Nice puzzle

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Jonathan
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Nice puzzle

Post by Jonathan » Mon Mar 16, 2009 6:14 pm

Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6 the game is over. Sue rolls first, if she doesn't roll a 6, Bob rolls the die, if he doesn't roll a 6, Sue rolls again. They continue taking turns until one of them rolls a 6.

Bob rolls a 6 before Sue.

What is the probability Bob rolled the 6 on his second turn?
Source: http://blag.xkcd.com/2009/02/11/a-math-problem-2/ (spoiler alert, especially comments) / http://lispclub.com/

It isn't that difficult, but many get it wrong, yet are absolutely convinced they're right. Don't be like them if you can help it. Note that it's legal to be bombed by the club with my answer, but not the other way around. :)
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Post by rumpole » Tue Mar 17, 2009 2:04 am

if they are asking what I think they are asking it be (bob 5/6) * (sue 5/6) * (bob 1/6)

but they maybe just asking for 1/6 ? it all depends on how you interpret the question
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Post by Z-Man » Tue Mar 17, 2009 8:22 am

The question asked in a way to eliminate interpretation: What is the implied probability of bob winning on his second throw, given that he won the game? Or P(bob wins on his second throw|bob wins).

Because, you know, the way the question is asked originally, the answer is really 0 or 1, depending on what happened.
Anyway, both your answers are wrong. The first one doesn't even make sense. What would the meaning of bob and sue be in the value?

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Post by Lucifer » Tue Mar 17, 2009 8:43 am

I believe the odds are 1/6 because the number that shows up on the die is not dependent on previous rolls.

(Showing my very rudimentary understanding of statistics and probability here)
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Post by Word » Tue Mar 17, 2009 9:16 am

I just guessed and of course it was completely wrong:
i thought when Bob has only 2 turns and there are 6 possibilities because of the die his chance is still 1/6

(5/6)^3 x(1/6) should be right.

Edit: :idea: !!

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Post by Lackadaisical » Tue Mar 17, 2009 9:35 am

Word wrote: (5/6)^3 x(1/6) should be right.
I don't think that's right.. i think you haven't accounted for bob winning yet?

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Post by Word » Tue Mar 17, 2009 10:05 am


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Post by Z-Man » Tue Mar 17, 2009 10:29 am

Good thing the xkcd page has more wrong answers as right ones. Lack is correct, you forgot something. You answered a different, easier, question.

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Post by Word » Tue Mar 17, 2009 11:52 am

lol k now i got it too xD

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Post by Jonathan » Tue Mar 17, 2009 12:12 pm

I hope so. I'll upload the answer now.
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Post by rumpole » Tue Mar 17, 2009 3:44 pm

Is that correct ? please explain why
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Post by Jonathan » Tue Mar 17, 2009 4:30 pm

What we want to know is the probability Bob ended the game on his second turn ("Bob2"), given that he ended it ("Bob"). That probability is probability(intersection(Bob2, Bob))/probability(Bob) (AKA P(Bob2|Bob) like Z-Man said). Bob2 is Bob ending the game on his second turn; Bob is Bob ending it on any of turn. Bob2 implies Bob per definition, so this simplifies to probability(Bob2)/probability(Bob).

Moving along, we see that the probability of ending on any turn is 1/6, and that this scales the probability of all subsequent turns by 5/6. This makes the probability of a game having ended on turn n (1/6)*(5/6)^n, where n=0 is the first. Probability(Bob2) is then easily shown to be 125/1296. Probability(Bob) is the sum of all (1/6)*(5/6)^n for positive odd n, which is 5/11 (somebody else show that). Now plug them in and you get (125/1296)/(5/11)=275/1296.
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Post by rumpole » Tue Mar 17, 2009 10:36 pm

rumpole wrote:if they are asking what I think they are asking it be (bob 5/6) * (sue 5/6) * (bob 1/6)
So I was correct then, finally i get something right :)
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Post by Jonathan » Tue Mar 17, 2009 10:56 pm

Um? That's a different answer.
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Post by rumpole » Tue Mar 17, 2009 11:08 pm

Then surely the other answer is wrong
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