# Thread: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but...)

1. ## Arguing about Probability and Statistics (From: DM's who want you to Roleplay but...)

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.

2. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

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.)

3. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by Domochevsky
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.

4. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

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.

5. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

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.

6. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

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.

7. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by Arbane
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.

9. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by Lord Torath
Looks like it has ended, but thanks anyway.

10. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

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..

11. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by JusticeZero
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...

12. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

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.

13. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by jaydubs
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?

14. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by Gavran
... 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.

Originally Posted by Gavran
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.

15. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by Gavran
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?
Originally Posted by jaydubs
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! It's around 60-70% if I remember correctly.

16. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by jaydubs
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. 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).

Originally Posted by The Random NPC
Birthday paradox! It's around 60-70% if I remember correctly.
70.6%, actually. It hits near certainty around 50.

17. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by jaydubs
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.

?

18. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

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

[BP(Both children are boys and at least one child is a boy) = P(Both children are boys) = P(the elder child is a boy and the younger child is a boy) = P(the elder child is a boy) * P(the younger child is a boy) = 1/4[/B]

So

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)

19. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by hymer
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

20. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by hymer
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.

21. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

@ SpectralDerp: Sorry, that's all Greek to me.

@ 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?

22. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by Lord Torath
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.
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%.

23. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by hymer
@ SpectralDerp: Sorry, that's all Greek to me.

@ 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.

24. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

@ 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.

25. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

To reiterate
Originally Posted by Lord Torath
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?

26. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

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.

27. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

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.

28. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by Lord Torath
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.

29. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Originally Posted by Lord Torath
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%

30. ## Re: Arguing about Probability and Statistics (From: DM's who want you to Roleplay but

Okay, I think I can get my head around that. Thanks for the help, ladies and gents!

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