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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.

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3 Answers 3

up vote 2 down vote accepted

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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@Didier: You may have missed "Player who loses starts new game." –  Henry Mar 18 '12 at 12:53
    
Indeed I did. Thanks. –  Did Mar 18 '12 at 13:05
    
@Henry +1ed and marked as answer. Thanks! –  jerrymouse Mar 18 '12 at 13:36

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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Thanks for your answer. The recursion function that is giving me required answer is: PN+1=PN(1−P)+P(1−PN) However I am unable to solve it for PN. Please help. –  jerrymouse Mar 18 '12 at 12:27
    
I mean assuming P=0.2, these should be the answers for N-- 1: 0.2000, 2: 0.3200, 3: 0.3920, 4: 0.4352 –  jerrymouse Mar 18 '12 at 12:52
    
I am getting these answers from the second recursion formula, not from the first. These are standard answers given to me. I need to solve for other Ns –  jerrymouse Mar 18 '12 at 12:52
    
@jerrymouse: Indeed I misread the rules of the game and first thought that the player winning a game was starting the new game. Answer modified: the technique to compute $P_N$ is again to consider the simpler recursion formula on $P_N-\frac12$. –  Did Mar 18 '12 at 13:03
1  
You may want to check the $+$ in your final line –  Henry Mar 18 '12 at 13:06

@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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If you want detailed calculations, see Problem Sets 2 and 5 (and their solutions) on this course website –  Dilip Sarwate Mar 18 '12 at 12:55
    
Here's a working link for Dilip Sarwate's broken link: courses.engr.illinois.edu/ece313/sp2009/Problems.html –  raindrop May 31 '13 at 5:22

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