.

Now, ask 200 people the question you've chosen to ask. Give every other person you ask the information, "at least one is [side]," and the rest the information, "the [color] one is [side]." Again, you choose which coin to tell them about, just be consistent across all whom you tell.

Count how many of the 100 you gave one kind of information to get it right. Count how many of the other 100 people get it right.

If your hypothesis is correct, you should see the "at least one" group get a 1/3:2/3 split on right/wrong answers (which is right and which is wrong depends, theoretically, on whether the coins match or not), and the "the [color] one" group get it right 1/2 the time (regardless of whether the coins match or not).

If you see a problem in the methdology with which the experiment is set up, please explain. The experiment is designed, here, to simulate you walking up to random people after having gotten the single sample, "the Joneses," and asking them questions about whether their children are the same sex based on the information you give about the sex of at least one/the older child.

huttj509
2015-01-28, 11:19 AM
Alright. Perform the experiment. Since you said we have no populations to worry about, we'll just do it with one pair of coins, one green and one blue. I will flip them and hand you the results.

You decide whether you're going to ask about heads or tails, and pick one of the coins which matches that result to give information about. The question needs to be in yes/no form, not "what is the probability" form.

Now, ask 200 people the question you've chosen to ask. Give every other person you ask the information, "at least one is [side]," and the rest the information, "the [color] one is [side]." Again, you choose which coin to tell them about, just be consistent across all whom you tell.

Count how many of the 100 you gave one kind of information to get it right. Count how many of the other 100 people get it right.

If your hypothesis is correct, you should see the "at least one" group get a 1/3:2/3 split on right/wrong answers (which is right and which is wrong depends, theoretically, on whether the coins match or not), and the "the [color] one" group get it right 1/2 the time (regardless of whether the coins match or not).

If you see a problem in the methdology with which the experiment is set up, please explain. The experiment is designed, here, to simulate you walking up to random people after having gotten the single sample, "the Joneses," and asking them questions about whether their children are the same sex based on the information you give about the sex of at least one/the older child.

The difference is in how often you can ask which question.

"At least one is heads, are they both heads" can be asked after 3/4 of the initial flips (HH, HT, TH)
"The Green one is heads, are they both heads" can be asked after half the initial flips (HH, HT)
"The Blue one is heads, are they both heads" can be asked after half the initial flips (HH, TH)

And similar for tails.

So how many questions can you ask for each flip?

HH: (at least H, green H, Blue H) = 3
HT: (at least H, Green H, Blue T, at least T) = 4
TH: (at least H, Blue H, Green T, at least T) = 4
TT: (at least T, Green T, Blue T) = 3

How often do you ask each question, if it's chosen randomly after the initial flip?

At least 1 Heads: (1/3)(1/4) + (1/4)(1/4) + (1/4)(1/4) = 1/12 + 1/16 + 1/16 = 10/48
Green Heads: (1/3)(1/4) + (1/4)(1/4) = 7/48
Green Tails: 7/48
Blue Tails: 7/48
At least 1 Tails: 10/48

Total: 48/48, good, math checks for that part.

So the odds of each question being applicable to be asked are not the same.

Pinnacle
2015-01-28, 11:26 AM
There being exactly one case, there is exactly one correct answer.
The amount of times any one answer will be correct is either 0 or all.

The question has nothing to do with how likely people are to give an answer to itself, it's about how likely an answer is to be correct

obryn
2015-01-28, 11:49 AM
You don't need any of that for an experiment, Segev.

Get a randomly generated sample of ten thousand pairs of coin flips.

ask, "At least one of these coins came up Heads. What are the odds that both are heads?"

Eliminate all tt results accordingly.

Count HH vs HT vs TH. You will find (obviously) that 1/3 of the remainder are HH.

Segev
2015-01-28, 12:05 PM
There being exactly one case, there is exactly one correct answer.
The amount of times any one answer will be correct is either 0 or all.

The question has nothing to do with how likely people are to give an answer to itself, it's about how likely an answer is to be correctYou cannot determine probabilty emperically without repeated experiments. The only way to verify that the math is being applied correctly is through empirical tests. If you find that your empirical tests do not match the hypothesis established by your math, then you have either set up the experiment wrong, or you have the math wrong.

I am arguing that the math is wrong for the case presented as the Joneses' family, because the math presented is for a different problem. One set up differently, under different assumptions than those provided in the problem itself.

You don't need any of that for an experiment, Segev.

Get a randomly generated sample of ten thousand pairs of coin flips.

ask, "At least one of these coins came up Heads. What are the odds that both are heads?"

Eliminate all tt results accordingly.

Count HH vs HT vs TH. You will find (obviously) that 1/3 of the remainder are HH.Except that that isn't what the "Jones question" presents.

It does not say, "We selected the Jones because we can tell you truthfully that at least one of their 2 children is a girl."

It says (or at least heavily implies by its wording), "Having selected the Joneses, we can tell you that at least one of their two children is a girl."

If you declare that the implication is not present, then you cannot at the same time declare that the implication is present that the first is the case.

Which means that, if you deny the implication, the question literally lacks sufficient information to be answered accurately.

Note: you are running your experiment by pre-determining the question and rejecting runs which generate combinations which are invalid with the question.

The claim I am making is that this order of decision-making - question first, reject families which do not match this question - is not in any way implied by the "Jones question" as presented. Because of this, it is a bad question, and does not illustrate the point it is trying to get across. Particularly if the point involves revealing "common misconceptions about probability."

Segev
2015-01-28, 12:09 PM
The difference is in how often you can ask which question.

"At least one is heads, are they both heads" can be asked after 3/4 of the initial flips (HH, HT, TH)
"The Green one is heads, are they both heads" can be asked after half the initial flips (HH, HT)
"The Blue one is heads, are they both heads" can be asked after half the initial flips (HH, TH)

And similar for tails.

So how many questions can you ask for each flip?

HH: (at least H, green H, Blue H) = 3
HT: (at least H, Green H, Blue T, at least T) = 4
TH: (at least H, Blue H, Green T, at least T) = 4
TT: (at least T, Green T, Blue T) = 3

How often do you ask each question, if it's chosen randomly after the initial flip?

At least 1 Heads: (1/3)(1/4) + (1/4)(1/4) + (1/4)(1/4) = 1/12 + 1/16 + 1/16 = 10/48
Green Heads: (1/3)(1/4) + (1/4)(1/4) = 7/48
Green Tails: 7/48
Blue Tails: 7/48
At least 1 Tails: 10/48

Total: 48/48, good, math checks for that part.

So the odds of each question being applicable to be asked are not the same.

All of this is accurate, and part of my point.

But what you're missing, I think, is that I am asserting that the way the "Jones question" is presented, the implication is that the above experiment is the one you've performed, and you've provided me with the information based on the Joneses family composition. That you would have presented the Joneses to me even if you had to tell me "at least one of them is a boy," and that if the children are of different sexes, there was a chance you'd have told me "at least one of them is a boy" rather than what you did tell me.

This is because the question as presented implies that you are telling me something about the Joneses, specifically, rather than telling me something about how you selected the Joneses to ask me about.

Again, if the question were worded such that it made clear that the probability you're asking about is regarding a single sample from a population pre-determined to have the given information be true, it would be fine. As-is, it does not suggest this to be the case. At best, you can say there is not enough information, because you have not been given the information about HOW it was determined what information to give nor what question (both boys/both girls) to ask.

SpectralDerp
2015-01-28, 12:44 PM
I am arguing that the math is wrong for the case presented as the Joneses' family, because the math presented is for a different problem.

So, in terms of conditional probability, if the problem doesn't ask us to calculate

P(Both children are boys | At least one child is a boy)

then what are we supposed to calculate instead?

Segev
2015-01-28, 01:17 PM
So, in terms of conditional probability, if the problem doesn't ask us to calculate, P(Both children are boys | At least one child is a boy), what is it asking about instead?

The problem, as presented, is asking us to calculate (by implication that the Joneses were chosen and the question based on their family):

P(Both children are boys | At least one child is a boy ^ [(there is a boy and a girl) c P(I told you at least one child is a boy rather than at least one child is a girl)=0.5]).

In natural language: "What is the probability that both children are boys, given that at least one child is a boy and that, if one child is a boy and the other is a girl, I had a .5 probability of telling you about the boy and asking about boys rather than telling you about girls and asking you about girls?"

That is, rather than having the textbook's claimed possibility set of:

BB
BG
GB
GG

We have an actual possibility set of:

From "at least one child is a boy," we reduce the possible cases to:

There are two cases out of the ones your information does not eliminate wherein there are two boys, and two cases out of that same set where they are not.

Probability = 0.5

Pinnacle
2015-01-28, 01:32 PM
You cannot determine probabilty emperically without repeated experiments.

I know. That's the problem.

Although I do think writing out all the possibilities and counting them's pretty good proof.

One set up differently, under different assumptions than those provided in the problem itself.
Assumption: 50% of all children are boys. The other 50% are all girls.
Assumption: Whether a child is a boy or girl has no impact on the gender of another child.
Inference: Approximately 25% of pairs of children have two girls, 25% have two boys, 25% have an older boy and younger girl, and 25% have an older girl and younger boy.

I don't think "Not assuming the questioner is screwing with when he asks the question to mislead us" is really an assumption.

Now, it is intended to throw you off a bit, by asking two similar questions one after the other. That's the point of it, it's a puzzle/brain teaser thing.

In natural language: "What is the probability that both children are boys, given that at least one child is a boy and that, if one child is a boy and the other is a girl, I had a .5 probability of telling you about the boy and asking about boys rather than telling you about girls and asking you about girls?"
That's not at all the question in the problem, though. That's your alternative problem with different results that you presented.
See how you had to add that extra information to get your version? That's not present in the original problem, and nor is it correct in that problem.

SpectralDerp
2015-01-28, 01:36 PM
What does the c mean? No matter how I read it, the statement

(there is a boy and a girl) c P(I told you at least one child is a boy rather than at least one child is a girl)=0.5

doesn't really make sense to me.

Edit: I have chosen to wait on the answer to the above question before I go further.

Segev
2015-01-28, 01:48 PM
Assumption: 50% of all children are boys. The other 50% are all girls.
Assumption: Whether a child is a boy or girl has no impact on the gender of another child.
Inference: Approximately 25% of pairs of children have two girls, 25% have two boys, 25% have an older boy and younger girl, and 25% have an older girl and younger boy.

That's about it, isn't it?So far, so good.

I don't think "Not assuming the questioner is screwing with when he asks the question to mislead us" is really an assumption.That's the problem, as...

Now, it is intended to throw you off a bit, by asking two similar questions one after the other. That's the point of it, it's a puzzle/brain teaser thing.You say right here it's meant to, when you said just a moment ago it's not right to assume that's the case.

If you assume that the "Jones Question" as presented is not a trick question, with no hidden assumptions, then what I outlined is what is being asked.

The idea that you decided, before accepting the Joneses as your example, that you would only tell me "at least one is a boy," and therefore that you would look for a family wherein this is true, is non-intuitive.

The idea that you looked at the Joneses, then decided to tell me about one of the children, and ask me if both children were that child's sex, on the other hand, is quite intuitive, given the manner in which the question is phrased.

That's not at all the question in the problem, though. That's your alternative problem with different results that you presented.Yes, but that problem is indistinguishable, given the information presented. Moreover, it is the more intuitive scenario based on how the question is asked.

See how you had to add that extra information to get your version? That's not present in the original problem, and nor is it correct in that problem.It is no less present in the original problem than is P(both children are boys | at least one child is a boy).

This is because there is an unstated assumption - BY CONVENTION, so it's fine - in that probabilty notation. Specifically, the unstated assumption is, "at least one child is a boy ^ I selected this sample because this is so."

To those uninitiated in probability theory, that is not an obvious assumption from the way the problem is worded. And from the wording of the problem, assuming that that is the probability equation is not a correct translation.

Maybe I can better illustrate this by translating it back to natural language:

P(both children are boys | at least one child is a boy) = What is the probability that both children are boys if I sought out a 2-child family where at least one is a boy?

One more time, the "Jones Question" is presented thusly:

The Joneses have two children. At least one is a girl. What is the probability that both are girls?

The first piece of information we are given is that there is a family, called the Joneses, that has two children. This implies that the person telling us about them has pulled them up as a singular sample. He did not have any acceptence nor rejection criteria which would have caused him to discard them and show us a different family instead.

The second thing we're told is that at least one is a girl. Because we do not know that he would always have told us that, and would have rejected any families which he could not truthfully say that of (finding a different one to show us), we are left not knowing that he would have told us about the girl if there was both a boy and a girl.

That is why this has the possibilities I outlined. The way the problem is worded, we are left to assume he has chosen an arbitrary child about which to give us information, NOT a family which allows him to give that specific information truthfully.

Segev
2015-01-28, 02:01 PM
This mistake was among the first things people explained: Just because you give me the information that at least one of the children is a boy, does not mean that there is a specific child you give me information about.Except that the problem as worded does not state that the question was decided on first. "The Joneses were chosen because at least one of their two children is a girl; what is the probability that both are girls?" is the question that you actually mean to be asking to get the 2/3 chance that both are not girls.

It's the order of operations vs. the order the problem is presented in. If you establish that you were going to tell me at least one is a girl, and would show me only families which allow you to tell me at least one is a girl, that works.

But the way it's presented, with the Joneses as an arbitrary family about which you are giving me information, that implies that you looked at the Joneses' children and chose one at random about which to give me information.

Even with the "at least" language, you have, for a single sample, reduced it to "the child I know about" and "the child I do not know about." It's irrelevant whether the child I know about is older or younger. I have only two children. I do not have a population of pairs of children which satisfy that condition, and you are not asking me about the odds of one such pair from that population having both be girls.

I suppose a better way to put it is... you have given me a single sample. You tell me nothing about the population from which it was taken. I am left, therefore, to assume it is a population of all two-child families. You then give me information about that one sample. Assuming you are not lying, I am left to assume that you would have given me information about one of the children, no matter which of the 4 possibilities (BB, BG, GB, GG) it turned out to be, and that you determined arbitrarily which child to give me information about.

It matters whether you selected the Joneses, then gave me information that fit them, or whether you selected the information you would give me, and then selected the Joneses because they fit the information. The problem as given implies that you did the first.

But please tell me, what does the c mean? No matter how I read it, the statement

(there is a boy and a girl) c P(I told you at least one child is a boy rather than at least one child is a girl)=0.5

doesn't really make sense to me.

Sorry, it was the best I was able to do for the symbollic "implies." That is...

(there is a boy and a girl) c P(I told you at least one child is a boy rather than at least one child is a girl)=0.5

...means, "If I there is a boy and a girl, then the probability that I would have told you that at least one child is a boy rather than that at least one child is a girl is 0.5."

Basically, it's stating that, if it had been BG or GB, there is only a 50% chance I'd have told you "at least one child is a boy," because I chose arbitrarily which to tell you about.

Oh! Let's try the coin experiment thusly:

I flip two coins and hide them in a box. I do not look at them. You open one side of the box and see what one of the coins came up as. Do you have a 50% chance of correctly guessing whether they both came up as that, or not?

This is the experiment that is being implied by the question with the joneses. You see one of the children. At least one, therefore, is of a known sex.

The PROPER way to phrase the question is to make it clear, somehow, that the Joneses were selected BECAUSE they fit the criteria, "at least one child is a girl."

Otherwise, you're leaving it as either "calculate also the odds of being asked and told about girls vs. boys" or "one sample, which means I can reduce it to 'child I know about' and 'child I don't,'" which is functionally the same as "older child."

obryn
2015-01-28, 02:13 PM
Oh! Let's try the coin experiment thusly:

I flip two coins and hide them in a box. I do not look at them. You open one side of the box and see what one of the coins came up as. Do you have a 50% chance of correctly guessing whether they both came up as that, or not?
Yes, because the coins are independent, and you know one specific coin has come up with that result.

It's a different scenario than "at least one coin came up heads" because you have precise knowledge of one specific coin rather than generalized knowledge of both coins. In order for it to be comparable, you would have to be told that at least one is heads, rather than yourself observing that this specific one is heads. You have definite knowledge of one coin rather than knowledge about the combination of coins.

This is the experiment that is being implied by the question with the joneses. You see one of the children. At least one, therefore, is of a known sex.
No.

It's a different question again. "I know this person in front of me is a girl; what are the odds that their sibling is a girl?" is a different question than, "at least one of two siblings is a girl - are both girls?"

The PROPER way to phrase the question is to make it clear, somehow, that the Joneses were selected BECAUSE they fit the criteria, "at least one child is a girl."
No, you just need to know that the statement you're given is true.

SpectralDerp
2015-01-28, 02:21 PM
But the way it's presented, with the Joneses as an arbitrary family about which you are giving me information, that implies that you looked at the Joneses' children and chose one at random about which to give me information.

Or maybe I randomly phoned families, asked "Do you have exactly two children?" and "Are all your children girls?", selected the first family where the answers were "Yes" and "No" respectively and then asked myself "How likely is it, that both of these children are boys?" and asked the internet for help.

Sorry, it was the best I was able to do for the symbollic "implies." That is...

(there is a boy and a girl) c P(I told you at least one child is a boy rather than at least one child is a girl)=0.5

...means, "If I there is a boy and a girl, then the probability that I would have told you that at least one child is a boy rather than that at least one child is a girl is 0.5."

Then the answer should be 1/3 again. Will rigorously prove it later.

obryn
2015-01-28, 02:47 PM
Basically, it's stating that, if it had been BG or GB, there is only a 50% chance I'd have told you "at least one child is a boy," because I chose arbitrarily which to tell you about.
Expanding...

As SpectralDerp said and I alluded to, this is pretty irrelevant. You still have the same probabilities so long as the person giving you the information made a true statement.

If BB, I must ask, "...at least one is a boy"
If BG, I can ask, "...at least one is a boy" or "...at least one is a girl"
If GB, I can ask, "...at least one is a boy" or "...at least one is a girl"
If GG, I must ask, "...at least one is a girl"

For each question, there are 3 possible scenarios with equal probability. So long as the information you're given is true, the probabilities are unchanged.

Segev
2015-01-28, 03:07 PM
Or maybe I randomly phoned families, asked "Do you have exactly two children?" and "Are all your children girls?", selected the first family where the answers were "Yes" and "No" respectively and then asked myself "How likely is it, that both of these children are boys?" and asked the internet for help.Not implied by the way the question was asked. You are, when you ask it of yourself, operating on additional knowledge not provided in the original "Jones Question."

Specifically, you know you settled on only looking at the first family wherein at least one child was a girl.

The question you have effectively asked is: "If the Joneses are the first 2-child family you find where at least one child is a girl, what is the probability that both children are girls?"

That is a different question, with more information given, than, "The Joneses have two children. At least one of them is a girl. What is the probability that both are girls?"

Then the answer should be 1/3 again. Will rigorously prove it later.It isn't, as we've shown. This is because we're operating with a situation where the question was formed based on the results, rather than results formed to suit the question.

It's about which is possible to be rejected.

If I can reject a piece of information being offered because it might be invalid, and thus will determine what question I ask based on the result I get, the probability is 1/2.

If I can reject a result because I cannot offer a given piece of information about it, and thus select the sample such that it conforms to the information, the probability is 1/3.

Segev
2015-01-28, 03:17 PM
It's a different question again. "I know this person in front of me is a girl; what are the odds that their sibling is a girl?" is a different question than, "at least one of two siblings is a girl - are both girls?"The issue is taht the problem as presented: "The Joneses have two children. At least one of them is a girl," implies that you started with the Jonses, and were going to tell me something about one of their children no matter what. Then you tell me some information about the children, implied to be information you decided to share based on something you observed about the Joneses.

Thus, it IS amounting to "this child in front of me is..."

Moreover, even if you switched it up, "this child in front of me" is no different from "at least one child."

If you knew that the Joneses were chosen because at least one child was a girl, and they showed you one of the Joneses' children who is a girl, you still have a 2/3 chance of being right if you guess the other one is a boy. Because you know that the sample from which the Jonses were chosen was 1/3 BG, 1/3 GB, and 1/3 GG.

However, the problem as given doesn't specify that the Joneses were chosen specifically because at least one child was a girl. It in no way hints at selection criteria, which is what actually makes the textbook answer true.

To get the textbook answer, you must be rejecting samples that do NOT match your criteria.

In short, your scenario quoted in this post is a better way to phrase the Jones Question.

"If I call families until I find one with two children, at least one of which is a girl, what is the probability that both are girls?"

The answer to that one is, in fact, 1/3. Because it tells you that you selected the family BASED ON the information, rather than the information based on the family.

The latter is implied by the way the question was worded.

If I wait until I see at least one of the children of the Joneses, and then tell you "at least one child is a [that child's sex]," that remains a true statement. However, the probability that both children are that child's sex remains 1/2, not 1/3.

Segev
2015-01-28, 03:24 PM
Expanding...

As SpectralDerp said and I alluded to, this is pretty irrelevant. You still have the same probabilities so long as the person giving you the information made a true statement.

If BB, I must ask, "...at least one is a boy"
If BG, I can ask, "...at least one is a boy" or "...at least one is a girl"
If GB, I can ask, "...at least one is a boy" or "...at least one is a girl"
If GG, I must ask, "...at least one is a girl"

For each question, there are 3 possible scenarios with equal probability. So long as the information you're given is true, the probabilities are unchanged.

Ah, but there is a 0.5 chance at a given question being asked across all scenarios.

See, the two that are mixed, you have to make a coin-flip to decide which you will ask.

So really, you have:

If BB and heads, I must say, "...at least one is a boy."
If BB and tails, I must say, "...at least one is a boy."
If BG and heads, I must say, "...at least one is a boy."
If BG and tails, I must say, "...at least one is a girl."
If GB and heads, I must say, "...at least one is a boy."
If GB and tails, I must say, "...at least one is a girl."
If GG and heads, I must say, "...at least one is a girl."
If GG and tails, I must say, "...at least one is a girl."

Thus, BB "...at least one is a boy" and GG "...at least one is a girl" have 25% chances of coming up, since the coin-flip is irrelevant to the results. Meanwhile, the remaining 4 results have a 12.5% chance of coming up, each.

By providing me the information, "...at least one is a boy," I know now that it can only be:

BB and tails

Even if I don't know you used a coin-flip, I am forced by the structure of the scenario to assume you chose arbitrarily which sex to tell me about in the cases GB and BG, because I know no better and as far as I can possibly know you had no reason to prefer one over the other.

Khedrac
2015-01-28, 04:15 PM
The problem, as presented, is asking us to calculate (by implication that the Joneses were chosen and the question based on their family):

P(Both children are boys | At least one child is a boy ^ [(there is a boy and a girl) c P(I told you at least one child is a boy rather than at least one child is a girl)=0.5]).

In natural language: "What is the probability that both children are boys, given that at least one child is a boy and that, if one child is a boy and the other is a girl, I had a .5 probability of telling you about the boy and asking about boys rather than telling you about girls and asking you about girls?"

The bolded bit is where you are going wrong: - you have absolutely no information on on how the decision to ask about boy or girl has been determined.

In the absence of any other information you have to assume the questioner is only asking about boys.

Looking at your toin coss with random determination of which question - it's a different problem because you specified the method of determination.

Without knowing the method of determination you cannot assume anything about it - it could be any of:
etc.

So, mathematically, because one knows nothing about the determination method one has to exclude it.

You might think they had a 50% chance of asking about either boys or girls, but I suspect that most people have an unconscious bias to ask about one or the other (just look at the way professions magicians can "force" people to think of a concept by planting the idea). Some people would be 100% ask about boys given at least one boy, only ask about girls given 2 girls, some would be the exact reverse, and other would be somewhere in between.
So - even in the real world, assuming 50% child distribution (not quite true), you are not going to have a 50%-50% distribution in answer to the question.

Segev
2015-01-28, 04:27 PM
The bolded bit is where you are going wrong: - you have absolutely no information on on how the decision to ask about boy or girl has been determined.This is precisely the problem.

In the absence of any other information you have to assume the questioner is only asking about boys.Quite the contrary! I have to assume the questioner has examined the Joneses and given me information about them. That is all I've been told: that the questioner has singled out a particular 2-child family, and that he has obtained a piece of information about them which he is sharing with me.

To assume that he is ONLY sharing with me families that match that information is to assume facts not presented.

At best, I can assume that he gave information, and would have given information regardless.

If I cannot assume that, I can only state that there is insufficient information provided in the "Jones Question" to decide, because I do not know whether the questioner decided on the Joneses, then formulated information+question about them, or decided on the information+question, and then chose the Joneses to fit.

Looking at your toin coss with random determination of which question - it's a different problem because you specified the method of determination.

Without knowing the method of determination you cannot assume anything about it - it could be any of:
etc.

So, mathematically, because one knows nothing about the determination method one has to exclude it.Except that one cannot exclude it. Even the "textbook" answer includes an unspoken determination method: "I have rejected results wherein at least one child was not a girl."

So, again, if we can make NO assumptions, it is a badly formulated problem because it has insufficeint information to answer the question it asks.

You might think they had a 50% chance of asking about either boys or girls, but I suspect that most people have an unconscious bias to ask about one or the other (just look at the way professions magicians can "force" people to think of a concept by planting the idea). Some people would be 100% ask about boys given at least one boy, only ask about girls given 2 girls, some would be the exact reverse, and other would be somewhere in between.
So - even in the real world, assuming 50% child distribution (not quite true), you are not going to have a 50%-50% distribution in answer to the question.Yes, but under the same assumptions wherein one assumes 50/50 boy/girl distribution in the population at large, one assumes "arbitrary" means "equal chance" for purposes of deciding.

Regardless, all you've done here is illustrate that the problem is still a badly worded one, but now because there is not sufficient information to answer at all, even by implication.

(I disagree; I think the implication is clear based on how people DO answer it, but since my point is the easier-to-prove "the question is a badly worded trick question with unspoken assumptions that those who don't know the answer wouldn't have reason to make," I can stop with your proof.)

SpectralDerp
2015-01-28, 04:50 PM
I have to assume the questioner has examined the Joneses and given me information about them. That is all I've been told: that the questioner has singled out a particular 2-child family, and that he has obtained a piece of information about them which he is sharing with me.

Funny how you are the only one who thinks that way and everyone else believed that they were supposed to consider a probability space (X,S,P) on which there are two identically distributed random variables A and B that take values in the set {Boy, Girl} such that their pushforward measure is the discrete uniform distribution on the power set of {Boy, Girl} and that they were supposed to calculate P(A=B=Girl|A=Girl) and P(A=B=Boy|A=Boy or B=Boy). It's almost as if it makes a difference if you understand conditional probabilities or just make up your own rules for how math questions are supposed to be interpreted and expect people to take you seriously.

Segev
2015-01-28, 05:10 PM
Funny how you are the only one who thinks that way and everyone else believed that they were supposed to consider a probability space (X,S,P) on which there are two identically distributed random variables A and B that take values in the set {Boy, Girl} such that their pushforward measure is the discrete uniform distribution on the power set of {Boy, Girl} and that they were supposed to calculate P(A=B=Girl|A=Girl) and P(A=B=Boy|A=Boy or B=Boy. It's almost as if it makes a difference if you understand conditional probabilities or just make up your own rules for how math questions are supposed to be interpreted and expect people to take you seriously.

The point is that yes, if you ALREADY KNOW THE ANSWER, you correctly interpret the badly-worded question, because the badly-worded question uses unspoken assumptions which are known to the people who ALREADY KNOW THE ANSWER.

This has been amply illustrated by my own efforts to create a game based on the puzzle, which ran into trouble because it revealed that simply sharing that information does not, in fact, change the odds of a person being asked "are they both [x]?" getting the answer right or wrong.

It matters, entirely, how the Joneses vs. the question asked about them was generated.

Using the knowledge-expert-specific language to make it sound like you know more than I do is not proving your point, though it could be an attempt at ad hominem by implying that I am ignorant and you are not.

The truth is, I have understood every bit of the conditional probability discussion (though I did make some errors which were corrected earlier; I will note that I make errors in my own work that I have to go back and correct when I realize something's off, and having a second set of eyes that know what they're looking at helps with such things; this does not make me stupid or inexpert in my field, let alone unable to discuss it intelligently).

No, I am not a statistician, but I do understand conditional probability. I also understand semantics.

The implied assumptions of the Jones Question as presented are not the unspoken assumptions that those who know it's a trick question meant to "illustrate that most people don't understand conditional probability" know to apply.

All I have stated is that the question could be worded better to unambiguously present the required assumptions, and that as stated, it does not do so. Therefore, it is either actively implying a different set of assumptions (which leads to a different answer than the "textbook" one), or it provides too little information because you cannot know the required assumptions (which means you cannot answer it).

Seriously, if I did not understand conditional probability, I would not have been able to converse in the language we've used, let alone constructed the table of options that we actually have to deal with if we make the assumption I still feel is implied in the way the question is asked.

To conclude: it is badly worded. To ask it in a way that is not a badly worded question - a trick question which fails to demonstrate what it seeks to by trying to fool the uninitiated into making bad assumptions - all you ahve to do is this:

Question 1: The Smiths were the first 2-child family we could find where the older child is a girl. What is the probability that both children are girls?

Question 2: The Joneses were the first two-child family we could find where at least one child is a girl. What is the probability that both children are girls?

These have the "textbook" answers of 1/2 and 1/3 for the reasons that we are attempting to illustrate, and they unambiguously reveal that "the older child is a girl" and "at least one child is a girl" are criteria which the families meet, rather than information which we gave about a family which could have had any of the BB/BG/GB/GG combinations and we would give information about one of the children after we determined their composition.

Pinnacle
2015-01-28, 05:35 PM
You say right here it's meant to, when you said just a moment ago it's not right to assume that's the case.
Y'know, I tried to be careful about that wording there.
Have you ever heard this one? It's not a logic problem, more like... wordplay, I guess.

Q: What's the spirit of a dead person that says "BOO!" and walks through walls?
A: Ghost.
Q: What's the area of the land near the ocean?
A: Coast.
Q: Who's the man who invited us to his home?
A: Host.
Q: If I have more cookies than everybody else, I have the...
A: Most.
Q: What do you put in a toaster?

People tend to say "toast", which is not correct. The pattern of the questions threw them off, but it didn't make the last question a trick question.

Here's another one:
A boy and his father are in a car accident and rushed to a hospital. A surgeon comes in to treat the boy, and upon seeing him immediately says "We need to find another surgeon; I can not treat this boy, he is my son." How can this be when the boy's father was also a victim?
The intended answer is that the surgeon is the boy's mother, but the audience overlooks it. More modern audiences sometimes answer "The surgeon is the boy's other father"--they may be more cognizant of the existence of gay couples, but are they not also more cognizant of the existence of female surgeons?
There's a number of contributing factors to this one, but I think the fact that the questioner put the idea that the surgeon could be the father into the listener's head before rejecting it is an important part of it.

The first question here puts the wrong answer in your head so you overlook the distinction. That's far different from deceptively leaving out information that invalidates your basic assumption and changes the answer. Also, the correct answer here is incompatible with your extra information that's not included in the question, so it probably wasn't implied.

Students learning this stuff being asked questions in a textbook or by a teacher might see problems where they don't have enough information to solve, since recognizing that fact is part of what they're learning.
With a puzzle, though, you can generally assume that all else is equal, because "I don't know" isn't very entertaining or tricky to figure out.

If you assume that the "Jones Question" as presented is not a trick question, with no hidden assumptions, then what I outlined is what is being asked.
So the question would naturally be interpreted that way, and also if you were going to write the question so that it would be interpreted that way you'd have to add the excess information (which, again, is incorrect additional information for the original problem).
I don't see how both can be true.

It is no less present in the original problem than is P(both children are boys | at least one child is a boy).
Something that you added that you weren't meant to assume and wasn't correct is no less present than the only clue present?

To those uninitiated in probability theory, that is not an obvious assumption from the way the problem is worded.
I don't think that's true, but regardless that wasn't the intended audience.

And from the wording of the problem, assuming that that is the probability equation is not a correct translation.
Well, it's incorrect.
Also there's nothing to translate.

That is why this has the possibilities I outlined. The way the problem is worded, we are left to assume he has chosen an arbitrary child about which to give us information, NOT a family which allows him to give that specific information truthfully.
So we could...
A) ...assume all possibilities are equally likely.
B) ...make no assumption, and state that there is insufficient data.
C) ...assume that the questioner used an elaborate random system to determine the data she presented and the questions she asked, then write out a set of rules that we made up and assume they're the ones she used.

I'm pretty sure that A is the most intuitive. B is technically the most correct.
C is pretty absurd, I have to say.
What if I assume a different set of rules than you do?

If BB, I must ask, "...at least one is a boy"
If BG, I can ask, "...at least one is a boy" or "...at least one is a girl"
If GB, I can ask, "...at least one is a boy" or "...at least one is a girl"
If GG, I must ask, "...at least one is a girl"

For each question, there are 3 possible scenarios with equal probability. So long as the information you're given is true, the probabilities are unchanged.
Well, not quite.
There are three scenarios for either question, but two of the three are half as likely.
Break this one down completely, and you'll get 50/50 odds.
This was the breakdown that I think initially confused Segev, because it doesn't match the correct answer to the puzzle.

Moreover, even if you switched it up, "this child in front of me" is no different from "at least one child."
Whoa there, that's wrong.
"This child in front of me" is quite different from "at least one child," because the latter also includes the case where the other child is the only one that matches the clue.
"This child in front of me is X" means that it could be This X/That X or This X/That Y, whereas "at least one child is X" allows for This X/That X, This X/That Y, and This Y/That X.
The latter is less data.

If you knew that the Joneses were chosen because at least one child was a girl, and they showed you one of the Joneses' children who is a girl, you still have a 2/3 chance of being right if you guess the other one is a boy.
There is only one other child. Assuming 50% chance that any child is a boy, there's only a 50% chance that he's a boy.

Because you know that the sample from which the Jonses were chosen was 1/3 BG, 1/3 GB, and 1/3 GG.
But you've identified that the first one is a girl. That eliminates one of those three possibilities... BG.
Now if you want to add age into this, that becomes an entirely different question: This child could have either a brother or sister, and either an older or younger sibling. If I ask what are the odds that the eldest is a girl, the odds are 75%. The eldest is a girl in any situation where this child has a sister or younger sibling, which is the case in three of the four possibilities.
But that's a completely different question.

However, the problem as given doesn't specify that the Joneses were chosen specifically because at least one child was a girl. It in no way hints at selection criteria, which is what actually makes the textbook answer true.
It in no way hints at selection criteria, so why are you making such detailed assumptions about it?

If I wait until I see at least one of the children of the Joneses, and then tell you "at least one child is a [that child's sex]," that remains a true statement. However, the probability that both children are that child's sex remains 1/2, not 1/3.
You said "at least one," but you were only talking about one of them. You wouldn't have even been aware of the situation where that child was a boy and the other was a girl.
The questioner in a puzzle is omniscient, because they're not a person but a narrator of a hypothetical. If everything they say isn't accurate and complete, you were presented with wrong information. That's something you have to recognize as possible in reality, but with a puzzle it just means it doesn't work and is unsolvable.

Question 1: The Smiths were the first 2-child family we could find where the older child is a girl. What is the probability that both children are girls?

Question 2: The Joneses were the first two-child family we could find where at least one child is a girl. What is the probability that both children are girls?

I think if you want to be perfectly unambiguous you still want to state the assumptions:
"Assuming that 50% of all children are boys and 50% are girls, and that one child's gender does not impact another's"

It may be obvious we're intended to make those assumptions for workable probability, but they're still assumptions.

SpectralDerp
2015-01-28, 05:56 PM
These have the "textbook" answers of 1/2 and 1/3 for the reasons that we are attempting to illustrate, and they unambiguously reveal that "the older child is a girl" and "at least one child is a girl" are criteria which the families meet, rather than information which we gave about a family which could have had any of the BB/BG/GB/GG combinations and we would give information about one of the children after we determined their composition.

Of course the "textbook" (a bad word!) answers treat given information as criteria when dealing with conditional probabilities. That's because the conditional probability of an event A given the event B is literally defined as the probability that event A occurs among the events that fit the criteria of type B. You might as well object that the result of "What is 2+2?" depends on the not at all obvious assumption that "+" refers to addition, instead of multiplication. Which is not only nonsense because everyone who understands basic arithmetic knows that the plus sign refers to addition, it also makes no difference because everyone who understands basic arithmetic knows that the results would be the same.

If jaydubs arbitrarily selected the Jones family from the group of all possible two-child families such that the sexes of their children have the distributions we have assumed throughout this thread, the possible options for the sexes of the children, now arbitrarily named A and B, are

(1) A is boy and B is a boy
(2) A is a boy and B is a girl
(3) A is a girl and B is a boy
(4) A is a girl and B is a boy

If jaydubs then determines which one of these options actually described A and B and would then give us the information "A or B is a girl", we can rule out one of these four options, namely the first one. The probability that the fourth option is then correct is 1/3.

Guess who gave this exact argument in this thread already? Everyone who understands conditional probability.

Segev
2015-01-28, 06:16 PM
I apologize, Pinnacle, but these posts are getting long enough that I have trouble responding to everything, and thus am liable to miss something. Here is my best effort.

Have you ever heard this one? It's not a logic problem, more like... wordplay, I guess.

Q: What's the spirit of a dead person that says "BOO!" and walks through walls?
A: Ghost.
Q: What's the area of the land near the ocean?
A: Coast.
Q: Who's the man who invited us to his home?
A: Host.
Q: If I have more cookies than everybody else, I have the...
A: Most.
Q: What do you put in a toaster?

People tend to say "toast", which is not correct. The pattern of the questions threw them off, but it didn't make the last question a trick question.It's definitely a head-game. Regardless, it's not the real problem with the questions. Even independent of the first one, the second question implies that the information given is derived from the nature of the Joneses as a family. It does not in any way suggest that the information given is a criterion by which the Joneses were selected, which is the requirement for the 1/3 answer to be correct.

The first question here puts the wrong answer in your head so you overlook the distinction.No, it really doesn't. I know, because I carefully examined the semantics of both for the distinction between the two. In fact, I was prepared to agree with the textbook answer, as I had heard the problem before (and just forgotten the answer).

It was attempting to gamefy it for further illustrative purposes that revealed the flaw in the wording to me.

That's far different from deceptively leaving out information that invalidates your basic assumption and changes the answer. Also, the correct answer here is incompatible with your extra information that's not included in the question, so it probably wasn't implied.Begging the question.

"The correct answer is 'blue,' so your assumption that the extra information is "it's the Sun around which the Earth orbits' when I asked what color the Sun is is incompatible with the correct answer. Obviously, you made a bad assumption that was nowhere in evidence in the question, since I was asking about Rigel."

Students learning this stuff being asked questions in a textbook or by a teacher might see problems where they don't have enough information to solve, since recognizing that fact is part of what they're learning.
With a puzzle, though, you can generally assume that all else is equal, because "I don't know" isn't very entertaining or tricky to figure out.And, if we assume "all is equal," then the way the problem is worded EITHER implies that the Joneses were examined and the information chosen based on that examination, OR does not provide enough information.

There is nothing in the way it is worded, EXCEPT TO THOSE WHO ALREADY KNOW THE ANSWER BEING SOUGHT, that implies that the Joneses were selected based on the criteria that at least one of their two children is a girl.

So we could...
A) ...assume all possibilities are equally likely.
B) ...make no assumption, and state that there is insufficient data.
C) ...assume that the questioner used an elaborate random system to determine the data she presented and the questions she asked, then write out a set of rules that we made up and assume they're the ones she used.

I'm pretty sure that A is the most intuitive. B is technically the most correct.
C is pretty absurd, I have to say.
What if I assume a different set of rules than you do?
If we assume all possiblities are equally likely, we have to consider all possible ways that the Joneses could have been selected and all possible ways that the question could have been selected.

Not only is that a far larger problem than anybody doing puzzles like this tends to expect, but it runs into the problem of, "Okay, did they decide they were going to ask about boys UNLESS there were no boys? Would they have asked me about girls even if both were boys? How did they decide to tell me that there is at least one boy if there is one boy and one girl, and not to tell me that there is at least one girl, instead?"

Serioulsy, "all possible" covers a lot of ground. And it really leads us to B).

C is not absurd; C is, perhaps, overthinking it, and we came to it because I attempted to construct a method for generating these questions and results in possible combinations wherein the INFORMATION GIVEN determined the chance of correct guesses.

The problem arises because the IMPLICATION of the problem as presented is that you're giving information abotu the Joneses, when what the textbook answer is assuming is that you're giving the criterion by which the Joneses were selected.

We've demonstrated that this makes all the difference.

Whoa there, that's wrong.
"This child in front of me" is quite different from "at least one child," because the latter also includes the case where the other child is the only one that matches the clue.
"This child in front of me is X" means that it could be This X/That X or This X/That Y, whereas "at least one child is X" allows for This X/That X, This X/That Y, and This Y/That X.
The latter is less data.Which, ironically, leads to the ablity to make a more accurate guess.

There is only one other child. Assuming 50% chance that any child is a boy, there's only a 50% chance that he's a boy.This is the fundamental problem with presenting it as "These are the Joneses. Here is information about them." Which is what the problem given does.

It reduces it to a single sample taken from the entire population of all 2-child families (which, by assumption, have equal odds of being BB, BG, GB, or GG). Because it is presented this way, the "at least one is a boy" information tells us that we know something about exactly one of the children. AGain, this may as well be "the child in front of me." This again reduces the problem to a 1/2 chance, because there is only "the child we know about" vs. "the child we do not know about."

It comes down to how you construct the experiment to prove your point.

You want to prove that the correct answer to the Jones Question is 1/3. To do this, you have to define the experiment to be set up in a way that is not presented in the problem.

Therefore, we're back to b) insufficient information.

To reitterate: Every time you assert that you are making no assumptions when you look at the Jones Question and come up with 1/3, you are wrong. You are making the unstated assumption that the information given is a criterion by which the Joneses were selected, rather than information generated based on examining the Joneses. I assert that the phrasing of the problem implies it is the latter, but acknowledge that technically, we cannot know for certain, so deliberate (or sloppy) misimplication can lead to this misunderstanding. Denotatively, we do have only B) as the correct situation: there is not enough information.

It in no way hints at selection criteria, so why are you making such detailed assumptions about it?That's the point: it in no way implies that there WERE selection criteria.

Because of this, the implication is that the Joneses were selected with no such criteria.

Therefore, the information given must have been generated based on the composition of the Jones family.

You said "at least one," but you were only talking about one of them. You wouldn't have even been aware of the situation where that child was a boy and the other was a girl.True. That's a good way to put it.

It is also why the way the problem is phrased implies the wrong assumption: It implies that, "At least one is a girl" is information about the Joneses. Thus, by implication, it suggests that we would not have even been aware of the situation where the child examined was a boy and the other a girl.

You said it yourself: the Jones Question never mentioned selection criteria.

In order for the assumption that the information, "At least one is a girl," to be a selection criterion (the only way that the 1/3 answer is correct) can be accurate is if it's somehow indicated in the phrasing of the problem.

As phrased, it implies either that it is information gleaned from examination, or it makes no implication at all and there is not enough information presented to knwo the answer.

The questioner in a puzzle is omniscient, because they're not a person but a narrator of a hypothetical. If everything they say isn't accurate and complete, you were presented with wrong information. That's something you have to recognize as possible in reality, but with a puzzle it just means it doesn't work and is unsolvable.And I agree. If you assume that there is no implication that can be drawn, and thus no assumption about how the Joneses were chosen nor the information presented selected nor the question formulated, then the question has too little information to be answered.

And that again makes it a badly worded problem, since the intended answer is "1/3."

Fiery Diamond
2015-01-29, 01:45 AM
I've been reading this thread with great interest, and I hate to break it to you guys, but I've come to the conclusion that Segev is right.

There are three possible ways of looking at the problem as presented.

1) We lack sufficient information, as we do not know which came first, the chicken or the egg the question or the family. Unless we know whether the information is/are criterion/criteria or whether we started with a family and then decided the question, we cannot know the odds.

Nobody can argue with you if you choose this approach, but it is super boring.

2) This is a textbook word problem and/or mind puzzle. Clearly, then, it must be simply trying to put a basic probability into non-math language. Therefore, the question is the originator, and it's asking about what percentage of families that meet the criteria have both girls. The answer is then 1/3.

This is what the people arguing against Segev are insisting is the obvious and intuitive and CLEARLY CORRECT interpretation of the problem. However, remove the first sentence and replace it with...

3) This is a real-life scenario. There is no omniscient question-asker. The majority of cases in which this question would arise (and therefore the clear and obvious and intuitively correct interpretation of the question) would be something along the lines of "My co-worker Sally Jones has two kids. I saw one of them come to visit her the other day, and the kid was a girl, so I know at least one is a girl. What's the chance they both are?" In this case, the wording "The one I saw is a girl" and "(So I know) at least one is a girl" are identical, making the case exactly the same as "the older one is a girl" and giving an answer of 1/2.

This is actually the most intuitive reading of the question, which would explain why people "wrongly" answer 1/2 frequently.

As Segev said, it's just a badly worded question.

SpectralDerp
2015-01-29, 02:35 AM
1) We lack sufficient information, as we do not know which came first, the chicken or the egg the question or the family. Unless we know whether the information is/are criterion/criteria or whether we started with a family and then decided the question, we cannot know the odds.

This is simply wrong and has been explained several times already.

If jaydubs arbitrarily selected the Jones family from the group of all possible two-child families such that the sexes of their children have the distributions we have assumed throughout this thread, the possible options for the sexes of the children, now arbitrarily named A and B, are

(1) A is boy and B is a boy
(2) A is a boy and B is a girl
(3) A is a girl and B is a boy
(4) A is a girl and B is a boy

If jaydubs then determines which one of these options actually described A and B and would then give us the information "A or B is a girl", we can rule out one of these four options, namely the first one. The probability that the fourth option is then correct is 1/3.

3) This is a real-life scenario. There is no omniscient question-asker. The majority of cases in which this question would arise (and therefore the clear and obvious and intuitively correct interpretation of the question) would be something along the lines of "My co-worker Sally Jones has two kids. I saw one of them come to visit her the other day, and the kid was a girl, so I know at least one is a girl. What's the chance they both are?" In this case, the wording "The one I saw is a girl" and "(So I know) at least one is a girl" are identical, making the case exactly the same as "the older one is a girl" and giving an answer of 1/2.

This is also simply wrong and has also been explained several times already.

The scenario you describe here fits problem 1 because problem 1 specifies a child and then specifies the sex of the child. Problem 2 does not do so. Your scenario contains more information than provided in problem 2 and therefore they don't match up.

A scenario that problem 2 actually models is "My co-worker Sally Jones has two kids. I asked her if both of them were boys and she said 'No'".

The information "at least one of the two children is a girl" does not give you the information that there is a distinguished child (in this example, the child that visited Sally) and that the distinguished child is a girl.

This is why people get this question wrong. They think the question boils down to the sex of the other child, because problem 1 actually had a distinguished "other child". Problem 2 has no "other child".

If there is a distinguished child, we can call it A and the other child B, look at the possible options

(1) A is boy and B is a boy
(2) A is a boy and B is a girl
(3) A is a girl and B is a boy
(4) A is a girl and B is a boy

and disregard the first two options because they don't meet the criterion "A is a girl". The probability of the fourth option is then 1/2. However, if we don't know the sex of a particular distinguished child, only the first option can be disregarded and so the probability of the fourth option is 1/3.

Gavran
2015-01-29, 04:34 AM
I've been reading along myself, and while I'm not going to insist that the question is wrong, I am going to say that the question is designed to mislead and that is in some ways the same thing. It seems to me this is meant as a learning opportunity / reminder to carefully consider your data, which is not at all a bad thing, but to pretend the question is completely straightforward seems silly to me. To mangle Jay R's words, it is a bad question but that's what makes it a good riddle.

Further, I'd just like to point out that you've been basically repeating yourselves for two pages now. I propose that maybe it's time to peacefully move on, especially as the conversation has at times been rather dismissive and rude.

Perhaps someone has a new problem? Otherwise perhaps the thread should be allowed to sink. I don't know, I'm just posting because "Funny how you are the only one who thinks that way" rubs me the wrong way when I'm here quietly thinking similar thoughts.

SpectralDerp
2015-01-29, 07:42 AM
I've been reading along myself, and while I'm not going to insist that the question is wrong, I am going to say that the question is designed to mislead and that is in some ways the same thing. It seems to me this is meant as a learning opportunity / reminder to carefully consider your data, which is not at all a bad thing, but to pretend the question is completely straightforward seems silly to me.

The question is not designed to mislead. The context of the question is designed to mislead. The first question is supposed to make you think the second question gives you information about a particular child, despite the fact that it does not. Similarly how "What do you put in a toaster?" is not misleading, the answer is bread and not toast, the previous questions have a pattern where the answers rhyme with "oast" but the last question does not give off this impression.

The allegations that the result changes given a different reading have already been disproven, multiple people have shown how the math looks like in these scenarios and gotten the same result. The primary concern is about whether or not the phrase "at least one" specifies an object and the straightforward answer is no. Critis of this question are claiming that they would mean something else when they would ask it that way because a common feature of the english language is misleading them, they phrase specific information in an unspecific fashion and think it can be recovered when it can not. As a result, they are simply comitting the logical fallacy known as "affirming the consequence" and think that the well-defined scenario contains additional information when it trivially does not.

And no, if people want to continue to spread misinformation, I will not let them have the last word.

Segev
2015-01-29, 09:09 AM
SpectralDerp, you're wrong.

We have actually shown that, if I choose the Joneses out of all possible 2-child families, the part you have right is that the possible results are:

Child A is a boy and child B is a boy
Child A is a boy and child B is a girl
Child A is a girl and child B is a boy
Child A is a girl and child B is a girl

(Incidentally, you typoed and had the fourth option be girl/boy again, but it's clear you meant the above.)

If jaydubs them goes and finds out information about them and tells us, "at least one is a girl," we've shown the math, in building this precise scenario as our 2-coin game, that shows that either:

We assume that, in the case of A xor B is a girl, he chose which to tell us about with 0.5 probability of either, or
we assume that, in the case of A xor B, we do not know by what algorithm he chose to tell us about the girl instead of the boy.

If we assume that he, himself, was exposed to only one of the children, and thus does not know, himself, if it's BG, GB, or GG, we again have the 0.5 probability. You can arrive at this conclusion either by saying, "we know information about one specific child, but not about the other," and equating it to the "the older child" problem, or you can arrive at it by spelling out ALL the possibilities again (BBH, BBT, BGH, BGT, GBH, GBT, GGH, GGT; H/T determines which child jaydubs saw).

I also dispute that it's a good riddle. Good riddles can be tricky and can require odd ways of looking at something, but they should never have a separate, different way of looking at it provide a different answer that fits the riddle-as-presented.

While I know SpectralDerp and others dispute that any other possible interpretation of the puzzle-as-presented is right, their insistence does not make it so.

I believe I have shown that the issue lies in the fact that colloquial, natural-language interpretation would not assume that "at least one is a girl" was a selection criterion, but rather was a piece of information derived from observation of the arbitrarily-selected Jones family. And whether it is a criterion or an observation makes all the difference.

Because of this inclarity, there are at least 2, and possibly 3-to-infinite depending on what assumptions you think are reasonable to make, possible valid answers to the "riddle," which makes it a poor riddle.

Segev
2015-01-29, 09:14 AM
The allegations that the result changes given a different reading have already been disproven, multiple people have shown how the math looks like in these scenarios and gotten the same result.Er...no? The different scenarios quite explicitly produce 1/2 if it's not a criterion but an observation, and 1/3 if it is a criterion. This has been shown repeatedly.

If it's NOT a criterion, the 2-coin game is what we're effectively playing. The math on that was shown multiple times to come to 1/2.

The primary concern is about whether or not the phrase "at least one" specifies an object and the straightforward answer is no.No, the primary concern is whether it specifies a criterion for selecting the Joneses, or specifies information observed about the Joneses.

You can derive from the latter option that it implies "at least one" child has been observed, which, if it is just one child, creates the "this child vs. that child" situation for 1/2, and, if it is both children, creates the "there's probably a 1/2 chance he would have told me about the other child and asked about that child's sex" situation for 1/2.

Critis of this question are claiming that they would mean something else when they would ask it that way because a common feature of the english language is misleading them, they phrase specific information in an unspecific fashion and think it can be recovered when it can not. As a result, they are simply comitting the logical fallacy known as "affirming the consequence" and think that the well-defined scenario contains additional information when it trivially does not. No, you're begging the question. "You cannot assume information not present, but you should assume that it is a criterion because that creates the right answer, and so is obviously the right assumption to make."

And no, if people want to continue to spread misinformation, I will not let them have the last word.Nor will I. We may be at this for a while. :P

Cazero
2015-01-29, 10:19 AM
I used generalization to try solving the problem, and I think I nailed it.
The Joneses have N kids. At least N-1 of them are boys. What is the probability that N are boys?

First method
The question specified the gender of N-1 of the kids, and the fact you don't know wich ones are set is irrelevant since the question's answer depends of the only variable left, the gender of the unspecified kid. There are 2 equally probable genders, the probability of the remaining kid to be a boy is 1/2.

Arguable, but straightforward.

Second method.
If you put the kids gender in an ordered list, there are N+1 possible combinations of gender with at least N-1 boys in it, N with a girl and one with only boys. The probability that all kids are boys is 1/(N+1). Right?
Wrong.

If you put the kids gender in an ordered list, there is N! (1*2*3*...*(N-1)*N) different kid orders possible, and for any given order, the position of the unspecified gender kid is set, leaving two different possibilities for each position.
Considering you have a 1/N! probability to be in any given order, you add up your N! different possibilities to get only boys, wich each have a probability of 1/(N!*2), for a final probability of 1/2. There is one girl in the mix, or there is none. Order is irrelevant, and the first method is a simpler way to obtain the same result.

If you give more information (like saying the N-1 older kids are boys) without changing the question (there are still N kids and at least N-1 boys), you are merely specifying a list of "valid" ordering methods, replacing the N! from the equation with something else, wich still simplifies with itself, leaving a probability of 1/2.

With N=2
At least one is a boy. The other is either a boy or a girl. We have the following possibilities :
one boy, one girl
one boy, one boy
Now, changing the unspecified but existing kid order, we get the additional possibilities :
one girl, one boy
one boy, one boy
There is another "one boy, one boy" in the list of possible results.

Now, with N=3
At least 2 are boys. Possibilities :
one boy, one boy, one girl
one boy, one boy, one boy
one boy, one boy, one girl
one boy, one boy, one boy
one boy, one girl, one boy
one boy, one boy, one boy
one boy, one girl, one boy
one boy, one boy, one boy
one girl, one boy, one boy
one boy, one boy, one boy
one girl, one boy, one boy
one boy, one boy, one boy
Exactly 1/2 probability to have only boys. Most lines seem redundant, but actually contain swapped gender-guaranteed boys.

SpectralDerp
2015-01-29, 12:34 PM
I used generalization to try solving the problem, and I think I nailed it.
The Joneses have N kids. At least N-1 of them are boys. What is the probability that N are boys?

Assuming that sexes of the children are independent and each equally likely to be either boy or girl and there are no other options, basic math reveals that

P(All N children are boys | At least N-1 of them are boys)
= P(All N children are boys and at least N-1 of them are boys) / P(at least N-1 of them are boys)
= P(All N children are boys) / P(At least N-1 of them are boys)
= P(All N children are boys) / P(All N children are boys or exactly N-1 children are boys)
= P(All N children are boys) / [P(All N children are boys) + P(Exactly N-1 children are boys)]

P(All N children are boys)
= P(Child 1 is a boy and ... child N is a boy)
= P(Child 1 is a boy) * ... P(Child N is a boy)
= (1/2) * ... (1/2)
= (1/2)^N

P(Exactly N-1 children are boys)
= P(Child 1 is a girl and the children not numbered 1 are boys or ... child N is a girl and the children not numbered N are boys)
= P(Child 1 is a girl and the children not numbered 1 are boys) + ... P(Child N is a girl and the children not numbered N are boys)

P(Child M is a girl and the children not numbered M are boys)
= P(Child M is a girl) * P(the children not numbered M are boys)
= (1/2) * (1/2)^(N-1)
= (1/2)^N

=> P(Exactly N-1 children are boys) = N * (1/2)^N

=> P(All N children are boys | At least N-1 of them are boys)
= (1/2)^N / [(1/2)^N + N * (1/2)^N]
= (1/2)^N / [(1/2)^N {1+N}]
= 1 / [1+N]

Once again, math gives you one result, falsely interpreting that one child has been specified gives you another.

You know what, I'll amend my earlier statement that I won't let people get away with spreading misinformation. If I have proven you wrong already, I am not going to repeat myself. Please refer to my earlier Q.E.D.s that you are wrong. New people may still submit nonsense for rebuttal, but the only acceptable format is .tex

Segev
2015-01-29, 01:35 PM
You know what, I'll amend my earlier statement that I won't let people get away with spreading misinformation. If I have proven you wrong already, I am not going to repeat myself. Please refer to my earlier Q.E.D.s that you are wrong. New people may still submit nonsense for rebuttal, but the only acceptable format is .tex

a) the guy you responded to in this particular math discussion is not me; I thought his approach interesting, but wasn't going to take the time to dissect it, myself, as I figured you'd probably do so.

b) You keep saying you've proven me wrong, but you keep arguing a point with which I don't disagree and then claiming it proves something else entirely.

c) I have repeatedly proven you wrong when it comes to the actual case being argued.

d) The sole point I am trying to get across is that this problem is badly worded

as a teaching tool, because it relies on an "insider's" interpretation of it to get the correct answer
It therefore fails to demonstrate what it wants to: that people don't understand probability the way they think they do
because it instead demonstrates that the people who get it "wrong" just don't know the unspoken assumption that is not at all clear from the language used
as a riddle, because it has at LEAST 2 possible correct answers (probably 3, possibly infinite) depending on the assumptions you make about what is "really" being asked
And one of those answers is "there is not enough information because there are unspoken assumptions and I do not know which ones you're using"
As a word problem, because the semantic natural language implies different unspoken assumptions than are actually used for the "textbook" answer.

The only way this problem is successful is as a way to "prove" that the one asking it is "more informed" because they happen to be a part of an informal intellectual club which already knows the answer and therefore the underlying unspoken assumptions.

I have repeatedly offered ways to phrase it such that it serves as the required teaching tool without being a trick question that only those who already know the desired answer could get right by virtue of knowing the answer and using that to divine the unspoken assumptions being used.

I do not dispute that, if the Joneses are the first family with 2 children that you come across wherein at least one of those children is a girl, the probability that both are girls is 1/3.

All I dispute is that, if you tell me you have found the Jonses, and that at least one of their children is a girl, I am able to know that you were looking for a family that met that criterion, rather than gave me information about the first family you found. And that distinction matters to determining the probability that both children are girls.

In the latter case, it doesn't matter if you tell me "at least one" or "the older one" or "this one." It's a 1/2 chance that both are girls. In the former case, based on the fact that you selected them so that at least one is a girl, it doesn't matter if you go on to tell me that the older one IS a girl or you introduce me to one of them that IS a girl. The odds remain 1/3 that both are.

It doesn't matter what you tell me. It matters how the selection was made. If you want what you tell me to be sufficient for me to accurately state whether there is a 1/3 or 1/2 chance that both are girls, you must tell me whether you used "at least one is a girl" as a selection criterion, or just looked at them after selection and determined it.

Cazero
2015-01-29, 01:50 PM
Assuming that sexes of the children are independent and each equally likely to be either boy or girl and there are no other options, basic math reveals that

P(All N children are boys | At least N-1 of them are boys)
= P(All N children are boys and at least N-1 of them are boys) / P(at least N-1 of them are boys)
= P(All N children are boys) / P(At least N-1 of them are boys)
= P(All N children are boys) / P(All N children are boys or exactly N-1 children are boys)
= P(All N children are boys) / [P(All N children are boys) + P(Exactly N-1 children are boys)]

P(All N children are boys)
= P(Child 1 is a boy and ... child N is a boy)
= P(Child 1 is a boy) * ... P(Child N is a boy)
= (1/2) * ... (1/2)
= (1/2)^N

P(Exactly N-1 children are boys)
= P(Child 1 is a girl and the children not numbered 1 are boys or ... child N is a girl and the children not numbered N are boys)
= P(Child 1 is a girl and the children not numbered 1 are boys) + ... P(Child N is a girl and the children not numbered N are boys)

P(Child M is a girl and the children not numbered M are boys)
= P(Child M is a girl) * P(the children not numbered M are boys)
= (1/2) * (1/2)^(N-1)
= (1/2)^N

=> P(Exactly N-1 children are boys) = N * (1/2)^N

=> P(All N children are boys | At least N-1 of them are boys)
= (1/2)^N / [(1/2)^N + N * (1/2)^N]
= (1/2)^N / [(1/2)^N {1+N}]
= 1 / [1+N]

Once again, math gives you one result, falsely interpreting that one child has been specified gives you another.

You know what, I'll amend my earlier statement that I won't let people get away with spreading misinformation. If I have proven you wrong already, I am not going to repeat myself. Please refer to my earlier Q.E.D.s that you are wrong. New people may still submit nonsense for rebuttal, but the only acceptable format is .tex

All of this looks good, but it is not applicable to the case I studied.
The question specifies that at least N-1 kids are boys. Wich means P(At least N-1 children are boys)=1. It changes everything.

P(at least N-1 children are boys) = 1
P(the specific child excluded by the N-1 is a girl) = 1/2 since it is independant from other specified genders.

P(Exactly N-1 children are boys) = P(At least N-1 children are boys) * P(the specific child excluded by the N-1 is a girl)
= 1 * 1/2
= 1/2

P(N children are boys)= P(at least N-1 children are boys) - P(Exactly N-1 children are boys)
= 1 - 1/2
= 1/2

That is my reasoning. Now if I'm wrong, I'd like you to show me where my mistake is.

edit : breaking down the case listing of 2 children using my previous demonstration
1) child A is a boy, child B is a boy
2) child A is a boy, child B is a girl
3) child A is a girl, child B is a boy
4) child A is a girl, child B is a girl

At least one is a boy. It does not simply exclude 4. It's more complicated. You can't remove an odd number of possibilities, because the number of different possibilities associated to a single variable parameter (gender of one kid) is 2.
Either at least child A is a boy, and it exclude both 3 and 4, or at least child B is a boy, and it exclude both 2 and 4. You have a 1/2 probability to have a 1/2 probability, and a the other 1/2 probability gives you a 1/2 probability. 1/4+1/4 = 1/2, the final probability is 1/2.

To get both the boy then girl and girl then boy entry, you have to add possibilities assossiated to the reverted AB order. You get 8 possibilities.
1) child A is a boy, child B is a boy
2) child A is a boy, child B is a girl
3) child A is a girl, child B is a boy
4) child A is a girl, child B is a girl
5) child B is a boy, child A is a boy
6) child B is a boy, child A is a girl
7) child B is a girl, child A is a boy
8) child B is a girl, child A is a girl
Same reasoning than before. Setting one parameter (the gender of either child) removes 4 possibilities. Either at least A is a boy, excluding 3, 4, 6 and 8, or at least B is a boy, excluding 2, 4, 7 and 8. The final probability is 1/2.

Jenerix525
2015-01-29, 02:24 PM
All I dispute is that, if you tell me you have found the Jonses, and that at least one of their children is a girl, I am able to know that you were looking for a family that met that criterion, rather than gave me information about the first family you found. And that distinction matters to determining the probability that both children are girls.

I believe the bolded part is where your interpretation differs from the textbook/formally-trained-professional interpretation.

You see it as being told information.
Textbooks see it as knowing information.

I'm also wondering why your earlier examples selected by age not gender. It discounted the possibility of

BB - "Neither child is a girl"
GG - "Neither child is a boy"

Segev
2015-01-29, 02:36 PM
I believe the bolded part is where your interpretation differs from the textbook/formally-trained-professional interpretation.

You see it as being told information.
Textbooks see it as knowing information.

I'm also wondering why your earlier examples selected by age not gender. It discounted the possibility of

BB - "Neither child is a girl"
GG - "Neither child is a boy"

The actual problem(s) that started this particular conversation were roughly thus:

The Smiths have two children. The older one is a girl. What is the probability that both are girls?

The Joneses have two children. At least one is a girl. What is the probability that both are girls?

The first is why "age" entered it at all. It's technically irrelevant; it's used in the first problem to try to create a way of saying "THIS child" versus "THAT child."

If we were being consistent and trying to illustrate the point that is supposed to be illustrated by these questions, they could be better worded:

The Smiths are part of a group of families with two children wherein the older child is a girl. What is the probability that both are girls?

The Joneses are part of a group of families with two children wherein at least one of the children is a girl. What is the probability that both are girls?

This formulation tells you that you know this information about the family because of selection criteria, rather than because you examined (or were told about) at least one of the children to learn it.

Cazero
2015-01-29, 05:31 PM
So, I have found something on this specific problem on Wikipedia, and apparently it's a paradox (http://en.wikipedia.org/wiki/Boy_or_Girl_paradox).
There are perfectly valid demonstrations for both 1/2 and 1/3 probability. If there is a fallacy somewhere, I can't see it.

Segev
2015-01-29, 05:48 PM
Fascinating. I'm mildly pleased to see that my own conclusions are supported by others' extensive research. Also, the otehr details in that article are interesting.

Gavran
2015-01-29, 05:49 PM
So, I have found something on this specific problem on Wikipedia, and apparently it's a paradox (http://en.wikipedia.org/wiki/Boy_or_Girl_paradox).
There are perfectly valid demonstrations for both 1/2 and 1/3 probability. If there is a fallacy somewhere, I can't see it.

Thank you for this.

Many people argued strongly for both sides with a great deal of confidence, sometimes showing disdain for those who took the opposing view. Think the author was reading this thread?

Segev
2015-01-29, 06:06 PM
Thank you for this.

Of people reading that article, at least one has written in this thread. </ducks and covers>

SpectralDerp
2015-01-29, 06:11 PM
All of this looks good, but it is not applicable to the case I studied.
The question specifies that at least N-1 kids are boys. Wich means P(At least N-1 children are boys)=1.

That depends on what you consider P to be.

In my notation, the value P(A) refers the "unconditioned" probability of an event A. Meaning that it is calculated by only making assumptions about "probabilities, not events". Meaning I am assuming that for each child, the probability of that child being a boy or a girl is 1/2 each and that the sexes of the children are independent. I am not assuming anything about what their sexes actually are, just how they are distributed.

[Notice that the knowledge of "The Joneses have exactly N children" actually falls neither in the category of probabilities, nor events. It cannot even have a probability in our case. Understanding this requires presumably less that a Bachelor's degree, but I really don't want to get into that]

The value P(A|B) refers to the conditional probability of an event A, under the assumption that an event B is true.
This value is interesting, because it is calculated by only making assumptions about "events, not probabilities". The reason for that is because of the equation

P(A|B) = P(A and B) / P(B)

This equation tells us that if we want to calculate a conditional probability, we have to calculate two "unconditional" probabilities and divide them.

Now, what does "at least N-1 kids are boys" mean?
Technically, there is no such thing as an objective probability. Probabilities are numerical values that express plausibility. They are dependent on knowledge. If I know that I have two siblings, a brother and a sister, the I can evaluate the plausibility of "SpectralDerp has a brother" as 1, if you know that I have two siblings, you can evaluate it as 3/4 and if you don't know nothing about me, you can barely evaluate it at all (what's the probability that I have N siblings?).

Your own personal evaluation of "All N of the Joneses children are boys" is P(All N of the Joneses children are boys | B), where B is the event that describes that the things that you know about the sexes of the Joneses children and other events in the realm of out "probability space" actually apply to the sexes of the Joneses children.

Given that you know that "at least N-1 kids are boys" (and nothing that goes beyond that), your can now evaluate the plausibility of "All N of the Joneses children are boys" to be equal to the value of

P(All N children are boys | At least N-1 children are boys)
= P(All N children are boys and at least N-1 children are boys) / P(At least N-1 children are boys)
= P(All N children are boys) / P(At least N-1 children are boys)

In my last post, I have done the math. The correct result is 1/[N+1].

P(the specific child excluded by the N-1 is a girl) = 1/2 since it is independant from other specified genders.

To specify the sex of a child, you have to first specify a child and then say "the child I have just specified has the sex suchandsuch". The statement "At least N-1 children are boys" does not specify any child. This has been explained multiple times already. If there are two children, the options are (Boy,Boy), (Boy,Girl), (Girl, Boy) and (Girl,Girl). The statement "At least one child is a boy" is compatible with the first three option and incompatible with the fourth options. The statement "There is a child with the distinguishing feature X that the other child does not have and the child with the distinguishing feature X is a boy" allows you to sort the original options in such a way that the child with the distinguishing feature X is always the first one in the pair and then discard the options where the first child in the pair is not a boy. It is therefore compatible with two options and incompatible with two others.

If you are not comfortable with this, try to properly calculate P(All N children are boys and at least N-1 children are boys) / P(At least N-1 children are boys). Or just try to understand the calculation I have done above, the correctness of each step follows near-trivially from the axioms of probability.

SpectralDerp
2015-01-29, 06:33 PM
Please take the effort to properly display (http://en.wikipedia.org/wiki/Help:Displaying_a_formula) mathematical notation when you write on Wikipedia.

There are perfectly valid demonstrations for both 1/2 and 1/3 probability.

The article correctly demonstrates that

P(Both children are boys | at least one child is a boy) = 1/3

and that

P(Both children are boys | A specific child has been sampled and turned out to be a boy) = 1/2

Given that the results are different, they describe different problems.

Since "at least one child is a boy" is exactly the background information in the second question, the correct way to calculate the answer to the second question is the first one.

Fiery Diamond
2015-01-29, 08:18 PM
Please take the effort to properly display (http://en.wikipedia.org/wiki/Help:Displaying_a_formula) mathematical notation when you write on Wikipedia.

The article correctly demonstrates that

P(Both children are boys | at least one child is a boy) = 1/3

and that

P(Both children are boys | A specific child has been sampled and turned out to be a boy) = 1/2

Given that the results are different, they describe different problems.

Since "at least one child is a boy" is exactly the background information in the second question, the correct way to calculate the answer to the second question is the first one.

Except that, in the common parlance, "At least one child is a boy" can mean "A specific child has been sampled and turned out to be a boy." People who don't use probability theory in their ordinary lives don't make the distinction in their speech, and should not be expected to assume that a distinction is being made when the question is posed. In order to be perfectly clear in wording, you have to tell the listener (assuming the listener is not already a student of probability) that "At least one child is a boy" is broader than "a specific child has been sampled and turned out to be a boy." An average joe that hears you say "at least one child is a boy" can interpret that to mean "the person telling me this has determined the identity of one of the children but not the other." And - and this is the part you keep wrongly arguing against - that doesn't make the listener wrong. It just means he applied a different basic assumption to what the language used implied. Saying "oh, but interpreting it that way is wrong," is wrong. It is, for better or worse, due to the way in which we communicate in non-rigorous settings, semantically ambiguous. Arguing that it is not is a failure to understand how language works outside of academic settings.

SpectralDerp
2015-01-30, 01:53 AM
Except that, in the common parlance, "At least one child is a boy" can mean "A specific child has been sampled and turned out to be a boy." People who don't use probability theory in their ordinary lives don't make the distinction in their speech, and should not be expected to assume that a distinction is being made when the question is posed.

"People who don't use probability theory in their ordinary lives" are primarily dead, comatose or don't have functioning brains, most regular people have to evaluate likelihoods on daily basis. The meaning of "at least one child is a boy" is not only completely independent from any particular of probability theory and would almost never align with "a specific child has been sampled and turned out to be a boy". In fact, had the questioner meant to express "a specific child has been sampled and turned out to be a boy", there are a variety of ways to express that to make it clear that this is what actually happened. And since conditional probabilities are functions on knowledge, the exact knowledge you assume makes all the difference, so a person who is familiar with it is going to be very precise with it.

Should there have been a hint? I don't think so. The context of the question was

Problem 1: Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Problem 2: Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
and the reader should immediately notice the difference in phrasing and thus consider the exact meaning of that phrase. A disclaimer of "Substituting a phrase for what people might want to colloquially express with it is going to make a difference" still wouldn't have helped, as the difference was explained several times throughout the thread and people still made this mistake. The simple introspection "Does this information entail the information I intend to extract from it?" would have fixed that.

" An average joe that hears you say "at least one child is a boy" can interpret that to mean "the person telling me this has determined the identity of one of the children but not the other." And - and this is the part you keep wrongly arguing against - that doesn't make the listener wrong.

"Just because the listener's interpretation of what the speaker said is wrong does not mean the listener's interpretation of what the speaker said is wrong".

I give up.

Math and logic are acceptable boogeymen, let's blame them and not the inability of people to just think logically. They should have known that people weren't going to take what they said to be what they meant, it's all their fault!

Cazero
2015-01-30, 02:39 AM
Please take the effort to properly display (http://en.wikipedia.org/wiki/Help:Displaying_a_formula) mathematical notation when you write on Wikipedia.
I did not write anything on Wikipedia. I just found the page.

The article correctly demonstrates that

P(Both children are boys | at least one child is a boy) = 1/3

and that

P(Both children are boys | A specific child has been sampled and turned out to be a boy) = 1/2
Yes. I would add that "at least one child is a boy" is strictly equivalent to "a unspecified child has been sampled and turned out to be a boy", because it is the only way to assert knowledge with certainty rather than keeping it in the blur cloud of probability space. And then, logically, your unspecified sampled child becomes specific by virtue of having been sampled (nothing stops you from putting an arbitrary order on the children with the child you know about first, just like saying the older one is a boy is ordering them by age), and both propositions are logically equivalent.
And yet the conditionnal probability calculations that you made remain perfectly valid and give an answer of 1/3, while some other modelisations get mixed results of 1/2 or 1/3. Wich is why we have a paradox.

SpectralDerp
2015-01-30, 04:01 AM
Yes. I would add that "at least one child is a boy" is strictly equivalent to "a unspecified child has been sampled and turned out to be a boy", because it is the only way to assert knowledge with certainty

{scrubbed} What you think is equivalent to "an unspecifed child has been sampled and turned out to e a boy" is not "at least one child is a boy", it's "somebody at one point knew that at least one child is a boy".

I have already disproven an analogous case of this misconception, namely that "I know that at least one of the two Jones children is a boy" is equivalent to "I have investigated the sex of one of the Jones children and learned that it is a boy". The first proposition does not logically entail the second one, because I could have asked Mr. Jones the question "Is at least one of your children a boy?" and habe received the answer "Yes".

Notce that this is irrelevant with respect to the answer of problem 2. It does not say "Mr. Jones has two children and told me that at least one of his children is a boy, what is the probability that both are boys?" because introducing any additional information changes the answer. The correct answer begins by assumng exactlx the information provided, which is that "at least one child is a boy".

Cazero
2015-01-30, 07:04 AM
{scrubbed}This is an ad hominem. I will now ignore it.

What you think is equivalent to "an unspecifed child has been sampled and turned out to e a boy" is not "at least one child is a boy", it's "somebody at one point knew that at least one child is a boy".
You are saying that sampling a random child and finding out it is a boy implies that at least one child is a boy, and not the other way around. In probabilities, it translates as P(at least one child is a boy)≥P(sampling a random child and discovering it is a boy). Wich is perfectly true. 3/4≥1/2. I'm not arguing against that.

What I am arguing is that once the sampling happened, the event of sampling a random child and discovering it is a boy happened. It is no longer a random child, it is now a specific child, and P(the sampled child is a boy)=1. But we still have P(at least one child is a boy)≥P(the sampled child is a boy), and P cannot be higher than 1. Then P(at least one child is a boy)=1. But it is true that we might not be in that case. We can assume that for the thought experiment, some magical mojo gave us information without any sampling happening at all. However, we still are in a case where P(at least one child is a boy)=1, because we have certainty on that information. It doesn't matter how we got it, the question is based on the given fact that at least one child is a boy.

And the funny part is that none of what is above matters. The only thing that matter is if the children genders are independant or not.
First possibility : the genders are independant. This assumption is supported by the simplifications used to keep the problem simple. The probability of the other child being a boy is 1/2.
Second possibility : the genders are dependant. This assumption is supported by the theoric sampling we use, wich is based on the probability of gender distribution among the children given the definition of the problem. Given at least one child is a boy, the probability of the other child being a boy is 1/3.

The paradox is "solved" when we can discard an assumption with certainty (P(the children genders are independant)=0, or P(the children genders are independant)=1), or if both give the same result. For example, if the older child is a boy, the sample in the second possibility also gives a probability of 1/2, and if Mr Jones have twins, the gender are clearly not independant.

I have already disproven an analogous case of this misconception, namely that "I know that at least one of the two Jones children is a boy" is equivalent to "I have investigated the sex of one of the Jones children and learned that it is a boy". The first proposition does not logically entail the second one, because I could have asked Mr. Jones the question "Is at least one of your children a boy?" and habe received the answer "Yes".
It is true that the first proposition does not logically entail the second one, but your example fails to address that.

Asking Mr Jones about the gender of his kids is investigating the gender of his kids. Hearing Mr Jones speaking of the gender of his kids is equivalent to investigating the gender of his kids because you obtain the exact same information. That information is "Mr Jones just told me the gender of one of his children". Given the simplifications of the logical problem, Mr Jones does not lie about the gender of his kids. "Mr Jones just told me the gender of one of his children" implies "I know that at least one of the two Jones children is a boy" because Mr Jones doesn't lie. Receiving information in any form about the child gender would imply the same "I know that at least one of the two Jones children is a boy", and might contain added information.

This does not contradict your statement. Your statement says that "I know that at least one of the two Jones children is a boy" does not imply any sort of information transfer regarding a child gender. Wich is impossible in real life, but a perfectly valid statement in a logic problem. And I think this is where the paradox lies.

Notce that this is irrelevant with respect to the answer of problem 2. It does not say "Mr. Jones has two children and told me that at least one of his children is a boy, what is the probability that both are boys?" because introducing any additional information changes the answer. The correct answer begins by assumng exactlx the information provided, which is that "at least one child is a boy".
The way I obtained information about a child gender is also irrelevant in that case. What matters is the final amount of information and it's certainty, or P(this information is true). How I obtained it only change the certainty : if I saw a boy in Mr Jones garden, I know the probability he has at least a boy is equal or above (his other children can be boys too) the probability that the child I saw was one of his own, and if Mr Jones tells me that he has at least one boy, I know the probability he has at least one boy is equal to the probability he didn't lie to me. Here, the final information is about the gender of one child, specified in problem 1 (the older) or not specified in problem 2, and it's level of certainty is 1 because the problem would be too complicated otherwise.

SpectralDerp
2015-01-30, 09:29 AM
This is an ad hominem. I will now ignore it.

{scrubbed}

You are saying that sampling a random child and finding out it is a boy implies that at least one child is a boy, and not the other way around.

Yes, that is what I am saying. You are saying that that "finding out that at least one child is a boy" and "sampling a random child and finding out it is a boy" are strictly equivalent, which means that "that sampling a random child and finding out it is a boy implies that at least one child is a boy and the other way around."

In probabilities, it translates as P(at least one child is a boy)≥P(sampling a random child and discovering it is a boy). Wich is perfectly true. 3/4≥1/2. I'm not arguing against that.

Fun fact, "A implies B" does not actually imply "P(A) is greather than or equal to P(B)" for general events A and B. This is slightly relevant here, but the explanation is too complicated to give here.

What I am arguing is that once the sampling happened, the event of sampling a random child and discovering it is a boy happened. It is no longer a random child, it is now a specific child, and P(the sampled child is a boy)=1. But we still have P(at least one child is a boy)≥P(the sampled child is a boy), and P cannot be higher than 1. Then P(at least one child is a boy)=1. But it is true that we might not be in that case. We can assume that for the thought experiment, some magical mojo gave us information without any sampling happening at all. However, we still are in a case where P(at least one child is a boy)=1, because we have certainty on that information. It doesn't matter how we got it, the question is based on the given fact that at least one child is a boy.

You are equivocating the probability before the sampling with the probability given the result of the sampling, P(A) and P(A|the sampled child is a boy) are not the same thing, in our case we have that P(At least one child is a boy)=3/4 and P(At least one child is a boy|At least one child is a boy)=1.

I personally find it rather telling that all of my critics have so far been unable to properly use this notation.

And the funny part is that none of what is above matters. The only thing that matter is if the children genders are independant or not.
First possibility : the genders are independant. This assumption is supported by the simplifications used to keep the problem simple. The probability of the other child being a boy is 1/2.
Second possibility : the genders are dependant. This assumption is supported by the theoric sampling we use, wich is based on the probability of gender distribution among the children given the definition of the problem. Given at least one child is a boy, the probability of the other child being a boy is 1/3.

The joint distribution of the sexes of the children is not uniquely determined by the distributions of the sexes of the children unless the sexes of the children are independent. If the sexes of the children are not independent, the answer to this problem is unknowable without further knowledge of the probability space. Also, the bolded part is false, as above. There is no "other child".

The paradox is created because people think the posterior implies a posterior from which they can calculate the probability from 1/2 and it is solved by noticing that these people are wrong.

Let (X,S,P) be a probability space, B={0,1} and X and Y be identically distributed independent B-valued functions on X that are measurable when all subsets of B are measurable and the pushforward measure on B be the discrete uniform distribution. Calculate P(X+Y=2|X=1) and P(X+Y=2|X=1 or Y=1)

then the people who conflate unspecific information with specific information still would have said the answer is 1/2 in both cases, because they would still assume that they can re-label X and Y into A and B such that the event "X=1 or Y=1" entails "A=1".

It is true that the first proposition does not logically entail the second one, but your example fails to address that.

My example proves that the first proposition does not logically entail the second one, because it gives a possible scenario where the first proposition is true but the second proposition is false.

Segev
2015-01-30, 09:48 AM
Math and logic are acceptable boogeymen, let's blame them and not the inability of people to just think logically. They should have known that people weren't going to take what they said to be what they meant, it's all their fault![/COLOR]

I think you mean, "Those people should have known that the language they use in their daily lives means something else entirely to those of us who phrase this question this way precisely because we already know the answer."

Seriously, SpectralDerp, the article that was linked to wikipedia spells out our entire discussion and why the guy who originally formulated the question as you're defending it acknowledged it was poorly worded. For precisely the reasons I've been giving.

Nobody is saying that people don't misunderstand probability analysis, nor make the mistakes the question is trying to illustrate. They're just saying that the question, as worded, fails to illustrate this because the "incorrect" answer can be correctly given based on how the question was asked.

I know I've repeatedly offered ways to re-word it so it does exactly the job it's intended to. I don't know why you're still trying to claim that means people arguing with you "don't understand conditional probability."

Cazero
2015-01-30, 11:03 AM
-got scrubbed, anticipating the scrubbing of the quote-
Sorry for not understanding the word logic as in "the meticulously defined terms and rules of a specific sector of mathematics" rather than it's common use as "the ability to determine consequences from causes". But I don't remember contradicting myself. Maybe changing opinion due to reading something on wikipedia, but nothing else.

Yes, that is what I am saying. You are saying that that "finding out that at least one child is a boy" and "sampling a random child and finding out it is a boy" are strictly equivalent, which means that "that sampling a random child and finding out it is a boy implies that at least one child is a boy and the other way around."No, that is not what I am saying. I am saying that the difference between "finding out that at least one child is a boy" and "sampling a random child and finding out it is a boy" doesn't matter in the specific case we are studying, because we are only using the information that both statements implies : "at least one child is a boy". I don't care wether the implication works the other way around or not.

Fun fact, "A implies B" does not actually imply "P(A) is greather than or equal to P(B)" for general events A and B. This is slightly relevant here, but the explanation is too complicated to give here.Good thing that is not what I said then.
I said that "A implies B" imply that "P(B) is greater than or equal to P(A)". As in P(B) = P(A) + P(B and not A), wich is a property of "A implies B".
B is the knowledge there is at least one boy. A is the investigation proving that one child is a boy. That is what I said.

You are equivocating the probability before the sampling with the probability given the result of the sampling, P(A) and P(A|the sampled child is a boy) are not the same thing, in our case we have that P(At least one child is a boy)=3/4 and P(At least one child is a boy|At least one child is a boy)=1.
Ho. Well. I guess I'm wrong here then.

The joint distribution of the sexes of the children is not uniquely determined by the distributions of the sexes of the children unless the sexes of the children are independent. If the sexes of the children are not independent, the answer to this problem is unknowable without further knowledge of the probability space.
Maybe? I'm not a math teacher. I'm just some guy who believe he's smart enough to understand a wikipedia article.

Also, the bolded part is false, as above. There is no "other child".
Are you seriously claiming that Mr Jones had only one child all along? Sorry, couldn't resist.

The paradox is created because people think the posterior implies a posterior from which they can calculate the probability from 1/2 and it is solved by noticing that these people are wrong.

Let (X,S,P) be a probability space, B={0,1} and X and Y be identically distributed independent B-valued functions on X that are measurable when all subsets of B are measurable and the pushforward measure on B be the discrete uniform distribution. Calculate P(X+Y=2|X=1) and P(X+Y=2|X=1 or Y=1)

then the people who conflate unspecific information with specific information still would have said the answer is 1/2 in both cases, because they would still assume that they can re-label X and Y into A and B such that the event "X=1 or Y=1" entails "A=1".
Now that talks to me. Re-labeling, re-ordering, same thing. That's what I did when I concluded the probability should be 1/2.

When I re-order the set such as "X=1 or Y=1" entails "A=1 and (B=0 or B=1)", I am supposed to separate the case where both X and Y are 0 first. And this is where the difference is. I assumed you had to order first and separate later, but that actually changes the possible values of A.
Given that, you have 1/3 probability of having the order imposed by the fact only X is set at 1, 1/3 probability of having the order imposed by the fact only Y is set at 1, and 1/3 probability of order not mattering because both are at 1. 1/3 probability of both being at 1.

So maybe there is no paradox after all.

My example proves that the first proposition does not logically entail the second one, because it gives a possible scenario where the first proposition is true but the second proposition is false.
Now I just don't get it. Unless I'm seriously mistaken, your example proves you can get knowledge of the gender of the Joneses children without investigating about it because he can tell you. And that's just not important.
I don't care that "I know one of the Joneses children is a boy" does not imply that "I investigated the gender of one of the Joneses children and he's a boy". I'm not using that implication, I'm using the reverted implication, and I went to great length to say that what we used to imply that "I know one of the Joneses children is a boy" doesn't matter.

Segev
2015-01-30, 11:29 AM
It is accepted, per the wikipedia article, that the method by which the information is determined - that is, whether the information was determined by inspection of the sample or the sample was determined by using the information as criterion - determines whether it's 1/2 or 1/3.

Given that neither is explicitly stated in the problem-as-presented, and that colloquial natural language implies that the information was derived from inspection of the sample, the question is at best ambiguous and potentially implies the 1/2 case.

This is, again, covered in the article.

SpectralDerp
2015-01-30, 12:04 PM
Seriously, SpectralDerp, the article that was linked to wikipedia spells out our entire discussion and why the guy who originally formulated the question as you're defending it acknowledged it was poorly worded. For precisely the reasons I've been giving.

Neither Wikipedia, nor the guy it cites as having giving formulated this question at one point, are infallible authorities. You have also indicated that you have edited that article yourself, which also means that it is not neutral. After reading it, I have completely changed my mind.

[...]"at least one child is a boy" is strictly equivalent to "a unspecified child has been sampled and turned out to be a boy"

You are saying that that "finding out that at least one child is a boy" and "sampling a random child and finding out it is a boy" are strictly equivalent,

No, that is not what I am saying.

:annoyed:

Good thing that is not what I said then.
I said that "A implies B" imply that "P(B) is greater or equal than P(A)". As in P(B) = P(A) + P(B and not A).
B is the knowledge there is at least one boy. A is the investigation proving that one child is a boy. That is what I said.

The issue here (which I don't really want to elaborate on because it's rather tecnical) is that an event with a probability can imply an event without a probability. Yes, there are events without probability. Sometimes this is because the model doesn't account for them (but could have), other times this is because no model can account for them.

When I re-order the set such as "X=1 or Y=1" entails "A=1 and (B=0 or B=1)", I am supposed to separate the case where both X and Y are 0 first. And this is where the difference is. I assumed you had to order first and separate later.
Given that, you have 1/3 probability of having the order imposed by the fact only X is set at 1, 1/3 probability of having the order imposed by the fact only Y is set at 1, and 1/3 probability of order not mattering because both are at 1. 1/3 probability of both being at 1.

When you order first and separate later, you have a 1/2 probability of having X first with 1/2 probability of Y being at 1, and another 1/2 probability of having Y first with 1/2 probability of X being at 1. 1/2 probability of both being at 1.

All you need to do now is prove that ordering first and separating later is not allowed by the formal rules of logic. I guess it's just an operator priority property that I don't know about or something like that.

X and Y are functions, they take values at points. We take a point z with "X(z)=1 or Y(z)=1". This means that either "X(z)=1 and Y(z)=1" or "X(z)=1 and Y(z)=0" or "X(z)=0 and Y(z)=1". Each of these three events has probability 1/4, so for each of them, so the set of z such that "X(z)=1 and Y(z)=1" isn't empty and neither are the other two. This set remains the same even if change the roles of X and Y. The event "X(z)=1 or Y(z)=1" becomes "Y(z)=1 or X(z)=1", the sets of z that such that these events are true are the same. The event "X(z)=1 or Y(z)=0" becomes "X(z)=0 or Y(z)=1" and vice versa.

Let's call this set C. Let's now relabel X and Y as A and B. This means either
(1) X is relabeled as A and Y is relabeled as B,
or
(2) X is relabeled as B and Y is relabeled as A,
and there are no other options.

The set of all z such that "A(z)=1 or B(z)=1" is therefore the same set as the set of all z such that "X(z)=1 or Y(z)=1". We just called that set C.

Assume we did (1). Since C contains at least one point z such that "X(z)=0 and Y(z)=1", A(z)=1 does not hold for all z in C.
Assume we did (2). Since C contains at least one point z such that "X(z)=1 and Y(z)=0", A(z)=1 does not hold for all z in C.

Since there were no other options to relabel, we cannot simply relabel X and Y such that one of them is always 1 in all cases we have yet to rule out.

All of this might admittedly seem unnecessary convoluted. Why do I bring this up? Because at no point did I need to say what the points z are. Z could be the set {(1,1),(1,0),(0,1),(0,0)}, X could be the function that maps (x,y) to x and Y could be the function that maps (x,y) to Y. Z could also contain elaborate descriptions of Mr. Jones typical breakfast, whether or not Nazis went to Mars or what the lottery numbers are. The point is that it doesn't matter. No matter how I model it, as long as (Z,S,P), X and Y have at least the properties I specified, the unspecified properties don't change anything

Cazero
2015-01-30, 12:35 PM
For completion's sake.

Yes. I would add that "at least one child is a boy" is strictly equivalent to "a unspecified child has been sampled and turned out to be a boy"

Yes, that is what I am saying. You are saying that that "finding out that at least one child is a boy" and "sampling a random child and finding out it is a boy" are strictly equivalent, which means that "that sampling a random child and finding out it is a boy implies that at least one child is a boy and the other way around."

No, that is not what I am saying.
You added an implication (bolded for emphasis) that I never stated, because I don't speak in rigorous formal logic. The strict equivalence I'm talking about is a logical conclusion of the fact that wether said implication exists or not doesn't matter, wich implies that we don't need to prove it to use it. Yes, in rigorous logic, it might not be true. My later claim was about that added implication. Wich is why I immediatly added the following specification of my concern.

I am saying that the difference between "finding out that at least one child is a boy" and "sampling a random child and finding out it is a boy" doesn't matter in the specific case we are studying, because we are only using the information that both statements implies : "at least one child is a boy". I don't care wether the implication works the other way around or not.
I consider this reasoning is sufficient to claim that "at least one child is a boy" is strictly equivalent to "a unspecified child has been sampled and turned out to be a boy" in the specific context of the problem, because how we found out that "at least one child is a boy" is true doesn't matter.

Also, I thought more about the problem and edited some parts of my post since then. You might want to do the same.

SpectralDerp
2015-01-30, 12:42 PM
Given that neither is explicitly stated in the problem-as-presented, and that colloquial natural language implies that the information was derived from inspection of the sample

- Colloquial "natural" language does not imply that the information was derived from inspection of the sample. The proof is as easy as my ability to ask you "Mr. Jones has two children and at least one of them is a girl, what's the probability that both of them are girls?" and then answer the question "How do you know that at least one of them is a girl?" with "Mr. Jones told me". Do you not like that example because it involves Mr. Jones? Then I can just switch to an equivalent problem without children.

"I have a device in my brain that can be remote-controlled to detonate, which then kills me. If you want to know why, help me figure something out. I know that there is a computer somewhere in the world that, on every New Year's day, randomly selects either "Life" or "Death" with 1/2 chance each and as a fail safe, does so twice and independently. I know that this causes the device in my brain detonaes if and only if the computer selects "Death" twice. Given the fact that I am alive, I know that this year, the computer selected "Life" at least once. Given what we now know, what's the probability that it selected "Life" twice? It matters to me, because I know hotline I can call and I when I call during a year where "Life" was selected twice , they safely disarm the device. Otherwise, they detonate the device. I would like to know if my odds are at least 50:50".

What colloquial language implies is that the information might have been derived from inspection of the sample and it might not have been derived from inspection of the sample. "Given that neither is explicitly stated in the problem-as-presented", I can't put "the information has been derived from inspection of the sample" into my background knowledge. All I can put into my background knowledge is the information I have been provided, which is that "at least one child is a girl". The probability of "both children are girls" given what I know is therefore P(Both children are girls|at least one child is a girl)=1/3.

You added an implication (bolded for emphasis) that I never stated.

You have not used this exact wording, but you have said

"finding out that at least one child is a boy" and "sampling a random child and finding out it is a boy" are strictly equivalent

which means that "that sampling a random child and finding out it is a boy implies that at least one child is a boy and the other way around".

It also isn't equivalent in our specific case. Our specific case is "at least one child is a boy" (it's hard to keep track in which discussion the question is about two girls and in which it's about two boys, so please excuse me if I get that mixed up) and since I have provided a scenario where the condition "at least one child is a boy" is true and one of the two statements (namely "at least one child is a boy, again) is true but the other statement (namely "an unspecified child has been sampled and turned out to be a boy") is false, I have already proven that it makes no difference.

Cazero
2015-01-30, 12:51 PM
What colloquial language implies is that the information might have been derived from inspection of the sample and it might not have been derived from inspection of the sample. "Given that neither is explicitly stated in the problem-as-presented", I can't put "the information has been derived from inspection of the sample" into my background knowledge.

Wait. The paradox answer is right there. We only need to calculate the probability that the information is derived from inspection of the sample.
And if that probability cannot be calculated, the answer is "we don't have sufficient data to get the correct answer with certainty, however, we can say it is comprised betwen 1/2 and 1/3".
Let's get that Nobel.

Gavran
2015-01-30, 01:46 PM
Neither Wikipedia, nor the guy it cites as having giving formulated this question at one point, are infallible authorities. You have also indicated that you have edited that article yourself, which also means that it is not neutral. After reading it, I have completely changed my mind.

But you or your teachers are? The existence of that page is undeniable proof that the disagreement is larger than this thread. I'm not going to tell you that you need to agree with everyone else here, but surely you can at least accept the possibility that (as greater minds than ours are apparently unable to come to an agreement) you might be wrong.

Also, having read this whole thread I've read nothing remotely implicating someone edited that article* other than your earlier accusations. When I first read your accusation I went and checked the article history, and indeed, the last significant edits were days before it was linked to this thread. For someone who quotes Wikipedia policies as a passive aggressive way of accusing someone of vandalism, I'd have thought you'd also Assume Good Faith (http://en.wikipedia.org/wiki/Wikipedia:Assume_good_faith). That's good practice in any kind of debate too, not just Wikipedia...

*Here (http://www.giantitp.com/forums/showsinglepost.php?p=18735583&postcount=152) you can see Segev's initial reaction to the article and here (http://www.giantitp.com/forums/showsinglepost.php?p=18735678&postcount=154) you can see a little joke that you may have misconstrued. Which would be somewhat ironic... but ultimately forgivable.

Segev
2015-01-30, 02:04 PM
Neither Wikipedia, nor the guy it cites as having giving formulated this question at one point, are infallible authorities.Ah. Well, I can respect your desire to come to the conclusion based on your own analysis. I do, however, reserve the right to find your analysis entirely faulty, as you've failed to actually demonstrate that the formulation of the question is unambiguous and clearly means only what you claim it does.

You have also indicated that you have edited that article yourself, which also means that it is not neutral.

No, I indicated that I have read it. I can honestly say I have never edited a wikipedia article in my entire life. I don't have an account with them to do so.

Segev
2015-01-30, 02:19 PM
- Colloquial "natural" language does not imply that the information was derived from inspection of the sample. The proof is as easy as my ability to ask you "Mr. Jones has two children and at least one of them is a girl, what's the probability that both of them are girls?" and then answer the question "How do you know that at least one of them is a girl?" with "Mr. Jones told me". Do you not like that example because it involves Mr. Jones? Then I can just switch to an equivalent problem without children. This formulation still implies that you found Mr. Jones and asked him for information about his children, then formulated your question based on the information given. This is identical, for practical purposes, to having inspected his children.

To get the 1/3 probability, you would have had to have established that you would have gone on to find Mr. Smith, and ask him, and Mr. Johnson, and ask him, until you found one of them who could tell you that at least one was a girl.

Selection criterion vs. observed information.

If you ask Mr. Jones, "Is one of your children a boy?" and will use him iff he says "yes," you're right. It's 1/3 chance both are boys.

But nothing in the way you expressed the question here, nor the original problem, suggests that you would not be asking me about Mr. Jones unless he had "at least one boy."

"I have a device in my brain that can be remote-controlled to detonate, which then kills me. If you want to know why, help me figure something out. I know that there is a computer somewhere in the world that, on every New Year's day, randomly selects either "Life" or "Death" with 1/2 chance each and as a fail safe, does so twice and independently. I know that this causes the device in my brain detonaes if and only if the computer selects "Death" twice. Given the fact that I am alive, I know that this year, the computer selected "Life" at least once. Given what we now know, what's the probability that it selected "Life" twice? It matters to me, because I know hotline I can call and I when I call during a year where "Life" was selected twice , they safely disarm the device. Otherwise, they detonate the device. I would like to know if my odds are at least 50:50".

What colloquial language implies is that the information might have been derived from inspection of the sample and it might not have been derived from inspection of the sample. "Given that neither is explicitly stated in the problem-as-presented", I can't put "the information has been derived from inspection of the sample" into my background knowledge. All I can put into my background knowledge is the information I have been provided, which is that "at least one child is a girl". The probability of "both children are girls" given what I know is therefore P(Both children are girls|at least one child is a girl)=1/3.In your example, you have presented the fact that there is no way we would be having this conversation if it were not true that it had chosen Life at least once. That tells me selection criteria.

With the Mr. Jones problem, there is no indication that we would not, if Mr. Jones had told you, "I have a daughter," that we would not be having a conversation about the chances that both his children are girls.

There is, in your deadly example, no chance that the device chose "Death" twice, and that we'd be having a conversation about whether or not it chose "Death" once or twice.

To formulate the Mr. Jones problem similarly to yours, you'd have to do something like... "I met Mr. Jones in that city where they execute families with more than one girl. He has two children in school. What are the chances both are boys?"

In that one, we eliminate the possiblity that you could possibly have met Mr. Jones if he had two daughters, and thus could not possibly have this conversation with me in the case he has a daughter that we know about and that you want the odds of the second being a daughter, as well.

Jenerix525
2015-01-30, 04:53 PM
The first is why "age" entered it at all. It's technically irrelevant; it's used in the first problem to try to create a way of saying "THIS child" versus "THAT child."

So... What you're saying is that you used the context of "age" (aka first/second child) as part of your logic for the second question, when it was only provided by the first?

The problem, as presented, is asking us to calculate (by implication that the Joneses were chosen and the question based on their family):

P(Both children are boys | At least one child is a boy ^ [(there is a boy and a girl) c P(I told you at least one child is a boy rather than at least one child is a girl)=0.5]).

In natural language: "What is the probability that both children are boys, given that at least one child is a boy and that, if one child is a boy and the other is a girl, I had a .5 probability of telling you about the boy and asking about boys rather than telling you about girls and asking you about girls?"

That is, rather than having the textbook's claimed possibility set of:

BB
BG
GB
GG

We have an actual possibility set of:

From "at least one child is a boy," we reduce the possible cases to:

There are two cases out of the ones your information does not eliminate wherein there are two boys, and two cases out of that same set where they are not.

Probability = 0.5

I see no reason that the probability set cannot be

BB and I am telling you about boys "At least one is a boy"
BB and I am telling you about girls "Neither is a girl"
BG and I am telling you about boys "At least one is a boy"
BG and I am telling you about girls "At least one is a girl"
GB and I am telling you about girls "At least one is a girl"
GB and I am telling you about boys "At least one is a boy"
GG and I am telling you about girls "At least one is a girl"
GG and I am telling you about boys "Neither is a boy"

Reducing, as you did.

BB and I am telling you about boys "At least one is a boy"
BG and I am telling you about boys "At least one is a boy"
GB and I am telling you about boys "At least one is a boy"

The 'neither' statements are equivalent to 'both are the other'. Because of that, they will never be presented as a riddle, but they still exist as possibilities within the scenario.

Perhaps you didn't assume age was involved in the second question (in which case your terms first/second child are very misleading). That would mean that in addition to assuming someone else is telling you the information, you are assuming that said person is The Riddler, and would never provide complete information.

If you are, in fact, Batman then I'll stop contending the point.

Fiery Diamond
2015-01-30, 10:19 PM
"People who don't use probability theory in their ordinary lives" are primarily dead, comatose or don't have functioning brains, most regular people have to evaluate likelihoods on daily basis. The meaning of "at least one child is a boy" is not only completely independent from any particular of probability theory and would almost never align with "a specific child has been sampled and turned out to be a boy". In fact, had the questioner meant to express "a specific child has been sampled and turned out to be a boy", there are a variety of ways to express that to make it clear that this is what actually happened. And since conditional probabilities are functions on knowledge, the exact knowledge you assume makes all the difference, so a person who is familiar with it is going to be very precise with it.

Should there have been a hint? I don't think so. The context of the question was

and the reader should immediately notice the difference in phrasing and thus consider the exact meaning of that phrase. A disclaimer of "Substituting a phrase for what people might want to colloquially express with it is going to make a difference" still wouldn't have helped, as the difference was explained several times throughout the thread and people still made this mistake. The simple introspection "Does this information entail the information I intend to extract from it?" would have fixed that.

"Just because the listener's interpretation of what the speaker said is wrong does not mean the listener's interpretation of what the speaker said is wrong".

I give up.

Math and logic are acceptable boogeymen, let's blame them and not the inability of people to just think logically. They should have known that people weren't going to take what they said to be what they meant, it's all their fault!

You're making it very difficult to be civil with you. In order:

1. The vast majority of people do NOT use mathematically rigorous probability theory. They use heuristics. If you believe otherwise then I guess you've learned something today.

2. Here, you are patently wrong, and that is simply because you are carrying your own innate assumptions and asserting that they are facts of language. I'm sorry, but if you believe that the general usage of the English language follows rigorous math-like rules for how it is used, I'm ... well, I can't comprehend your lack of comprehension.

3. This is true. But - what the questioner meant to express isn't the point. I don't see why you can't understand that. The intent behind the wording is utterly irrelevant. The point is that it ISN'T unambiguously specified and can readily be interpreted in a way other than what your default is. If that is possible, then the questioner's wording is at fault.

4. Absolutely. But I don't care about the assumptions the questioner is making. I care about the assumptions the listener is making. If the listener can make different assumptions than the questioner without having what the questioner said and what the listener assumed be mutually incompatible, the fault is with the questioner, not the listener.

5. Let's rephrase that. "Just because the listener's interpretation of what the speaker said is not what the speaker meant does not mean that the listener's interpretation of what the speaker said is wrong." Yes, this is exactly what I'm saying. So long as the interpretation is a reasonable one to make based on the language used, yes, this is exactly what I'm saying. The burden of specificity and eliminating unwarranted assumptions falls SOLELY on the speaker.

Yaktan
2015-01-30, 11:07 PM
This started with a logic puzzle. You cannot reasonably complain that a logic puzzle expects you to think logically.

Straybow
2015-01-31, 12:59 AM
To get the 1/3 probability, you would have had to have established that you would have gone on to find Mr. Smith, and ask him, and Mr. Johnson, and ask him, until you found one of them who could tell you that at least one was a girl.

Selection criterion vs. observed information.
SpectralDerp is being too technical. To communicate effectively with somebody who isn't mathematically inclined and trained the language has to be translated into the vernacular.

"a specific child has been sampled and turned out to be a boy" means either:
"the older child is a boy" or "the younger child is a boy"

You're losing people by trying to determine the method of sampling (asking so-and-so). Sampling means someone chose the sample (older child or younger child) and then gave information specific to the sample. It doesn't matter how the information was gained (direct questioning, observation, magic spell, etc), only if it is specific.

If there is no sampling, then what you have is "one child is a boy." It doesn't matter how the information was gained, there is less information in the statement. The information provided is what determines the basis of probability analysis. The added information changes the probability from 1/3 to 1/2.

Cazero
2015-01-31, 05:17 AM
"a specific child has been sampled and turned out to be a boy" means either:
"the older child is a boy" or "the younger child is a boy"

You're losing people by trying to determine the method of sampling (asking so-and-so). Sampling means someone chose the sample (older child or younger child) and then gave information specific to the sample. It doesn't matter how the information was gained (direct questioning, observation, magic spell, etc), only if it is specific.

If there is no sampling, then what you have is "one child is a boy." It doesn't matter how the information was gained, there is less information in the statement. The information provided is what determines the basis of probability analysis. The added information changes the probability from 1/3 to 1/2.

When I look at it, "one child is a boy" means either "the older child is a boy" or "the younger child is a boy", exactly like the situation with sampling.
I don't see how "one child is a boy" can contain less useful information than "a specific child has been sampled and turned out to be a boy". If how the sampling was made doesn't matter, what additionnal information do I have with sampling?

Cazero
2015-01-31, 06:50 AM
Sorry for double post.
I detailed the method I used with tables and I think I solved my problem myself.

Considering the normal children gender distribution, we have the children set A, B as follows :

Child A
Child B

Girl
Girl

Girl
Boy

Boy
Girl

Boy
Boy

If we sample A or B, we can eliminate 2 rows.

Having knowledge of the gender of either A or B is equivalent to sampling, but not in that set.
We are sampling an element that is defined as A or B.
It can be represented with the theoric children set X, Y defined as follows :
-X is A if A is a boy and is B otherwise
-Y is B if A is a boy and is A otherwise
-when we are sampling a boy, we are sampling X in priority
-when we are sampling a girl, we are sampling Y in priority
Resulting in the following table :

Child X
Child Y

Girl
Girl

Boy
Girl

Boy
Girl

Boy
Boy

And sampling one element in the X,Y set only eliminate one row : the first row if we sampled a boy, and the last row if we sampled a girl.

The problem is actually all about defining the sampling method, and if said method allows to use the first table or force us to default to the second.
If we are told "at least one child is a boy", we don't have enough information to use the A,B set, and the probability to have two boys is 1/3. If we see one of the children and identify a boy, we can clearly identify the child we see as A and have enough information to use the A,B set, and the probability to have two boys is 1/2.

SpectralDerp
2015-01-31, 10:34 AM
I'm feeling kinda ill and caught myself typing "both girls are boys", assume that nonsense like that is due to bad proof-reading and trying to keep straight whether or not problem 2 is about boys or girls.

The paradox is that people think that

"one child is a boy" means either "the older child is a boy" or "the younger child is a boy"

People still make this mistake when dealing with the problem phrased in terms of random variables. Even people who have studied mathematics for several semesters often look at the problem

Let (X,S,P) be a probability space, B={0,1} and X and Y be identically distributed independent B-valued functions on X that are measurable when all subsets of B are measurable and the pushforward measure on B be the discrete uniform distribution. Calculate P(X+Y=2|X=1 or Y=1).

and then assume that [X=1 or Y=1] is equivalent to [A=1] after a relabeling of X and Y as A and B and present P(A+B=2|A=1)=1/2 as the answer. They give the wrong answer because they make a mistake, not because the problem is vague.

We only need to calculate the probability that the information is derived from inspection of the sample.
And if that probability cannot be calculated, the answer is "we don't have sufficient data to get the correct answer with certainty, however, we can say it is comprised betwen 1/2 and 1/3".
Let's get that Nobel.

We can actually make the calculation (where G is the number of girls) that

P(G=2 | we know that G>1)
= P(G=2 | we know that G>1 from inspection of the sample) * P(we know that "G>1" is derived from inspection of the sample | we know that G>1)
+ P(G=2 | we know that G>1 without having inspected the sample) * P(the information "G>1" is not derived from inspection of the sample | we know that G>1)

And then if we can calculate the terms on the right, we can calculate the term on the left. But it's possible that we can calculate the term on the left without knowing the terms on the right.
In fact, we don't know the terms on the right side, but people assume that

P(we know that "G>1" is derived from inspection of the sample | we know that G>1) = 1

and

P(the information "G>1" is not derived from inspection of the sample | we know that G>1) = 0

and get 1/2.

The existence of that page is undeniable proof that the disagreement is larger than this thread. I'm not going to tell you that you need to agree with everyone else here, but surely you can at least accept the possibility that (as greater minds than ours are apparently unable to come to an agreement) you might be wrong.

The article on Marilyn vos Savant (http://en.wikipedia.org/wiki/Marilyn_vos_Savant#Famous_columns) also mentions the disagreement that many people had with her (correct) answer to the Monty Hall problem, which was the initial problem in this thread, as well as the problem we are currently discussing, which is a possible reason for why it might have caught the attention of some posters. Also, note that the two (http://en.wikipedia.org/wiki/Boy_or_Girl_paradox) articles (http://en.wikipedia.org/wiki/Marilyn_vos_Savant#Famous_columns) don't even agree with each other. So no, while there are things I would take as an indicator that I am wrong, Wikipedia articles aren't among them.

"Just because the listener's interpretation of what the speaker said is not what the speaker meant does not mean that the listener's interpretation of what the speaker said is wrong." Yes, this is exactly what I'm saying.

And you are wrong. "The listeners interpretation of what the speaker said" is what the listener believes what the speaker meant. That's what an interpretation is. And if the listener believes "the speaker meant X", but "the speaker did not mean X", then the listener is wrong.

I can honestly say I have never edited a wikipedia article in my entire life. I don't have an account with them to do so.

And when you wrote

Of people reading that article, at least one has written in this thread. </ducks and covers>

what were you saying?

Also, you don't need an acoount to edit Wikipedia.

Pinnacle
2015-01-31, 12:36 PM
He probably meant that he had both posted in this thread and read the article. Which is what he said.
'Course, he only said that "at least one" person had done both, so he could have easily been referring to the person who posted the link. Or you.

Anyway, that's the opposite.

Fiery Diamond
2015-01-31, 08:16 PM
I'm feeling kinda ill and caught myself typing "both girls are boys", assume that nonsense like that is due to bad proof-reading and trying to keep straight whether or not problem 2 is about boys or girls.

The paradox is that people think that

People still make this mistake when dealing with the problem phrased in terms of random variables. Even people who have studied mathematics for several semesters often look at the problem

Let (X,S,P) be a probability space, B={0,1} and X and Y be identically distributed independent B-valued functions on X that are measurable when all subsets of B are measurable and the pushforward measure on B be the discrete uniform distribution. Calculate P(X+Y=2|X=1 or Y=1).

and then assume that [X=1 or Y=1] is equivalent to [A=1] after a relabeling of X and Y as A and B and present P(A+B=2|A=1)=1/2 as the answer. They give the wrong answer because they make a mistake, not because the problem is vague.

We can actually make the calculation (where G is the number of girls) that

P(G=2 | we know that G>1)
= P(G=2 | we know that G>1 from inspection of the sample) * P(we know that "G>1" is derived from inspection of the sample | we know that G>1)
+ P(G=2 | we know that G>1 without having inspected the sample) * P(the information "G>1" is not derived from inspection of the sample | we know that G>1)

And then if we can calculate the terms on the right, we can calculate the term on the left. But it's possible that we can calculate the term on the left without knowing the terms on the right.
In fact, we don't know the terms on the right side, but people assume that

P(we know that "G>1" is derived from inspection of the sample | we know that G>1) = 1

and

P(the information "G>1" is not derived from inspection of the sample | we know that G>1) = 0

and get 1/2.

The article on Marilyn vos Savant (http://en.wikipedia.org/wiki/Marilyn_vos_Savant#Famous_columns) also mentions the disagreement that many people had with her (correct) answer to the Monty Hall problem, which was the initial problem in this thread, as well as the problem we are currently discussing, which is a possible reason for why it might have caught the attention of some posters. Also, note that the two (http://en.wikipedia.org/wiki/Boy_or_Girl_paradox) articles (http://en.wikipedia.org/wiki/Marilyn_vos_Savant#Famous_columns) don't even agree with each other. So no, while there are things I would take as an indicator that I am wrong, Wikipedia articles aren't among them.

And you are wrong. "The listeners interpretation of what the speaker said" is what the listener believes what the speaker meant. That's what an interpretation is. And if the listener believes "the speaker meant X", but "the speaker did not mean X", then the listener is wrong.

And when you wrote

what were you saying?

Also, you don't need an acoount to edit Wikipedia.

Once again, what the speaker meant is utterly irrelevant. The problem is in isolation from any particular truth being conveyed. An ambiguously worded statement in this circumstance HAS NO SINGLE CORRECT INTERPRETATION. The burden of specificity and eliminating plausible other interpretations lies on the speaker. Look back at that "color of the Sun" analogy. You sound EXACTLY like the questioner in that analogy.

SpectralDerp
2015-02-01, 03:45 AM
He probably meant that he had both posted in this thread and read the article. Which is what he said.
'Course, he only said that "at least one" person had done both, so he could have easily been referring to the person who posted the link. Or you.

Anyway, that's the opposite.

and then "of the people reading that article" became "of the people writing that article" in my mind.

Sorry, my mistake.

An ambiguously worded statement in this circumstance HAS NO SINGLE CORRECT INTERPRETATION.

To correctly interpret a statement is to arrive at a belief about what the speaker/author meant to express with that statement and for that belief to be true. Therefore, there is a single correct interpretation, namely the one where the interpreter comes to believe "the speaker/author meant X", where is X is what the speaker/author meant.

And the statement "at least one child is a boy" is not ambiguously worded. And people think it means "Child X is a boy" for a particular child X because they think it unambiguously says that. This is what several people in this thread have described their reasoning as. People make the same mistake when the problem is being posed in terms of random variables, where no colloquialisms exist.

Fiery Diamond
2015-02-02, 12:36 AM

and then "of the people reading that article" became "of the people writing that article" in my mind.

Sorry, my mistake.

To correctly interpret a statement is to arrive at a belief about what the speaker/author meant to express with that statement and for that belief to be true. Therefore, there is a single correct interpretation, namely the one where the interpreter comes to believe "the speaker/author meant X", where is X is what the speaker/author meant.

And the statement "at least one child is a boy" is not ambiguously worded. And people think it means "Child X is a boy" for a particular child X because they think it unambiguously says that. This is what several people in this thread have described their reasoning as. People make the same mistake when the problem is being posed in terms of random variables, where no colloquialisms exist.

Do you understand what "ambiguous" means? It means that there is more than one interpretation of the words for which the words are true. We are arguing that that's why there is no single correct interpretation. I can't speak for those who have trouble when using variables (I don't, since I at least understand very basic probability), but for those who trip up with the wording... (oh, and you insist it unambiguously does not mean for a particular child. Both you and that hypothetical other person are wrong. It could mean either, which is why it's ambiguous.)

Let me see if I can explain why it is ambiguous in normal parlance in terms that will make it clear.

We have a set which I will call Set A.
Set A is composed of Set A' and Set A''.
Set A'' is further composed of individuals such as A''1 and A''2.

Would you agree that when I say: "I have a case which fits Set A" that I'm being ambiguous about whether this case fits Set A' or Set A'' by not specifying?

This is precisely the case for our example, and the reason you're so resistant to the idea is that in the lingo of probability, the words used for Set A and Set A' are identical, but Set A'' must be specified if intended. However, to those who speak the normal parlance rather than the lingo of probability, Set A doesn't automatically mean Set A', and in fact I would think that most people unfamiliar with probability lingo would default to assuming Set A''. In order to be unambiguous when asking the question of those people, you must find some way of specifying Set A' rather than simply Set A (which really isn't hard to do). Now, there are surely many people who don't understand conditional probability and will still get it wrong because they don't know why there's a difference between Set A' and Set A'', but that doesn't change the fact that the wording is ambiguous.

For the record:

Set A: At least one kid is a girl
Set A': "At least one kid is a girl" is a criterion; we did not arrive at this knowledge by sampling.
Set A'': We know "at least one kid is a girl" because we sampled one.
Examples of items in A'':
A''1: The older child is a girl
A''2: The child I saw is a girl
etc.

Regardless of how much you want to deny it, this IS how the problem wording is able to correctly be interpreted in common parlance, people who don't understand conditional probability not realizing there is a difference between A' and A'' notwithstanding.

Just because the speaker speaks the lingo of probability and therefore clearly means Set A' when he says "at least one kid is a girl" does not make it the only meaning of "at least one kid is a girl." His intentions are irrelevant to the ambiguity of the statement.

SpectralDerp
2015-02-02, 02:14 AM
Would you agree that when I say: "I have a case which fits Set A" that I'm being ambiguous about whether this case fits Set A' or Set A'' by not specifying?

No, I don't agree. Otherwise, the statement "the older child is a girl" would also be ambiguous, because it doesn't specify "the older child is a girl and the younger child is a girl" or "the older child is a girl and the younger child is a boy".

The correct thing to do when being told A and asked to calculate the probability of X is to calculate P(X|A). If A is the disjoint union of A' or A'', then maybe that helps me calculate P(X|A) because I can calculate all the terms in P(X|A') * P(A') + P(X|A'') * P(A''), which is identical to P(X|A).

Segev
2015-02-02, 11:54 AM
This started with a logic puzzle. You cannot reasonably complain that a logic puzzle expects you to think logically.And that's not what I'm "complaining" about. I am objecting to the claim that the problem, as presented, is unambiguously equivalent to, "Of all 2-child families wherein at least one child is a girl, the Joneses are one such family. What is the probability that both are girls?"

That is one possible way to read it, but it is not the only one, and it is certainly not the way that everybody will interpret it even after rigorous analysis of the semantics.

This is shown by both this thread and by the wikipedia article, wherein it is revealed that this argument is older than any of us here thought and that the man credited with the original formulation agreed, after discussion with objectors, that it was poorly worded.

To reiterate, the implication I took from it, which I hold is as valid as the one intended by the "textbook" answer, is, "I have selected the Jonses out of all possible two-child families. Having done so, I can truthfully tell you that at least one of their two children is a girl. What is the probability that both of their children are girls?"

As demonstrated repeatedly in this thread, that version of the experiment yields a 1/2 probability (all else being equal), because it includes the possibilities that the question-asker would be asking a different question if both children were boys, and has an even chance of asking a different question if at least one of the Joneses' children is a boy.

If we reject the idea that there is any way to assume a specific chance that the asker would choose to ask about boys or girls in the two cases where one is a boy and the other is a girl, the answer becomes, "There is not enough information to calculate that probability."

Even if you assume that it is utterly ludicrous to assume a specific probability regarding the information given in the cases where either kind of information could be given, that leaves the question open as to whether you COULD have asked us, in the cases BG, GB, or BB, about boys, what the odds were that you would have asked us what you did. Since we cannot know from how the question was asked if you could possibly have asked us about boys, because there's no indication as to whether you would have been capable of asking us about the Joneses if they had two boys, the only correct answer remains "There is not enough information."

If you allow enough assumption power to assume "the person asking this question obviously selected the Joneses because they fit the criterion of 'at least one girl,'" then you must also allow enough assumption power to assume "the person asking this question obviously, in the cases of one boy and one girl, must have chosen arbitrarily and, all else being equal, even odds, which to ask us about, and would have asked us about boys if both were boys."

That sounds like a mouthful, and it is, in natural language, because it's so basic an assumption that we rarely need to put it into words.

We can actually make the calculation (where G is the number of girls) that

P(G=2 | we know that G>1)
= P(G=2 | we know that G>1 from inspection of the sample) * P(we know that "G>1" is derived from inspection of the sample | we know that G>1)
+ P(G=2 | we know that G>1 without having inspected the sample) * P(the information "G>1" is not derived from inspection of the sample | we know that G>1)I will assume here that each instance of "G>1" is really "G>=1" since "G>1" is not information we're given. (We know it is 1 or 2, and not 0.)

And then if we can calculate the terms on the right, we can calculate the term on the left. But it's possible that we can calculate the term on the left without knowing the terms on the right. Technically, the standard assumption when dealing with probability notation would be the assumption you want us to apply to the problem given. If you assume, therefore, that "we can solve the left hand side of the equation without solving the right," then you are assuming that we're using the standard assumptions, which is begging the question (i.e. assuming the point you're trying to make as part of the proof that your point is true).

For the equation you've written to be a meaningful breakdown, you actually have to abandon that standard assumption. Otherwise, you're a priori declaring P(we know that "G>=1" is derived from inspection of the sample)=0 and P(we know that "G>=1" is not derived from inspection of the sample)=1.

Since, to make the point that we must assume that the problem as presented can only have arisen if "G>=1" was not derived from inspection of the sample, you must not assume a priori those two probabilities, we must abandon the assumption that we can give P(G=2 | we know G>=1) based on knowledge that this notation directly implies that we have not inspected the sample.

In other words, having moved from natural language to mathematical notation, it is incumbant upon us to be certain that the assumptions of the mathematical notation do really reflect the assumptions we're allowed to make from the natural language. Your right-hand side of the equation is actually the correct mathematical notation for the natural language problem with which we're presented. That the left-hand side, with its attendent assumptions, is equivalent is what you must prove. You cannot, theref

In fact, we don't know the terms on the right side, but people assume that

P(we know that "G>1" is derived from inspection of the sample | we know that G>1) = 1

and

P(the information "G>1" is not derived from inspection of the sample | we know that G>1) = 0

and get 1/2.That is the natural-language interpretation of the problem, actually. Because, in natural language, the assumption is that the first bit of information you're given (sans qualifiers) is given without criteria for its acceptance.

"The Joneses have 2 children."

That's the first thing we're told as the problem is presented. There is no indication given that the Joneses were in any way pre-screened for more than being a family with 2 children. Heck, for all we know, they weren't screened at all, and you'd have told us about them if they had a different number of children.

"At least one child is a girl."

This second piece of information is not presented as a criterion as to why the Joneses were selected. Absent that presentation as criterion, it is natural (by natural language convention) to assume that it is obtained by observing the children.

So yes, people assume

P(we know that "G>1" is derived from inspection of the sample | we know that G>1) = 1

and

P(the information "G>1" is not derived from inspection of the sample | we know that G>1) = 0

and get 1/2.

This is because that's the natural language interpretation.

However, let's see what happens if we assign other values to those two probabilities. For simplicity, let's look at both being 1/2. This would be the assumption that we don't know whether it was determined by inspection or not, and that all else being equal, there's a 50% chance it was a criterion vs. being information gleaned by examination.

That gives us 4 cases from the first interpretation and 3 from the second, for a total of 7. Each of those first 4 cases has a probability of 1/8 (1/2 * 1/4), and the second 3 cases has a probability of 1/6 (1/2 * 1/3).

In two of the first 4 cases, both children are girls: 1/4
In one of the second 3 cases, both children are girls: 1/6

1/4 + 1/6 = 3/12 + 2/12 = 5/12 chance that both are girls.

Which is neither 1/2 nor 1/3.

So, it depends on what probability we assign to the chances that the information is gleaned by inspection or by criterion. I would venture that it likely is a sliding scale from 1/3 to 1/2, depending on the probabilities assigned to each of them.

Once again, natural language assumptions assign the probabiliy one way and those who have the specific lesson in mind assign it the other. Because natural language is actively misleading, rather than merely allowing people to make poor guesses despite having complete knowledge, it is a poorly-worded

The article on Marilyn vos Savant (http://en.wikipedia.org/wiki/Marilyn_vos_Savant#Famous_columns) also mentions the disagreement that many people had with her (correct) answer to the Monty Hall problem, which was the initial problem in this thread, as well as the problem we are currently discussing, which is a possible reason for why it might have caught the attention of some posters. Also, note that the two (http://en.wikipedia.org/wiki/Boy_or_Girl_paradox) articles (http://en.wikipedia.org/wiki/Marilyn_vos_Savant#Famous_columns) don't even agree with each other. So no, while there are things I would take as an indicator that I am wrong, Wikipedia articles aren't among them.Given that the point being argued is that there is disagreement and that there is no clear answer based on the problem as written, I think you've conceded it here.

If you're claiming that people make mistaken assumptions about probability, and don't innately understand conditional probabilities, then nobody is arguing with you.

If you're claiming that "The Joneses are the first 2-child family we found wherein at least one is a girl; what is the probability that both chldren are girls?" has a correct answer of 1/3, nobody is arguing with you.

It's only if you're claiming that, "The Joneses have 2 children. At least one is a girl. What is the probability that both children are girls?" is unambiguously correctly answered "1/3," and that nobody who understands conditional probability could possibly come up with any other answer, then you're wrong.

(If, on the other hand, you're claiming that nobody who understands conditional probability and already knows that this problem is designed to demonstrate that others don't understand it could fail to reason out that the correct interpretation is counter to natural language, you've got more of a point. But that's getting into "no true scottsman" territory: It says that only those who already know that this problem is a trick question AND the assumptions that are intended despite being opposite those given by natural language interpretation understand conditional probability. Since knowing this specific problem, in this formulation, is not required to understand conditional probability, it is a fallacious claim.)

And you are wrong. "The listeners interpretation of what the speaker said" is what the listener believes what the speaker meant. That's what an interpretation is. And if the listener believes "the speaker meant X", but "the speaker did not mean X", then the listener is wrong.By that logic, "I am the President of the United States," being interpreted by a listener to mean that I am claiming to be the head of the Executive Branch of the nation known as the United States of America is the fault of the listener if what I meant was, "I am at least 35 years old." It doesn't matter that what I meant is not actually contained within the statement I made; failure to understand me is entirely the listener's fault.

Note, I am making this statement solely about what you wrote. You made no exception for the speaker being misleading or even outright nonsensical. Even if you hedge it by saying "well, the speaker has to have an understanding of what he's saying," I can counter by pointing out that it is required by law that the POTUS be at least 35, so the speaker could honestly be making the (particularly stupid, in this example) blackbird/crow fallacy. "If all Presidents of the United States must be at least 35, then all who are at least 35 must be Presidents of the United States."

It doesn't matter how stupidly flawed that logic is; by your logic, it is entirely the listener's fault that he misunderstood what the speaker meant.

He probably meant that he had both posted in this thread and read the article. Which is what he said.
'Course, he only said that "at least one" person had done both, so he could have easily been referring to the person who posted the link. Or you.

Anyway, that's the opposite.

and then "of the people reading that article" became "of the people writing that article" in my mind.

Sorry, my mistake.Fair enough. Now, for purposes of this discussion, is there ambiguity in my statement that could have led to that incorrect interpretation of my meaning, or was it that you genuinely misread one word as another and thus arrived at a meaning that could not logically be derived, based on natural language reading, from what I wrote?

To correctly interpret a statement is to arrive at a belief about what the speaker/author meant to express with that statement and for that belief to be true. Therefore, there is a single correct interpretation, namely the one where the interpreter comes to believe "the speaker/author meant X", where is X is what the speaker/author meant.Again, does that mean that it doesn't matter what the speaker/author says, nor the meaning of the words he used; it is entirely the fault of the listener if he does not interpret it correctly?

And the statement "at least one child is a boy" is not ambiguously worded.No, but the entire context of the problem makes it ambiguous whether that information is derived by inspection, or is information about a criterion.

Otherwise, by what you're saying, the two-coin problem I presented a while back would have different results based solely on whether I said, "At least one coin is Heads," versus, "The blue coin is Heads."

And people think it means "Child X is a boy" for a particular child X because they think it unambiguously says that.If it is derived by inspection, it does say that. No, really, it does.

Again, take the two-coin problem. If you have inspected one or both coins, and tell me (truthfully), "At least one is heads," that tells me that there are two coins, each with a 1/2 probability of being heads, and that you have given me information about one of them. This means there is "a coin about which I know the face which is up," and "a coin about which I do not know this information."

Look at the Joneses again. Let's say that you did select the Joneses specifically because at least one of their two children is a girl. Let's say you tell me that. Then you go on to tell me, "In this case, their older child is a girl."

You now have told me three things: The Joneses are a family selected because at least one child is a girl, and that the older child is a girl, and that that second piece of information is "in this case," which means it's not something you would necessarily have told me if it weren't true and is not something you would have chosen a family specifically that this was true about.

The probability that both are girls remains 1/3, because the fact that, by inspection, you've determined that the older child is a girl is irrelevant.

This is what several people in this thread have described their reasoning as. People make the same mistake when the problem is being posed in terms of random variables, where no colloquialisms exist.Nobody is arguing here that people don't make mistakes about conditional probability.

What is being argued is that the problem as presented does not illustrate this misunderstanding, at least not accurately. It clouds the illustration by presenting a question which, interpreted via natural language, could lead somebody who understands conditional probability and does not know the answer already by virtue of being "in" on the unstated assumptions to still get it wrong.

I have repeatedly offered phrasings which unambiguously present the information and yet which could still lead to those who do not understand conditional probability making the desired mistake.

SpectralDerp
2015-02-02, 02:22 PM
I am objecting to the claim that the problem, as presented, is unambiguously equivalent to, "Of all 2-child families wherein at least one child is a girl, the Joneses are one such family. What is the probability that both are girls?"

Are these question also ambiguously worded?

"The Jones are a two child family, what's the probability that both children are girls?"

"The Jones are a two child family, what's the probability that at least one child is a girl?"

Segev
2015-02-02, 02:34 PM
Are these question also ambiguously worded?

"The Jones are a two child family, what's the probability that both children are girls?"

"The Jones are a two child family, what's the probability that at least one child is a girl?"

Natural language interpretations incoming. You tell me if I get what you meant by them.

"The Jones are a two child family, what's the probability that both children are girls?"

This one tells me that there is a two-child family. There do not seem to be any criteria restricting how you came by them, and it doesn't matter whether "two-child" is a restriction or an observation for purposes of determining this probability. Given that you have told me nothing about how you selected them, I am left to assume that you chose them at random from the population of all 2-child families.

It would seem therefore that they have the BB/BG/GB/GG possibilities, and therefore there is a 1/4 chance that both children are girls.

"The Jones are a two child family, what's the probability that at least one child is a girl?"

Here, we again are presented with a single sample of which we are given no information on restrictive criteria other than, perhaps, "two-child." As before, it wouldn't actually matter if "two-child" was a selection criterion or an observed fact.

We are still given no information beyond this, which means we must assume, since they come from the population of all possible 2-child families, that they again have the following possible arrangements with equal probability: BB/BG/GB/GG.

Therefore, we have a 3/4 probability of at least one child being a girl.

It is worth noting that, if you selected the Joneses by rejecting any families which did not have at least one girl, but did not tell me that, you could still ask both questions as you presented them. However, the answer to the first question would become "1/3," and to the second "1."

It would, however, be unreasonable for me to assume that to be the case from how the questions were presented.

SpectralDerp
2015-02-02, 03:53 PM
And what's (1/4)/(3/4)?

Because that's how conditional probability works. I do not need to assume that "What's the probability of A, given B?" means "What's the probability of A, given that I made a selection such that B for all possible options", any valid interpretation of the first question simply has to give the same answer as the second one. I am not working with a convention that they mean the same, there is no standard assumption that they mean the same, it's simply a provable fact about different models for sensible interpretations of probability and conditional probability that any interpretation of the first question has to give the same answer as the second one. They have to satisfy P(A|B) = P(A and B) / P(B). If your interpretation doesn't agree, then it goes contrary to how probability works.

Segev
2015-02-02, 04:20 PM
And what's (1/4)/(3/4)?

Where are you getting (1/4)/(3/4)?

You will have to demonstrate, to prove the point you seem to be trying to prove (that the two questions above, conflated, yield the 1/3 result and therefore the original formulation is unambiguous), that you can combine the two statements in a way that remains equally unambiguous and that this division is the correct mathematical representation of the combination of the two questions.

I think the hardest obstacle will be proving that the two questions combine as: "The Jones are a two child family, at least one of which is a girl; what is the probability that both are girls?"

In fact, to anticipate you (and hopefully avoid arguing in circles)...

"The Jones are a two-child family; what is the probability that both are girls?"

"The Jones are a two-child family; what is the probability that at least one are girls?"

"The Jones are a two-child family, at least one of which is a girl; what is the probability that both are girls?"

The opening statement to all three is the same: "The Jones are a two-child family."

All three, in natural language, begin with the assumption that this sample is presented without selection bias. Barring further statement of selection criteria, therefore, the possbile combinations of children the Joneses have is: BB, BG, GB, GG.

The first question then presents no further information, and asks straight-forwardly what the probability of both being girls is. From the above analysis of equally-probable combinations, exactly 1 of the 4 is "both girls," which gives us "1/4."

The second question also provides us no further information, but asks us something different. It asks us what the probability is of at least one girl. This, of course, gives us "3/4."

You have now asked, "What is (1/4)/(3/4)?" with the obvious intention of having the answer (1/3) being the answer to the third question.

Thus, you are saying that the first two combine to make the third in a way that (answer to first)/(answer to second) = (answer to third).

Given your penchant for switching to mathematical notation without analyzing whether it accurately reflects the natural language of the word problem, I will examine what it looks like in mathematical notation.

First question is asking, "What is P(both children are girls|two-child family selected from all possible two-child families)?"

Second question is asking, "What is P(at least one child is a girl|two-child familiy selected from all possible two-child families)?"

From your question in the post I quoted at the start of this one, I assume you are asserting that the third question is asking:

"What is [P(both children are girls|two-child family selected from all possible two-child families)/P(at least one child is a girl|two-child family selected from all possible two-child families)]?"

We now can convert "givens" to division:

[P(both children are girls)/P(two-child family selected from all posible two-child famlies)]/[P(at least one child is a girl)/P(two-child family selected from all possible two-child families)]

The two "P(two-child family selected from all possible two-child families)" cancel, leaving us with:

P(both children are girls|at least one child is a girl)

Your claim is that "What is the probability that both children are girls, given that at least one child is a girl?" is equivalent to "The Jones have two children, at least one of which is a girl; what is the probability that both are girls?"

However, there is a crucial problem in your claim: There are assumptions implied in the first which are not in the second.

Specifically, the assumption is that you know that "given" means "the sample is selected such that," rather than "it was determined by inspection."

The mathematical notation inherently makes that assumption. The phrasing based on it also makes that assumption.

This still goes back to the two-coin game: I walk up to you and tell you, "I have two fair coins in this box which were flipped. At least one of them is Heads." I then ask you, "Are they both Heads?"

I have not told you whether I flipped them until I had at least one Head, or if I just flipped them and looked through a window into in the box, and formulated my question based on having seen one coin that was Heads.

Do you have 1/2 or 2/3 chance of being right if you guess "no?"

If you are speaking in the language of probability and actually say, "What is the probability that the two coins I have in this box are both heads-side-up given that at least one is heads?" then you are implying, given the context of the language of probability, that you have not examined them after the fact, but determined a priori that you would flip them until you got at least one Head.

I offer you a possible way to prove me wrong, however, since there is something bugging me about the above analysis.

My claim is that the natural langauge way to read the original problem is: "The Joneses have 2 children. I have seen at least one of them and identified that she is a girl. If I had seen a boy and a girl, I might have told you about the boy, instead, with 50% probability. If I had seen both of them and they were both boys, I would definitely be telling you that at least one was a boy. If I have seen them both, and they are both girls, I would be telling you the same thing I am telling you now. If I had told you that at least one was a boy, I would be asking you the probability that both are boys. What is the probability that both are girls?"

Yes, that's quite a bit wordier than the original problem, but natural language carries a lot of implication and assumption when used. We usually don't think about it.

For reference, the way you ask us to interpret it (and the assumptions carried in the language of probability are): "The Joneses have two children. At least one is a girl. Because we are discussing this in the language of probability, you are to assume that this is a criterion for the Joneses' selection, and not that I have inspected the children in any way. There is no chance that I would have told you about boys nor asked the probability that both children were boys. What is the probability that both children are girls?"

The above translates to "What is P(both of two children are girls|at least one child is a girl)?"

How would you translate the version I spelled out as the natural language reading of the original problem into mathematical notation?

Segev
2015-02-02, 04:23 PM
And what's (1/4)/(3/4)?

Because that's how conditional probability works. I do not need to assume that "What's the probability of A, given B?" means "What's the probability of A, given that I made a selection such that B for all possible options", any valid interpretation of the first question simply has to give the same answer as the second one. I am not working with a convention that they mean the same, there is no standard assumption that they mean the same, it's simply a provable fact about different models for sensible interpretations of probability and conditional probability that any interpretation of the first question has to give the same answer as the second one. They have to satisfy P(A|B) = P(A and B) / P(B). If your interpretation doesn't agree, then it goes contrary to how probability works.

Except that, if all of that is true, then when I walk up to you with my two-coin game, without telling you how I decided what information to give you nor what question to ask, you should be able to "win" 2/3 of the time by answering "no."

But we've demonstrated that, because I choose what question to ask after inspecting the coins and randomize which I'll ask about if they're heterogenous, your odds are only 1/2.

Clearly, there is something wrong with making the claim that you need assume nothing. What in the Jones question is different than me walking up to you with my two coins in a box, from the information you are given? (You are not told how I came by the knowledge that at least one is Heads, nor whether I would ever ask you about Tails if either of the coins came up Tails.)

Cazero
2015-02-02, 04:30 PM
And what's (1/4)/(3/4)?

Because that's how conditional probability works. I do not need to assume that "What's the probability of A, given B?" means "What's the probability of A, given that I made a selection such that B for all possible options", any valid interpretation of the first question simply has to give the same answer as the second one.

The problem here is not the interpretation of the question. Everybody understands the question correctly.
The problem is the interpretation of B. There are two interpretations of B, one with P(B)=3/4 and the other with P(B)=1/2.
If the wording is ambiguous, both can be considered valid, and we have a problem that doesn't lie in the maths. At this point, you can stop ranting about the way conditional probability works.

This thread sounds suspiciously similar to RAW vs RAI threads. I find it fitting, yet disturbingly scary.

Segev
2015-02-02, 04:40 PM
Because that's how conditional probability works. I do not need to assume that "What's the probability of A, given B?" means "What's the probability of A, given that I made a selection such that B for all possible options", any valid interpretation of the first question simply has to give the same answer as the second one.

Actually... looking this over: that's the assumption you're making. The language of probability is such that the assumption is always made taht "given" means "for all possible options, and that I would never have told you the alternative because I eliminated the other options."

Again, that's presuming the conclusion, i.e. begging the question: "Because this assumption is always made, I do not need to make an assumption, and anybody who doesn't make that assumption doesn't get how probability works." The correct thing to say would be, "anybody who doesn't make that assumption is not speaking the language agreed upon by the zeitgeist of those who discuss probability."

Still, I confess that your example niggles at me for the inconsistency. It seems to cancel out any reason to care how the samples were selected, or from what population. And yet, the 2-coin game reveals that it does, in fact, matter.

As I type this, I am driven to recall that, however, we've hidden again our assumption: we've assumed that the "given" is a criterion, not an observation.

In fact, in my own mathematical analysis to try to see where your (1/4)/(3/4) came from, I forgot that we left out the probability that we would ask about something other than girls entirely.

And that's the problem with it: The natural language sentense that says "and at least one child is a girl" does not equate as obviously as it seems to with "given that at least one child is a girl" based on the two probabilty questions leading up to it.

Natural language shifts its contextual meaning when you introduce the Joneses and then give a piece of information about them.

The moment you do that, you are implying (whether you mean to or not) that this is information gleaned by observation. The natural language in which a criterion is given is different.

And I ramble because this kind of thing DOES bug me. There shouldn't be a distinction. But there is. If there weren't, then it wouldn't be equally valid for me to approach you after having looked at the Joneses, found out what the sex of both children was, chosen which sex to tell and ask you about based on that, and then come and given you exactly the same information and question as given in the original problem.

And you'd have a 2/3 chance of getting it right by saying "no," when we've demonstrated that that is not true.

And yet, I could 100% truthfully, with no intention of misleading, tell you that information and ask you that question.

Segev
2015-02-02, 04:47 PM
The problem here is not the interpretation of the question. Everybody understands the question correctly.
The problem is the interpretation of B. There are two interpretations of B, one with P(B)=3/4 and the other with P(B)=1/2.
If the wording is ambiguous, both can be considered valid, and we have a problem that doesn't lie in the maths. At this point, you can stop ranting about the way conditional probability works.

This thread sounds suspiciously similar to RAW vs RAI threads. I find it fitting, yet disturbingly scary.

Actually, as evidenced by my rambling, her posing of it in this form has me puzzled again. I recognize something is off, but her line of reasoning is compelling.

Can you identify where P(B) is ambiguous?

Her phrasing of the question about P(B) is, "The Jones have two children; what is the probability that at least one is a girl?"

hm....

"The Joneses have two children; what is the probability that, if you observe one of them, it will be a boy?"

"The Joneses have two children; what is the probability that, if you observe one of them, it will be a girl?"

Clearly, the answer to both of these cannot be "2/3." And it comes about from the nature of BB/GB/GB/GG - in the middle two cases, you have actually "GB, and I saw the boy/GB, and I saw the girl/BG, and I saw the boy/BG, and I saw the girl," which skews things.

Ahah! And that's where the ambiguity is. The natural language reading of the original problem implies that you're asking what the odds are that both are girls given that the inspector saw a girl upon examination of the family. Replace "and he saw the [boy/girl]" with "and he's choosing to tell me about the [boy/girl]." Whether he saw both or not, he's choosing which to tell us about. (Possibly a forced choice, if he saw only one; in which case which he saw is random.)

Cazero
2015-02-02, 05:04 PM
Can you identify where P(B) is ambiguous?

Sure thing.
Usual question, blah blah, at least one child is a girl. This is B.

P(B) is either :
Given that I observed one child and saw a girl, what was the probability for that event to happen?
or
Given that I observed both children and saw at least one girl, what was the probability for that event to happen?

The first is 1/2, the second is 3/4. Both allow you to claim that at least one child is a girl.

SpectralDerp
2015-02-02, 05:12 PM
Except that, if all of that is true, then when I walk up to you with my two-coin game, without telling you how I decided what information to give you nor what question to ask, you should be able to "win" 2/3 of the time by answering "no."

That's because the two-coin game translates into a scenario where more information is "given" than "at least one coin is [whatever]". The "given" information is "the result of the coin of which I inform you about is [whatever]", it specifies a coin, as opposed to "at least one coin is [whatever]", which does not.

Actually... looking this over: that's the assumption you're making. The language of probability is such that the assumption is always made that "given" means "for all possible options, and that I would never have told you the alternative because I eliminated the other options."

Actually, no. The assumption I am making is that probabilities and conditional probabilities have the property that "the probability of A given B" and "the probability of B" determine "the probability of (A and B)". Even the assumption that anyone is telling anything to anyone is not made.

The probability of a proposition can be interpreted to be a numerical value for the plausibility of that proposition. If someone is given the information "at least one child is a girl", that information elevates the plausibility of "Both children are girls" from 1/4 to 1/3. It doesn't matter if the person giving this information could give more information, what's important is the information that has actually been given.

All of this can be translated into scenarios involving games, just not the way you'd like it to.

Talakeal
2015-02-02, 08:46 PM
Sorry guys, I am still not getting the Green Eyed Dragons puzzle. I have read through several explanations, including those posted here, and while I more or less understand the logical elimination puzzle, I still don't see why your announcement makes it happen.

Unless the dragons never assume anything, they would have long since understood that everyone on the island new there were others on the island with green eyes.

On the other hand, if they never assume anything, how can they be absolutely certain that everyone on the island heard and understood the visitor's message, or that the visitor was completely correct and truthful?

The Random NPC
2015-02-02, 09:06 PM
Sorry guys, I am still not getting the Green Eyed Dragons puzzle. I have read through several explanations, including those posted here, and while I more or less understand the logical elimination puzzle, I still don't see why your announcement makes it happen.

Unless the dragons never assume anything, they would have long since understood that everyone on the island new there were others on the island with green eyes.

On the other hand, if they never assume anything, how can they be absolutely certain that everyone on the island heard and understood the visitor's message, or that the visitor was completely correct and truthful?

Every dragon starts off assuming that they don't have green eyes. The dragons only turn into sparrows when they find out they have green eyes. When you say at least one has green eyes, they all expect the other dragons to turn into sparrows, and when they don't, they realize that they have green eyes.

Talakeal
2015-02-02, 09:28 PM
Every dragon starts off assuming that they don't have green eyes. The dragons only turn into sparrows when they find out they have green eyes. When you say at least one has green eyes, they all expect the other dragons to turn into sparrows, and when they don't, they realize that they have green eyes.

I get that part. I am just not sure why telling them something they already know brings this about.

Yaktan
2015-02-02, 10:03 PM
Where are you getting (1/4)/(3/4)?

It is from Bayes' theorem, the heart of conditional probability. In our case, the 1/4 is the probability of both children being girls. 3/4 is the probability of one child being a girl. Bayes' theorem tells us that to find the conditional probability that both children are girls given that one child is a girl is 1/4 divided by 3/4. Since disputing Bayes' theorem is not going to work, to dispute the 1/3 probability you need to dispute that there is a 3/4 chance of one child being a girl. Though, this does seem to be what people are doing in a round about way.

Fiery Diamond
2015-02-03, 01:10 AM
No, I don't agree. Otherwise, the statement "the older child is a girl" would also be ambiguous, because it doesn't specify "the older child is a girl and the younger child is a girl" or "the older child is a girl and the younger child is a boy".

The correct thing to do when being told A and asked to calculate the probability of X is to calculate P(X|A). If A is the disjoint union of A' or A'', then maybe that helps me calculate P(X|A) because I can calculate all the terms in P(X|A') * P(A') + P(X|A'') * P(A''), which is identical to P(X|A).

Wow. There is so much wrong here I'm not sure if you're being serious. Take a step back, and forget about probability. This conversation hasn't actually been about probability for a long time - we all agree on the probabilities involved for any particular mathematical rendering. We're talking about language, and what a common parlance, or natural language as Segev is saying, interpretation of the language used is.

Firstly, as I said, divorce my statement from any talk of probability and you'll see how incorrect saying my statement about sets is unambiguous is.

Set A contains Set A' and Set A''. Is "this case is a Set A case" ambiguous as to whether I'm referring to Set A' or Set A''? Indisputably. Let's give some examples.

Joe: I bought a game yesterday.
Sally: Oh? What console was it for?
Joe: Huh? Don't be ridiculous, I meant that I bought a board game. If I'd meant a video game I would have said it.
Sally: I see. That's pretty ambiguous, though. It's just as reasonable for me to assume you meant a video game as a board game. In fact, people don't really buy board games that much any more, do they? So it's actually more reasonable to think you meant video game.
Joe: Don't be stupid. Clearly the only reasonable assumption is that I meant board game and your interpretation is clearly invalid. In fact, it's not even an assumption to think I meant board game. Just ask my friends; they know I don't play video games!
Sally: Maybe, and that's good for them. I didn't know that, though, and you can't have expected me to assume the same thing you did.
Joe: It's not an assumption, it's what it means! How can you even keep arguing this?

Joe: I completed a puzzle last night.
Sally: Oh? What was it a picture of?
Joe: What? Ugh, not this again. I said PUZZLE, not JIGSAW PUZZLE! What is with you making all this wrong assumptions instead of correctly understanding I was talking about a crossword puzzle!
Sally: Seriously?! Jigsaw puzzle is the default when you say puzzle unless you otherwise specify!
Joe: No, and only someone who doesn't understand puzzles would say that!

(Hint: you sound like Joe. In fact, you agree with Joe by claiming that Joe's statements were not ambiguous.)

In those examples:

1) Set A: Game
Set A': Board game
Set A'': Video game

2) Set A: Puzzle
Set A': Crossword Puzzle
Set A'': Jigsaw Puzzle

Also, what I've put in bold is where Segev and I are disagreeing with you. That's what the language means within the realm of those who speak the language of probability. That's NOT the only meaning in natural language. Being told "A is true, what's the probability of X" does NOT yield P(X|A) as the only interpretation in natural language! "A is true" does not only mean "|A"! That's not the only meaning in normal parlance! Why is this so hard to understand?

Now, in light of this...
Set A: "at least one child is a girl"
Set A': "|at least one child is a girl" or, in natural language "this family meets the criterion that at least on child is a girl, and that's how I chose this family"
Set A'': "a child has been sampled and is a girl; if both children have been sampled I randomly chose which sample to give you"

Also, on the topic of assumptions and people refusing to believe that words can have meanings other than what they insist is the only correct meaning...

Joe: What color is the sun, guys?
Alice the astronomy fan: White.
George the average dude: Yellow.
Henrietta the time-conscious: *checks sunset sky* Red.
Tracy the artist: Depends on the time of day. Yellow, Orange, or Red.
Sally the astronomy fan: Not enough info; you're not getting me this time Joe. After all, "the Sun" might be referring to something other than Sol and I'm not falling for that again.

NONE of the answerers here are wrong. They are answering different questions based on how they interpreted the question.

Alice's assumption: the color of a luminous object is the wavelengths of visible light it emits; the Sun means Sol
George: the color we perceive is how the color of an object is defined; the Sun means Sol, the color refers to the most frequently observed color
Henrietta: the color we perceive is how the color of an object is defined; the Sun means Sol, the color refers to the current perceived color
Tracy: the color we perceive is how the color of an object is defined; the Sun means Sol, the color refers to any of the colors we typically see
Sally's assumption: the color of a luminous object is the wavelengths of visible light it emits

Joe insisting that Alice's assumptions are the only valid ones makes him wrong.

SpectralDerp
2015-02-03, 01:20 AM
Set A contains Set A' and Set A''. Is "this case is a Set A case" ambiguous as to whether I'm referring to Set A' or Set A''? Indisputably.

You are unambiguously refering to set A, you are unambiguously not giving specific information about A' or A''. And this is what matters.

It is from Bayes' theorem, the heart of conditional probability. In our case, the 1/4 is the probability of both children being girls. 3/4 is the probability of one child being a girl. Bayes' theorem tells us that to find the conditional probability that both children are girls given that one child is a girl is 1/4 divided by 3/4. Since disputing Bayes' theorem is not going to work, to dispute the 1/3 probability you need to dispute that there is a 3/4 chance of one child being a girl. Though, this does seem to be what people are doing in a round about way.

That's actually not Bayes' theorem. Using Kolmogorov axioms for probability, P(A|B) = P(A and B) / P(B) is the definition of conditional probability. But it can be derived assuming P(A|B) and P(B) are sufficient to calculate P(A and B). So the issue here is a lot more fundamental than something like Bayes' theorem, it's the basic matter of "What does it mean for a value to be the probability of a proposition, given another?".

Fiery Diamond
2015-02-03, 01:23 AM
You are unambiguously refering to set A, you are unambiguously not giving specific information about A' or A''. And this is what matters.

Obviously. So you admit that you were wrong, then? Not giving specific information about A' or A'' is the definition of ambiguous, in case you hadn't noticed.

SpectralDerp
2015-02-03, 01:29 AM
Obviously. So you admit that you were wrong, then? Not giving specific information about A' or A'' is the definition of ambiguous, in case you hadn't noticed.

The question is whether or not it's ambiguous if the speaker if "case A" is specifying one of A' or A'', or not. And it's not. By the same logic, "the older child is a girl" is ambiguous too and "What's the probability that both children are girls, given that the older child is a girl?" is either 0 or 1.

You incorrectly assume that my calculation relies on the assumption of A' ("I didn't learn the sex of a particular child which I am now telling you about"), which is not the case and I said so from the very beginning. My calculation is for the case that neither A' or A'' are specified. I am not arguing the correct answer is P(X|A'), I am arguing it's P(X|A) and you can calculate it to be 1/3.

Fiery Diamond
2015-02-03, 01:57 AM
The question is whether or not it's ambiguous if the speaker meant to specify A' or A'', or not. By the same logic, "the older child is a girl" is ambiguous too and "What's the probability that both children are girls, given that the older child is a girl?" is either 0 or 1.

The speaker HAD to specify, or else the statement is void of enough information to make a determination. Look back. A includes both A' and A'', and either one gives a different answer than the other. If he doesn't, it's ambiguous by definition since either interpretation is a valid one. If both A' and A'' resulted in the same answer, it wouldn't matter that it was ambiguous. Also, I don't follow your assertion in this post. "The older child is a girl" is a member of our set A''. Are you saying that because each set is invariably composed of smaller sets, we either must ignore the smaller sets or continue on more specifically ad infinitum until we get to probability 1 or 0? Because... that's not the way language works. It may be the way MATH works, but it isn't the way language works. In language, we have something I'm going to call "an acceptable level of granularity." Take color in description. Here's an example:

I can talk about the man wearing robes that are a color.
I can talk about the man wearing red robes.
I can talk about the man wearing dark red robes.
I can talk about the man wearing crimson robes.

Three of those are acceptable to stop at, and I evaluate based on what I know and comfortably assign uncertainty to further detail. Those would be options 2 through 4. I don't HAVE to go all the way down to 4 just because I've decided to talk about the fact that his robes are a color. I DO, on the other hand, have the obligation not to give you the first line. That is not at an acceptable level of granularity. It is too ambiguous. If I say he had red robes, however, that's fine - the level of uncertainty and technical ambiguity that comes from not specifying whether they are light red or medium red or dark red is a level of uncertainty that I'm willing to accept and report on. But just "a color"? No, too ambiguous.

Where the line is drawn in how detailed you need to get is sort of arbitrary, but a lot of things about language are. That doesn't mean it isn't important. Our Set A, for our probability puzzle, is not at an acceptable level of granularity, specifically because the unsaid assumptions about which of A' and A'' it might be are going to have to be made in order to draw meaningful conclusions, and people WILL make assumptions, just like Sally did in our hypothetical examples (and just like Joe did).

I've gotta get to sleep, so I'll return to the conversation later.

SpectralDerp
2015-02-03, 02:04 AM
The speaker HAD to specify, or else the statement is void of enough information to make a determination.

Then the statement "The older child is a girl" would have the same problem. Is it because the older child is a girl and the younger child is a girl? Then the probability of both children being girls is 1. Otherwise it's 0. In fact, any conditional probability would either be 1 or 0. Mathematicians have found rules on how to calculate conditional probabilities such that "The older child is a girl" gives a well-defined conditional probability for "Both children are girls", it's 1/2. The correct way of dealing with probabilities is to apply these rules.

The fact of the matter is that one can calculate P(Both children are girls) and P(at least one child is a girl) and one can divide them. The result is 1/3 and is the probability of "Both children are girls" given that "at least one child is a girl". I don't need to make further assumptions. Questions involving A' and A'' are irrelevant.

Cazero
2015-02-03, 04:21 AM
I suggest forgetting binary for a moment. Binary systems are simples, wich means we do tons of weird assumptions about them.

I rolled a D4 and a D6. At least one of them rolled a 1. What is the probability that both rolled 1?
I don't want to actually do the calculations, because that's not important. What I want is to analyze the assumptions made.
We are in one of two situations.
1) I did not specify wich die rolled a 1 because I forgot to tell you. Silly me.
2) I did not specify wich die rolled a 1 because I'm being a jerk. Or writing the text of a math exercise, wich is kinda the same.
The natural assumption tends to be 2, and it becomes clear that it is 2 when you ask for more information and don't get any.

Now back in our binary boy/girl system.
The Joneses have two children. At least one of them is a boy. What is the probability that both are boys?
Transposing the assumptions from the dice situations, we have :
1) I did not specify that [I sampled exactly one child and saw a boy/I'm using a specific assumption about the selection of the Joneses that I didn't told you about/whatever else gets a final probability of 1/2] because it's complicated to explain and nobody wants to hear that.
2) I did not specify that [I sampled both children and saw at least one boy/I'm using a specific assumption about the selection of the Joneses that I didn't told you about/whatever else gets a final probability of 1/3] because it's complicated to explain and nobody wants to hear that. Also, I'm being a jerk and/or writing the text of a math exercise.
Here, the natural assumption is less clear, but tends to be 1, unless your natural assumption is that people are jerks/writing math exercises. The context of a math exercise should make you think it's 2, but some people simply don't alter their decisions on that. It's still ambiguous. It is also less likely to get cleared when asking for additional data and not getting any.

Segev
2015-02-03, 09:49 AM
Technically, SpectralDerp, you're right: "The older child is a girl" is ambiguous because we do not know if you chose the Joneses because they meet that criterion, or because you looked at the Joneses' children and gleaned that information and somehow decided the information you would tell us was that and only that.

However, the reason why it is not a problem in natural language is that the natural language assumption yields the same results as the "correct mathematical" interpretation.

That is the reason why people actually tend to respond to those two problems, as phrased, with, "Um, it's irrelevant that the older child is a girl. Since we know the sex of one child in both cases, we know that the remaining child's sex is independent and is thus 50/50 of being either."

When the experiment is conducted from the standpoint of examining the sample and giving information based on that examination, it's functionally equivalent to the "look in through a window and see only one of the two children" case. That is, you've determined by inspection the sex of one of the children. You now know unambiguously that THAT child is what you've observed it to be.

Technically, thinking about it, the natural language interpretation of both questions interprets to probability-language as, "What is the probability that both children are the same sex?"

This is because the hidden assumptions the natural-language interpretation makes - that is, that you've simply examined them and told us information about them - is that the question was always going to be asking whether both children are the sex of the child about whom we are given information.

The correct answer - 1/2 - is arrived at for that one either by naively assuming that one child's sex is independent of the other and we know one child's sex while we don't know the other's (thus the odds they're the same sex are the odds that the second child matches the known child's), or by examining all possibilities - BB/BG/GB/GG - recognizing the 50/50 odds of being asked about boys or girls in the middle cases, and thus coming to the "BBH/BBT/BGH/BGT/GBH/GBT/GGH/GGT" case.

Or you just start with the simplification: "He's really just asking me the probability that both children are the same sex, and telling me he's already determined one of their sexes, so is thus asking me if both of them are that sex."

Again, these are natural language interpretations.

One is very hard pressed to come to a natural language interpretation of "The Joneses have two children, at least one of whom is a girl," being "I predetermined I would ask you about girls, and made sure to pick the Joneses, who have at least one girl amongst their two children."

Yes, in the language of the probability community, it is clear what is meant. That community has its contextual shorthand, just as any other does. But teaching people about misconceptions about probability requires not presuming that they already speak the language in that fashion. Otherwise, you're just being smug that they couldn't read your mind.

To really demonstrate to a lay person the mistakes they make, you need to specify in the wording of the problem that this is not "sample, plus information gleaned from it," but really is "sample that meets these criteria."

Natural language helps you when you break it into the two sub-questions. There are no assumptions about how you gained information given that get in the way. The reason no assumptions get in the way is because you're not presenting it as information a priori.

The crux of it is that, when you give information about something in natural language, it seems to be by observation. When you ask (functionally) if another observation would give the same result (in this case, observing the second child), you imply your question is just about the odds of two things being the same.

You may not mean to, but you do.

Let's say that somebody who may or may not properly understand conditional probability tells you they don't think there's enough information in the problem to give an answer that is guaranteed to be correct. They ask you, "How do you know that at least one of the Joneses' children is a girl?" How do you answer them?

The Random NPC
2015-02-03, 10:30 AM
I get that part. I am just not sure why telling them something they already know brings this about.

It's easier to start with 2 dragons. One dragon expects the other to turn into a sparrow, because they assume that they don't have green eyes, but they know the other does. When that doesn't happen, they suddenly realize that the other dragon must be able to see a dragon with green eyes, and since they're the only dragon left, that must be them. Both dragons realize this on the same day, so they both turn into dragons on day 2. When you have three dragons, the third dragon expects the other two to turn into dragons on day 2, but when they don't, they realize that the other dragons must be able to see 2 other dragons with green eyes. So they all turn into sparrows on day 3. Basically, you add just enough information to allow all the dragons to realize that they themselves have green eyes.

SpectralDerp
2015-02-03, 12:48 PM
@ Segev,

I would try to start with the multiplicative nature of conditional probability and observe that both the probability of "two girls" and "at least one girl" are easily calculable.

Alternatively, I might be inclined to explain why interpretations of probability as a game can be misleading and the notion of a numerical calculus for plausibility is superior.

Segev
2015-02-03, 04:09 PM
So, I have two different issues. On the one hand, the dragons should have been able to make this conclusion long before the human made his announcement. To illustrate, I'll walk through some cases wherein all dragons have green eyes.

N=1: Dragon A is by himself and doesn't know his own eye color. He presumes non-green.

N=2: Dragon A knows dragon B has green eyes, and vice-versa, but neither knows the other is aware that there is at least one green-eyed dragon present.

N=3: Dragon A knows that dragons B and C have green eyes. Furthermore, A knows that B and C are both aware that there is at least one green-eyed dragon on the island, and that B and C both also know that A knows that there is at least one green-eyed dragon on the island. This information is symmetric to all dragons present.

But... that actually breaks down because beliefs-about-beliefs-of-others get screwy.

We can illustrate by making dragon C red-eyed. Now, he knows that the other two know that there is at least one green-eyed dragon, but the other two don't know that the green-eyed dragon of which they're aware knows that there is at least one green-eyed dragon.

As far as A and B are concerned, there might be just one, but he doesn't know. So him not disappearing on Night 1 is not surprising. Since A and B weren't expecting the other one to disappear, not expecting that he knew there was at least one green eyed dragon, they won't disappear on night 2, either.

Dragon C could make the erroneous assumption that this means they both saw at least two green-eyed dragons, and vanish on night 3. But that would really confuse the other two. They knew he wasn't green-eyed! And they failed to vanish because they thought the one green-eyed dragon didn't know there was at least one.

Extending this to more dragons doesn't seem to help.

With 3 green-eyed dragons, all of them not only know that there is at least one green-eyed dragon, but that all of them know this. But they can't, still, know that they're in the N=3 state rather than the N=2 state with them as the red-eyed. That is, they can't know that the other dragons know the one dragon they definitely know has green eyes is aware that at least one has green eyes.

And this spirals upwards to any number of dragons, as impossible chains of obviously false beliefs about what others believe are nonetheless logical.

This kind of cognitive dissonance bugs me.

I see the logic. I want to poke a hole in it because there's something off about the way it has to play out to work.

The key is that all dragons must know that all dragons know there is at least one green-eyed dragon.

If you can see at least three green-eyed dragons, you know that that condition must be fulfilled, don't you? Well, no. If you really do not have green eyes, the three only see two each, so while they know each dragon sees at least one green-eyed dragon, they don't know that all dragons know that all dragons know.

So... four?

Let's again assume that you really do have red eyes (as you would if you were one of 5 dragons, and saw all four others had green eyes). The other four each see three green-eyed dragons. As far as they're concerned, because they know you have red eyes, they're in the situation before, thinking they have red eyes and see three green-eyed dragons.

But here, we're definitely at the point of reductio ad absurdum, because we're assuming that somebody believes somebody believes that one of the dragons thinks there are no green-eyed dragons, even though we all know we all know that there is at least one green-eyed dragon.

I mean, again, assume you realy do have red eyes. You know there's at least one green-eyed dragon. You know each of the other four dragons knows there's at least one green-eyed dragon. You know each of the other four dragons knows that each of the other four dragons knows there's at least one green-eyed dragon.

You know there are at least 4. Everybody else must know there are at least 3. They have to know that, because they can count them. You know that the other dragons will either vanish on night 4, or that you'll have to vanish with them on night 5. If you don't want to vanish, you hope the others are all hoping they others vanish on night 3, because that would mean they each see 3 green-eyed dragons. You know they won't, because you know there are at least 4.

If they still don't vanish on the 4th night, that means they also saw 4 other green-eyed dragons, so you have to vanish, too.

I do think you have to see 4 green-eyed dragons for it to work. Maybe 3.

If you see two, you can't know that there are at least 3, so you can't know that the others DO know all dragons know there's at least one. If you see 3, you know each of those three can see at least two. You know that all dragons know that there is at least one green-eyed dragon.

But the problem remains that it's possible those three just see two each. (In fact, you hope so, as you don't want to be green-eyed.) If they do, their assumption would be that the vanishing happens on night 2 without them, or night 3 with them, depending on whether they have green eyes or not.

Nobody expects vanishings on night 1. If anybody sees only 2, and assumes everybody knows everybody knows there's at least one, they'll expect vanishings on night 2. If they don't happen, they'll know that the two they saw also saw 2, so vanish on the third.

Except...

If there WERE only 2 that had green eyes, both would have seen only the other one, and assumed that other one did NOT know there was at least one. So nobody would have vanished on night 2 not because they saw 2, themselves, but because they assumed the other one was the only one.

So night 3, the mistaken red-eyed dragon who saw two green-eyed dragons might vanish by himself!

All because he mistook the thought process of "maybe he the one green-eyed dragon I'm seeing doesn't know there's at least one green-eyed dragon" for "I'm seeing at least two green-eyed dragons."

Since that mistake is possible, it is not a foregone conclusion that nobody vanishing on night 2 means there's one more dragon than the two you can see with green eyes.

The weird bit is that it's not that everybody knows there's at least one, then. It's that everybody knows everybody knows everybody knows xN that there's at least one.

But I still feel like there's a point where it shouldn't be necessary to vocalize it. Just by observation, all should know there's at least one, and that should provice recurrent knowledge that all know all know "enought times."

And yet, it breaks down a bit when I examine it at any particular N. Still, 4 seems to...almost...work, the most.

If I have red eyes, and see 4 green-eyed, I would assume they'll all vanish on night 4. They each see only 3 green-eyed, so are waiting for vanishings on night 3.

Without an explicit statement that there is at least one, it is impossible that any condition would lead to a vanishing on night 1. N=1 would never know there was at least one. N=2 would be pretty certain the one he sees doesn't know he's green-eyed, and thus would not have reason to assume that him seeing or not seeing green eyes would matter, so no night 2 vanishings.

N=3 would therefore not vanish on night 3 based on not being certain night 2's failure to vanish was not due to uncertainty between the two green-eyes they can see. That uncertainty seems to propagate up, even though it seems like it shouldn't matter.

I get it, but it's still gonna bug me for a while. I've had this open too long, though, so I'm off for the day.

Talakeal
2015-02-03, 07:47 PM
It's easier to start with 2 dragons. One dragon expects the other to turn into a sparrow, because they assume that they don't have green eyes, but they know the other does. When that doesn't happen, they suddenly realize that the other dragon must be able to see a dragon with green eyes, and since they're the only dragon left, that must be them. Both dragons realize this on the same day, so they both turn into dragons on day 2. When you have three dragons, the third dragon expects the other two to turn into dragons on day 2, but when they don't, they realize that the other dragons must be able to see 2 other dragons with green eyes. So they all turn into sparrows on day 3. Basically, you add just enough information to allow all the dragons to realize that they themselves have green eyes.

I understand the process by which they turn into sparrows.

What I don't understand is why that process is not triggered until someone tells them something they already know.

Jenerix525
2015-02-03, 08:32 PM
I understand the process by which they turn into sparrows.

What I don't understand is why that process is not triggered until someone tells them something they already know.

Each dragon can see N dragons with green eyes.
If dragon 1 assumes it does not have green eyes, then dragon 2 is imagined to see N-1 dragons with green eyes.
If dragon 1 assumes dragon 2 assumes it does not have green eyes, then dragon 1 believes dragon 2 might imagine dragon 3 to see N-2 dragons.

This logic continues to the point of N-N=0; "he might think that this other guy might think that... ...that this last guy might think that there are no green eyed dragons."

The statement you make removes that last step. Which topples the tower of self-delusion.

The Random NPC
2015-02-03, 08:37 PM
I understand the process by which they turn into sparrows.

What I don't understand is why that process is not triggered until someone tells them something they already know.

Every dragon assumes that they don't have green eyes. They know that every other dragon has green eyes, but they think they are unique and have some other color eye. By saying that at least one has green eyes, every dragon is able to logic out that they have green eyes instead of some other color.

Yaktan
2015-02-03, 09:28 PM
Except with 3 or more dragons, they all already know that every other dragon knows that there is at least one other with green eyes, so telling them adds no new information.

EDIT: I just realized another way of expressing the disconect between the dragon situation and the mathematical induction used to explain it; to wit, the dragons are not an ordered set, so I do not think that induction works on them. (as we can see in the case of 3 dragons. Either they will each nod their head sagely and say "yes, everyone knew that already" or they will all have already turned into sparrows by now.

I have seen alternative presentations of a similar puzzle, which involve a method of ordering the participants, so you can actually apply induction to them.

EDIT2: Though, thinking through the puzzle I can see how the logic works. (I find red/blue hats seem to work better in my mind than green-eyed dragons) So now I am really puzzled, because of my above points. Hmmm.

Talakeal
2015-02-03, 10:39 PM
Ok, I think I got it.

So the dragons themselves never speak about eye color, so no one has ever made a global announcement before.

The dragons all know that there are green eyed dragons among them, and they all know that everyone else knows. But if you apply Princess Bride logic long enough (i.e. I know that you know that I know that you know that I know that you know...") you will get to a point where someone is lacking critical knowledge of someone else's knowledge without a global announcement.

Is that correct?

huttj509
2015-02-03, 11:48 PM
Ok, I think I got it.

So the dragons themselves never speak about eye color, so no one has ever made a global announcement before.

The dragons all know that there are green eyed dragons among them, and they all know that everyone else knows. But if you apply Princess Bride logic long enough (i.e. I know that you know that I know that you know that I know that you know...") you will get to a point where someone is lacking critical knowledge of someone else's knowledge without a global announcement.

Is that correct?

Yes. 3 dragons, Alice, Bob, and Charlie have green eyes.

A sees B and C. A knows B sees C, and C sees B. A knows at least one has green eyes.

If A did not have green eyes, then B would see C, and know at least one had green eyes, but B would assume C saw A and B as red eyes, and thus not know that at least one had green.

So A, B, and C all know that at least one has green eyes, but A thinks that B thinks that C does not know this (and thinks that C thinks B does not know this). By telling both B and C, A expects them to see each other, say "oh bother" and turn into sparrows after a couple of nights (first night would be B expecting C to sparrow, as C saw B and A red, and vice versa with C and B. Second night has A expecting B to realize C saw his green eyes, at the same time C realizes B sees his.) When this doesn't happen, A knows that B and C both saw her green eyes, and were expecting her to sparrow on night 2, so she sparrows on night 3.

Ow, I get it and my head still hurts.

Segev
2015-02-04, 10:53 AM
The problem is that the induction works on one level: you in theory must hypothesize that your fellow dragons hypothesize that one of their beliefs-about-beliefs of other dragons includes a belief that at least one dragon present thinks every single dragon has red eyes (where "red" is used as shorthand for "non-green").

So as long as you cannot be certain that that hypothesized imagined dragon in a multiply-projected hypothesis about other dragons' beliefs has not been told that there is at least one dragon with green eyes, you cannot assume the chain starts.

That is, in the 3-dragon example, if Bob really does have red eyes, while Alice and Greg have green, and Bob assumes that all dragons know at least one dragon has green eyes and that that's sufficient to start...

Bob will expect that Alice and Greg will both be waiting for the other to disappear on the first night. And that they will both disappear on the second. When they don't, Bob, thinking "all dragons know that at least one dragon has green eyes" is enough to start the chain, would assume that Alice and Greg are still here because both of them saw at least two green-eyed dragons, meaning he, too, must have green eyes.

So Bob would erroneously vanish on the third night.

Because Bob is a dragon, and dragons do not make logical errors, Bob would know that Alice and Greg, assuming Bob has green eyes, both see only one dragon with green eyes. Alice and Greg would assume they, themselves, have red eyes, and would therefore each would assume the other does not know there is at least one green-eyed dragon.

In short, Bob assumes that Alice assumes that Greg assumes all dragons present have red eyes.

With three, that's still...believable. After all, the chain of hypothesis is such that Bob cannot know it is impossible for any dragon present to doubt that all other dragons must see at least one other dragon with green eyes.

Let's go to 4.

To start with, we will again assume Bob really does have red eyes, while Alice, Greg, and Chuck all have green.

Alice, Greg, and Chuck are all assuming they, themselves, have red eyes. They're wrong, but they assume that, and also assume that the others and Bob assume the same about themselves (with only Bob being correct).

This means that Alice, Greg, and Chuck are in Bob's position from the 3-dragon problem: as long as they assume they, themselves, have red eyes, e.g. Greg assumes Alice assumes Chuck assumes there are no green-eyed dragons. Greg's belief is that he and Bob have red eyes, so Alice only sees Chuck's eyes as green, and therefore Greg thinks Alice thinks Chuck doesn't see any green-eyed dragons.

Bob, once again, understands Greg's belief. He believes Greg believes .... Chuck believes there are no green-eyed dragons.

Let's add Fred, a fourth green-eyed dragon, to the mix. We now have 5 total.

In theory, Fred is in Bob's old position, as far as Fred is concerned: he's a (self-believed) red-eyed dragon looking at 4 green-eyed dragons. He, like Bob, understands that Greg would believe Alice believes Chuck believes there are no green-eyed dragons.

Let's make Bob green-eyed, bringing us to 6 green-eyed dragons.

Now, the assumption chain becomes clearly silly.

Each dragon can see 5 green-eyed dragons. Even assuming they, themselves, are red-eyed, and that any dragon they choose to start the chain will believe himself to also be red-eyed, the most they can believe about that other dragon's beliefs is that that other dragon believes there are two red-eyed dragons and 4 green-eyed dragons.

The induction chain seems to hold if you just follow it blindly, but there is a new fact present: Each dragon knows that, even if two dragons present have red eyes (themselves and the erroneous assumption that the second dragon they're imagining beliefs for is red-eyed), all dragons present can clearly see at least 3 other green-eyed dragons.

Let me repeat: All dragons know that all dragons present can see at least three other green-eyed dragons.

This also means that, even if they, themselves have red eyes, and the dragon they're imagining beliefs for thinks they have red eyes, any dragon that dragon-whose-beliefs-are-being-imagined would imagine beliefs for knows, for certain, that all dragons know that at least one dragon present has green eyes.

It is therefore impossible, even if you assume that the dragon whose beliefs you're imagining thinks there are two red-eyed dragons out of the 6 dragons present, for the dragon whose beliefs you're imagining to believe that any other dragon would doubt that there is at least one green-eyed dragon present.

That is, Bob may think that Fred thinks there are two red-eyed dragons. Bob may think that Fred thinks Alice, therefore, sees only three green-eyed dragons. However, if Bob thinks Fred thinks Alice therefore would think that Chuck sees two, and would therefore think each of those two sees only one and believes the other to believe there are none...

Except Bob knows Fred knows Alice can see at least 3 green-eyed dragons. Bob knows Fred knows that Alice cannot possibly believe that any of those three green-eyed dragons she can see doubts that there is at least one green-eyed dragon.

Because Bob knows Fred knows this about Alice, Bob knows Fred knows this about all of them. In fact, he knows all dragons know this about all dragons.

Since all dragons must know that all dragons cannot possibly doubt that there is at least one green-eyed dragon, and all dragons know this about this essential knowledge, it is impossible to think that any dragon would think that any dragon would think that any dragon would think that any dragon would think that there are no green-eyed dragons present.

You have to be able to see 5 green-eyed dragons for this to work. This allows you to assume that another dragon is assuming that there are two more red-eyed dragons than you know there to be, and still know that that other dragon must know that all dragons know that all dragons know that at least one dragon has green eyes.

In other words, you must be certain that all green-eyed dragons you see can see at least 4 other green-eyed dragons.

But, as long as Bob can see at least 5 green-eyed dragons, Bob knows that it is imposible for any of those 5 green-eyed dragons to maintain a belief that any dragon could maintain a belief that any dragon is unaware that there is at least one green-eyed dragon.

It takes two green-eyed dragons for all green-eyed dragons to know there is at least one green-eyed dragon.

It takes three green-eyed dragons for all green-eyed dragons to know that all dragons know there is at least one green-eyed dragon. They can still believe that a green-eyed dragon might think that another green-eyed dragon would not know that all dragons know there is at least one green-eyed dragon, however.

If there are four green-eyed dragons, each knows that it is impossible for any dragon to believe that any given green-eyed dragon might believe there are no green-eyed dragons. However, it is still possible for a green-eyed dragon to believe that a green-eyed dragon might believe a third green-eyed dragon might not be aware that all dragons know there is at least one green-eyed dragon.

If there are five green-eyed dragons, each knows that it is impossible for any of the green-eyed dragons they can see to to believe that any dragon could believe any dragon doubts there is at least one green-eyed dragon present.

But it takes seeing five green-eyed dragons to know for a fact that all dragons you're seeing are unable to believe that any other dragon could harbor such a belief about any other dragon.

Therefore, you must see 5 green-eyed dragons...

...and I think I've talked myself into possibly needing to see 6 now. This has gotten too long. I'll come back again later.

Jenerix525
2015-02-04, 12:44 PM
as we can see in the case of 3 dragons. Either they will each nod their head sagely and say "yes, everyone knew that already" or they will all have already turned into sparrows by now.
And then promptly turn into sparrows. (A now knows that B could see green eyed dragons other than C.)

"everyone knew that already" is precisely the information gained each night when no dragon turns into a sparrow. Which builds up to what Talakeal succinctly described as Princess Bride logic.

I want to try a different way of describing my logic, just to see if it still makes sense.

Using G to mean the minimum number of green-eyed dragons any hypothetical dragon could believe.
And N again to mean the number of green-eyed dragons seen by the considering dragon.

If G=N+1, then the additional dragon must be the observer. The observer then transforms.
If a dragon does not transform, then N is at least equal to G.
When no dragons transform, they learn that every green-eyed dragon could see at least G dragons. Therefore there at least G+1 dragons with green eyes.
Each night with no transformation increases G by 1. Until one night there is a flock of sparrows on the island.

This only works if the island is operating on a collective minimum (which must be positive). Like, say, "at least one dragon has green eyes".

Segev
2015-02-04, 04:25 PM
I have yet to come to a contrary conclusion that I can prove, so I will not claim that I have found a provable flaw in this logic problem.

However, let's just take our arbitrary 100 dragons on the island, and we know that each of them can see 99 green-eyed dragons (GEDs).

Each assumes that all others see 98 GEDs.

Alright. That could start our "Princess Bride Logic" chain, but I want to pause there and think about this a moment.

The hypothesis is that, given that we know all 99 GEDs that we can see can see at least 98 GEDs, we therefore know that every dragon on the island is aware that there is at least one GED. Any one of us - GED or not - could say, "There is at least one GED on the island," and it would trigger the flock of sparrows event.

Each of us, being GEDs who can see 99 GEDs, would expect that the 99 we can see would disappear on the 99th night after one of us said that line. When that didn't happen, we'd know we all had to go the next night.

Now, note that the one change I've made is that there's no external human. Just one of us foolishly broke our unspoken taboo about not mentioning eye color.

Now, we dragons all know each other. We all know there is at least one GED on the island, and in fact all know that we each can see at least 98 of them. We know, therefore, that it is possible for any or all of us to truthfully say, "There is at least one GED on the island."

All it would take to trigger this knowledge and realization in our impeccably logical dragon brains is all of us meeting at once.

The question then arises: how many of us must meet simultaneously for us to actually trigger the knowledge that any of us could say, with certainty, "At least one of us has green eyes?"

This changes it from having to know that they know that he knows that she knows, to having only to ascertain that it is possible for any one dragon present to say, "at least one of us has green eyes."

Obviously, if I am a lonely GED who doesn't know he is one, I don't know that any of us could say that. In fact, I couldn't say it about my N=1 group with certainty.

If Talekeal joined me, I could say with certainty that at least one of us was a GED, but I wouldn't know that he could say it with certainty. (If I had red eyes, Talekeal would know only that there was 0-1 GEDs present. That is not at least one.)

However, if Pinnacle walked up (also a GED, of course), I would know that:

I could say "At least one of us has green eyes."
Referring to either Talekeal or Pinnacle.
Pinnacle could say, "At least one of us has green eyes."
I know she'd at least be referring to Talekeal.
Talekeal could say, "At least one of us has green eyes."
I know he'd at least be referring to Pinnacle.[/lsit]
Both of them know that I could say it.

What I don't know is that they know the other could say it.

That does mean that, to MY knowledge, not all of us know that all of us could say it. Saying it, then, would tell me that both of them now know the other can say it.

So let's have Jenerix525 join us. He is, of course, also a GED.

Now, I know:

I could say "At least one of us has green eyes."
[list] Referring to any of them.
Pinnacle could say, "At least one of us has green eyes."
I know she'd at least be referring to Talekeal or Jenerix525.
Talekeal could say, "At least one of us has green eyes."
I know he'd at least be referring to Pinnacle or Jenerix525.[/lsit]
Jenerix525 could say, "At least one of us has green eyes."
[list] I know he'd at least be referring to Pinnacle or Talekeal.
I know Jenerix525 knows all three of Pinnacle, Talekeal, and I could say it.
I know Jenerix525 knows I can see at least Talekeal and Pinnacle.
I know Jenerix525 knows Talekeal can see at least Pinnacle.
I know Jenerix525 knows Pinnacle can at least see Talekeal.
I know Pinnacle knows all three of Jenerix525, Talekeal, and I could say it.
I know Pinnacle knows I can see at least Jenerix525 and Talekeal.
I know Pinnacle knows Jenerix525 can at least see Talekeal.
I know Pinnacle knows Talekeal can at least see Jenerix525.
I know Talekeal knows all three of Jenerix525, Pinnacle, and I could say it. I know Talekeal knows I can at least see Pinnacle and Jenerix525.
I know Talekeal knows Jenerix525 can at least see Pinnacle.
I know Talekeal knows Pinnacle can at least see Jenerix525.

Therefore, I know that all of us know that any of us could say it.

Unfortunately, running this analysis with the assumption that I have red eyes, Pinnacle could not, for instance, know that Jenerix525 knows that all three of Talekeal, Pinnacle, and I could say it. Pinnacle would not know Jenerix525 knows of any GED that Talekeal can see.

This certainly feels like it's spiralling out to infinity, but I want to try at least one more. Let's add The Random NPC ("NPC" for short; I hope he doesn't mind).

Now, everything I said about what I know in the N=4 case, I can say about each of the other 4.

e.g....

The Random NPC knows:

The Random NPC could say "At least one of us has green eyes."
Referring to any of the others.
Pinnacle could say, "At least one of us has green eyes."
NPC knows she'd at least be referring to Talekeal or Jenerix525.
Talekeal could say, "At least one of us has green eyes."
NPC knows he'd at least be referring to Pinnacle or Jenerix525.[/lsit]
Jenerix525 could say, "At least one of us has green eyes."
[list] NPC knows he'd at least be referring to Pinnacle or Talekeal.
NPC knows Jenerix525 knows all four of NPC, Pinnacle, Talekeal, and I could say it.
NPC knows Jenerix525 knows I can see at least Talekeal and Pinnacle.
NPC knows Jenerix525 knows Talekeal can see at least Pinnacle.
NPC knows Jenerix525 knows Pinnacle can at least see Talekeal.
NPC knows Jenerix525 knows NPC can see at least Talekeal and Pinnacle.
NPC knows Pinnacle knows all four of Jenerix525, NPC, Talekeal, and I could say it.
NPC knows Pinnacle knows I can see at least Jenerix525 and Talekeal.
NPC knows Pinnacle knows NPC can see at least Jenerix525 and Talekeal.
NPC knows Pinnacle knows Jenerix525 can at least see Talekeal.
NPC knows Pinnacle knows Talekeal can at least see Jenerix525.
NPC knows Talekeal knows all four of Jenerix525, NPC, Pinnacle, and I could say it. NPC knows Talekeal knows I can at least see Pinnacle and Jenerix525.
NPC knows Talekeal knows NPC can at least see Pinnacle and Jenerix525.
NPC knows Talekeal knows Jenerix525 can at least see Pinnacle.
NPC knows Talekeal knows Pinnacle can at least see Jenerix525.

The above can rotate through the others, replacing NPC with Pinnacle, Talekeal, and Jenerix525 to prove that all of them know that all of them can say it.

Once again, we seem to be spiralling up, because the statement I can make now is that I know all of them know that all of them know all of us can say it.

Let's repeat what we know: If I see 4 GEDs, I know all of them know that all of us know that all of us can say, "At least one of us has green eyes."

Is it important that they all know that all of us know that all of us know all of us can say it? Or is it sufficient that all of us know that all of us know all of us can say it?

The point where all present know that all present know that all present can say, "At least one GED is present," is when there are at least 3 GEDs present.

Let's try building our hypotheticals, then.

In this case, you cannot tell if the minimum condition, "at least three GEDs present," has been met. Neither can the other two. None of you will assume that all of you can make the statement.

Here's the tricky bit. You know at least 3 GEDs are present. That's the hypothetical condition for assuming that all present know that the statement can be made. However, if you don't have green eyes, the other three won't know the condition has been reached.

And we're back to spiralling up, because even though you know for certain that all present are seeing at least 3 GEDs, you know know they know that about all present.

Although...

If you ARE a GED, then all present are absolutely positive that all present see at least 3 GEDs. They will assume the statement could be made by everyone present and that everybody knows it.

Everybody sees 4 other GEDs, so they're waiting for night 4; if everybody's still here, they'll all vanish on night 5.

If you ARE NOT a GED, however, only you are positive that everybody else sees at least 3 GEDs. If you assume that the statement is made, therefore, the others are looking around and seeing 3 GEDs. You would expect them to be waiting for the 3rd night to check, and to vanish on the 4th night. When they don't, your erroneous assumption that they all know that all knew the statement could be made meant they all knew everybody knew that would mean you'd vanish on night 5, alone, and needlessly.

Let's try one more, just for kicks and grins.

You come together with 5 other dragons, and they're all GEDs. You know that all present must be 100% certain that every single dragon here knows there are at least 3 GEDs present.

If you are NOT a GED, then the other 5 each see only 4, however. So while they all know that all must know they can make the statement, do they all know the others are not seeing only 3, and thus assuming the remainder see only 2?

Weirdly, we can pin down a number of GEDs that must be present for all to be able to say with certainty that every single dragon present knows that every single dragon present knows that any one of them could make the statement. It's 4.

The ambiguity on knowing whether you ARE the 4th dragon or not, and thus whether you make the rest aware that all are aware, creates the problem.

Yet, if you can see at least 4, you know they all know. Very weird.