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Please help me with this question:

  • Player "A" starts the first game.
  • Player who starts a game has probability "P" of winning that game.
  • Player who loses starts new game.

Assuming this series continues infinitely, whats the probability "A" will win Nth game.

Taking P=0.2, I initially thought for Nth probability as:

  • 1: 0.2
  • 2: 0.8*0.2
  • 3: (1-0.8*.02) * 0.2

But its wrong. Can someone please help me in right direction.

3 Answers 3

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@didier I think the recursion equation should be:

$ P_{N+1}=P_N(1-P) +P(1-P_N) $ This recursion equation gives me right answer. How do we solve this to get $P_N$ ?

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    Here's a working link for Dilip Sarwate's broken link: http://courses.engr.illinois.edu/ece313/sp2009/Problems.html2013-05-31
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Call $P_N$ the probability that player A wins game $N$ against player B. Then player A may win game $N+1$ either because player A won game $N$ and player B started game $N+1$ and lost it, which happens with probability $P_N(1-P)$, or because player A lost game $N$ and started game $N+1$ and won it, which happens with probability $(1-P_N)P$. Hence $P_1=P$, and, for every $N\geqslant1$, $ P_{N+1}=P_N(1-P)+(1-P_N)P. $ This recursion is equivalent to $P_{N+1}-\frac12=(1-2P)(P_N-\frac12)$ hence, for every $N\geqslant1$, $ P_N=\tfrac12\left[1-(1-2P)^N\right]. $ Note: As was to be expected, there is a loss of the initial conditions in the sense that $P_N\to\tfrac12$ when $N\to\infty$, for every $P$ in $(0,1)$.

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    @DidierPiau +1ed Thanks for putting me on right track. Cheers!2012-03-18
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As you say, the recursion is $P_{N+1}=P_N(1-P) +P(1-P_N)$ which, using Didier Piau's method, is equivalent to $P_{N+1}-\frac12=(1-2P)(P_N-\frac12)$ so so with $P_1=P$ $P_{N+1}-\frac12=(1-2P)^N (P_1-\frac12)=-\tfrac12(1-2P)^{N+1}$ and you get $P_N=\tfrac12-\tfrac12(1-2P)^{N}.$

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    @Henry +1ed and marked as answer. Thanks!2012-03-18