Probability

[Origomar]

Well ive been asking this question around forums and no one seems to know it and it does have a purpose just not a very interesting one and i can't find an equation (or one that i understand) that can answer this, and this is the most intelligent forum i know x.x


If three chances at something working and there is a 40% chance it will work each chance, what is the chance that it will work once.(if its over 100% ,but i doubt it is, for simplicity just put like 120%)


Ex. bobby has 3 apples and there is a 40% chance there is a worm inside each apple, what is the chance if he eats all 3 apples that he will eat a worm.

if you tell me 40% i will pop you on the head (that the chance for each individual one and im not asking how one will affect the others im asking what the overall effect is)

Edit: also sorry i have no idea where to put this thread

[Moff Chumley]

Well, It's logically impossible for it to be 120%, as just by thinking about it you can tell that there's less than a hundred percent chance. So adding is out, and multiplying is out, as that would find the probability of a worm being in EVERY apple.

I'm 90% sure that the answer remains 40%, as your chances of eating a worm don't increase with each successive apple, but I'm not positive.

EDIT: Read your edit. And the probability of one apple having a worm doesn't affect the probability of other apples having worms.

[Pharaoh's Fist]

Well ive been asking this question around forums and no one seems to know it and it does have a purpose just not a very interesting one and i can't find an equation (or one that i understand) that can answer this, and this is the most intelligent forum i know x.x


If three chances at something working and there is a 40% chance it will work each chance, what is the chance that it will work once.(if its over 100% ,but i doubt it is, for simplicity just put like 120%)


Ex. bobby has 3 apples and there is a 40% chance there is a worm inside each apple, what is the chance if he eats all 3 apples that he will eat a worm.

if you tell me 40% i will pop you on the head (that the chance for each individual one and im not asking how one will affect the others im asking what the overall effect is)

Edit: also sorry i have no idea where to put this thread

40%*40%*40% is 6.4%

There is a 6.4% chance that all three apples will have a worm in them.

1-6.4%=93.6%

There is a 93.6% chance that at least one apple will have a worm in it.

[Origomar]

40%*40%*40% is 6.4%

There is a 6.4% chance that all three apples will have a worm in them.

1-6.4%=93.6%

There is a 93.6% chance that at least one apple will have a worm in it.

thank you :)

[Flame of Anor]

40%*40%*40% is 6.4%

There is a 6.4% chance that all three apples will have a worm in them.

1-6.4%=93.6%

There is a 93.6% chance that at least one apple will have a worm in it.

Except I'm afraid that's wrong. It doesn't reflect on you, PhF, probability is hard, but the latter half of your answer is correct. The first half--the 6.4%--is right. The 93.6 chance, however, is that fewer than three apples will have a worm. This includes the chance of no worms at all.

You have to go at this from the other end: 60% chance of no worm. Just like the chance of all three having a worm was 40% cubed = 6.4%, the chance of all three having no worm is 60% cubed = 21.6% that no apple will have a worm.

P.S. A lot of probability errors could be avoided by reality checks. For the false 40% answer, just think: There's a 40% chance that the first apple has a worm. At-least-one-has-a-worm is already fulfilled four times out of ten--40%. For the probability to be stay 40% for if any apple at all has a worm, then all the other apples could not have a worm if the first didn't.

Moff Chumley, what you were thinking of with "the probability of one apple having a worm doesn't affect the probability of other apples having worms" is the probability of the next apple having a worm, not the probability of any apple having a worm.

[Lufia]

40%*40%*40% is 6.4%

There is a 6.4% chance that all three apples will have a worm in them.

1-6.4%=93.6%

There is a 93.6% chance that at least one apple will have a worm in it.

You got about it backwards. The complement to "at least one apple has a worm" is "none of the apples has a worm".

The complement to "all apples have worms" is "none have worms or one has a worms or two have worms".

You might be lacking a coefficient in your answer, too. But I'm braindead at this hour.

[Pharaoh's Fist]

I also seem to have phrased what I mean to say incorrectly. Let me retry.

40%*40%*40% is 6.4%

There is a 6.4% chance that there will be one worm in three apples.

60%*60%*60%=21.6%
There is a 21.6% chance that no worms are present

1-6.4%-21.6%=72%

There is a 72% chance that at least one apple will have a worm in it.

But that's not what the OP asked exactly, and is probably wrong somewhere...

[Flame of Anor]

I also seem to have phrased what I mean to say incorrectly. Let me retry.

40%*40%*40% is 6.4%

There is a 6.4% chance that there will be one worm in three apples.

60%*60%*60%=21.6%
There is a 21.6% chance that no worms are present

1-6.4%-21.6%=72%

There is a 72% chance that at least one apple will have a worm in it.

But that's not what the OP asked exactly, and is probably wrong somewhere...

Sorry to pick on you again, but, yes, it is wrong somewhere. You're very close to right, but you subtracted 6.4% from 100%. The 6.4% is the chance that all three have worms, right? And if all three have worms, then at least one has a worm, right? The 72% chance is the chance that at least one but not three apples will have a worm. 100% - 21.6% = 78.4% is the probability of at least one worm.

[Origomar]

Well the odds are still in my favor.

[Mr. Mud]

Well, It's logically impossible for it to be 120%, as just by thinking about it you can tell that there's less than a hundred percent chance. So adding is out, and multiplying is out, as that would find the probability of a worm being in EVERY apple.

I'm 90% sure that the answer remains 40%, as your chances of eating a worm don't increase with each successive apple, but I'm not positive.

EDIT: Read your edit. And the probability of one apple having a worm doesn't affect the probability of other apples having worms.

Sarcastic Answer: That depends on the species of worm and its behavioral habits .

Real Answer: But no, you're right.

[Recaiden]

It's 1-.6*.6*.6
The second half is the odds of there being no worms.

[billtodamax]

okay, there's a 0.4 chance that the first apple you eat will have a worm in it. Ergo, there's only a 0.6 chance that you will have to eat the second apple at all to find if there's a worm in it. so the next apple has a 0.4*0.6 chance that the next apple will have a worm and the first apple won't. Therefore the third apple has a 0.4*0.6*0.6 chance that it'll have a worm and the first two won't. So the total chance of an apple has a worm is:
0.4 + 0.4*0.6 + 0.4*0.6*0.6 = 0.784 or 78.4%

[Flame of Anor]

If anyone doubts my answer, note that billtodamax just did the problem a completely different way and got exactly the same answer.

Well the odds are still in my favor.

Well, if you like worms...

[billtodamax]

That's how my mother explained it too me, and she's a maths teacher.

[Jimorian]

Flame has it right.

To break it down further, you have 4 possible outcomes after eating all 3 apples.

0 worms (21.6% already figured from above)
1 worm
2 worms
3 worms (6.4% already figured from above)

For 2 worms exactly, you know that it will happen .4 x .4 of the time that the 3rd apple doesn't have a worm, or x .6 which gives us .096 0r 9.6% of the time. But, there are 3 different combinations of this depending on which apple doesn't have the worm, so you have to ADD all 3 combo together resulting in 28.8%

For 1 worm exactly, you again have 3 combos, this time of .4 against .6 x .6 giving us a 14.4% chance 3 times or 43.2% overall.

Did I do this right? Let's find out!

0 worms 21.6%
1 worm 43.2%
2 worms 28.8%
3 worms 6.4%

Add them all up for <drumroll> 100%. WOOT!

[Origomar]

if anyone was wondering i was trying to figure out what the chance is of me scoring a critical hit with a level 6 ranger that had improved criticals a rapier and kukri and two weapon fighting feat :D. ill probably suck if i find anything that is immune to crits...


edit:damn you prerequisites and your must have +8 BAB

[Lufia]

billtodamax isn't answering the same question, actually. He considers we are in a sequential game (one experience repeated, stop at a certain condition) while the question Flame answered doesn't ask for any kind of sequence.

The result is the same in this case but I wouldn't bet on it being the same for, say, four apples.

[billtodamax]

Well I suppose that it would've been easiest just to subtract the possibility of it not happening from 100%.

100%-21.6%=78.4%

EDIT: Actually, as the apples all have the same chance of having a worm in them, it doesn't matter what order you do the equation in. Basically as soon as you eat the first apple and it has a worm in it, you don't have to eat the second to know that one of the three apples has (have? I'm never sure about this one.) a worm.

[Flame of Anor]

Simply put: for problems of this type, if you have n identical randomizers, of which each has probability p of returning result x, then the probability p' of x being the case for all of the randomizers is:

p' = p^n

And the probability p'' of x being the case for none of the randomizers is:

p'' = (1-p)^n

And the clincher:

p'' != 1-p'

unless n = 1.

[averagejoe]

you don't have to eat the second to know that one of the three apples has (have? I'm never sure about this one.) a worm.

Has. You're referring to one apple, so you use the singular conjugation.

[billtodamax]

Has. You're referring to one apple, so you use the singular conjugation.

Thanks. I'm always so confused by this.

[Flame of Anor]

Thanks. I'm always so confused by this.

It's the same with "none" and "each".

[averagejoe]

It's the same with "none" and "each".

Well, unless I'm mistaken, I think the confusion came from the "one of the three," which refers to two different things, one singular and one plural, and which the conjugation was supposed to refer to.

[billtodamax]

Indeed it was.

[kamikasei]

Edit: also sorry i have no idea where to put this thread

I suggest you post in the Roleplaying Games forum, where the exact details of the rules you're using can be checked and the probability worked out for the specific problem you're dealing with - probability is a counterintuitive area and it's good to get eyes looking at the actual problem.

[Flame of Anor]

Well, unless I'm mistaken, I think the confusion came from the "one of the three," which refers to two different things, one singular and one plural, and which the conjugation was supposed to refer to.

Exactly. I'm just saying that the verb would be singular also in the cases of "none of the three" and "each of the three".