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Disclaimer: I am not looking for a solution to this, just a clue to help me arrive by myself to the right solution.

Two players are rolling a die, and the first one to get three ones, wins. So far, player A has gotten two ones, and player B has gotten one. What is the probability that player A wins?

What I am struggling with is the fact that the game can go on forever, so I am not sure of how to approach the problem. Any guidance is greatly appreciated.

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  • $\begingroup$ Have you heard of a geometric random variable? $\endgroup$ – Indefinite Sep 10 '17 at 12:20
  • $\begingroup$ "The game will go on forever" should translate to an infinite geometric series in your solution. $\endgroup$ – iamwhoiam Sep 10 '17 at 12:21
  • $\begingroup$ I have not heard of it @Indefinite , and through the research I've been doing, it seems a geometric series is not strictly necessary? $\endgroup$ – Bee Sep 10 '17 at 12:22
  • $\begingroup$ Letting $P_A(a,b)$ (resp. $P_B(a,b)$) denote the probability that $A$ wins given that $A$ has thrown $a$ ones and $B$ has thrown $b$ given that it is $A's$ turn (resp. $B's$ turn), we get a simple backwards induction. $\endgroup$ – lulu Sep 10 '17 at 12:23
  • $\begingroup$ Note: you did not specify whose turn it was. That is clearly relevant. $\endgroup$ – lulu Sep 10 '17 at 12:26
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It may be easier for you to cnsider 2 cases: 1) player 2 only tosses 0s before player 1 wins, 2) player 2 tosses one 1 before player 1 wins. Also for simplicity denote player 1 as A and player2 as B. Clearly the last step will be A. Since you did not specify who tossws first, i just assume it is A. So your possible outcomes are A, ABA, ABABA, etc to infinity. This is Geomerric series you can wasily compute, since p(a)=p(b)=0.5.

Can you handle the second case yourself? Since they are mutually exclusive, jyst sum the outcomes.

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