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So I found myself in a infinite loop while trying to do some probability. If A and B are independent, calculating P(A ∩ B) is as simple as P(A)P(B). However, how do I calculate P(A ∩ B) if they are dependent? I know P(A∣B)⋅P(B)=P(A∩B).

example: Charlie and Doug take turns rolling two dice, Charlie goes first. If Charlie rolls a sum of 6 before Doug rolls a sum of 7, Charlie wins. What is the probability Charlie wins.

Solution attempt:

P(Charlie wins) = P(Charlie rolls 6, 1st turn) + P(Charlie rolls 6 | Charlie and Doug don't win previous rounds)
= 5/36 + P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds)*P(Charlie and Doug don't win previous rounds)
= 5/36 + P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds)*P(Charlie doesn't win previous)P(Doug doesn't win previous)
= 5/36 + P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds)(31/36)*(30/36)

Now this is where I'm stuck, as I don't know how to calculate P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds). Can anyone give me some pointers?

jathu
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  • $P(A\mid B) \cdot P(B) = P(A \cap B)$. Of course that only helps if you know the conditional probability. – Daniel Fischer Jun 26 '13 at 23:30
  • I know P(A∣B)⋅P(B)=P(A∩B), as I used it above. However, I do not know the conditional probability, that is what I need to figure out essentially. – jathu Jun 26 '13 at 23:39
  • Ah, sorry. Couldn't see it for all the words. – Daniel Fischer Jun 26 '13 at 23:40
  • It's alright, my question layout is a little messy. Thanks for trying to help. – jathu Jun 26 '13 at 23:44
  • I think you should try a different angle. Let $X$ be the probability that Charlie wins when Charlie goes first, and $Y$ the probability that Doug wins when Doug goes first. Then you very quickly cycle back to where you started (with a few coefficients), and get something like $X = aX + bY + c$ and $Y = dX + eY + f$ if I'm not mistaken. – Daniel Fischer Jun 26 '13 at 23:47
  • I don't think the first equation is right. Note that Charlie rolling a 6 is independent of what happens previously. So (I'll abbreviate) we have $P(CR6|CaDdwpr)=P(CR6)$. So I think that second term should just be $P(CR6\cap CaDdwpr)=P(CR6)P(CaDdwpr)$. This doesn't really help solve the problem, because doing it with this method seems to set up an infinite sum. – Matt Jun 26 '13 at 23:57

1 Answers1

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Let $p$ be the required probability that Charlie wins the game. Charlie can win in two ways, fast or slow: (i) Fast: Charlie wins immediately, probability $\frac{5}{36}$; (ii) Slow: Charlie doesn't win immediately, neither does Doug on his first try, but Charlie ultimately wins, probability $\left(\frac{31}{36}\cdot \frac{30}{36}\right)p$. Thus $$p=\frac{5}{36}+\left(\frac{31}{36}\cdot\frac{30}{36}\right)p.$$ Solve for this linear equation for $p$.

Remark: Without explicitly saying so, we have used conditional probability. The probability that Charlie (ultimately) wins given that neither player wins on his first try is just $p$.

We can instead use your general strategy. The probability that Charlie wins on his first try is $\frac{5}{36}$. The probability that Charlie wins on his second try is $\left(\frac{31}{36}\cdot\frac{30}{36}\right)\frac{5}{36}$. The probability Charlie wins on his third try is $\left(\frac{31}{36}\cdot\frac{30}{36}\right)^2\frac{5}{36}$. And so on. For the probability Charlie wins, we add up the infinite geometric series with first term $\frac{5}{36}$ and common ratio $\frac{31}{36}\cdot\frac{30}{36}$. We can find the sum using the ordinary formula for the sum of an infinite geometric series.

André Nicolas
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  • How is it that you can just make P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds) = P(Charlie wins)? – jathu Jun 27 '13 at 00:20
  • I have added to the Remark at the end an alternate way to the answer which is closer in spirit to what you tried. But for my "faster" way, note that the probability Charlie ultimately wins given that Charlie didn't toss an immediate $6$, and Doug did not win on his first try, is just $p$, since in effect the game begins again, with again Charlie having first toss. – André Nicolas Jun 27 '13 at 00:29
  • Wow, thanks! :D – jathu Jun 27 '13 at 00:41