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Here is the problem: $A$ and $B$ roll a dice taking turns with $A$ starting this process. Whichever one rolls the first $6$, wins. Find the probability of $A$ winning.

I know how to solve this problem using a geometric sequence summation. I came across this problem which said "Hint:you will have to use a geometric series to find the answer".

I began to wonder if there was an alternative solution which doesn't use geometric series. I am not sure where to begin even. Does anyone know of such a solution? I am talking about an analytical solution not a simulation or a monte carlo method of doing it.

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  • $\begingroup$ It appears that the major step in any solution is to use conditional probability. If both Player A and Player B fail to roll a 6 on a given turn, then you're back at square 1, as far solving the problem. $\endgroup$ – Matt Rosenzweig Jun 27 '14 at 17:46
  • $\begingroup$ Yea, I see that, but not sure how to incorporate that into a solution. $\endgroup$ – user1775614 Jun 27 '14 at 17:48
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Let $p$ be the probability of $A$ winning. If $A$ doesn't win with the first move, it is as if $A$ and $B$ had swapped their roles. Thus, $p=\frac{1}{6}+\frac{5}{6}(1-p)$

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  • $\begingroup$ Great! This agrees with the geometric sequence that you would get sum(1/6 times (5/6)^(2k) ) with k ranging from 0 to infinity which evaluates to 6/11and agrees with your solution!! Thanks!! $\endgroup$ – user1775614 Jun 27 '14 at 18:29

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