Find Probability that a player will win Nth match 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.
 A: 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 $(0,1)$.
A: 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 $$P_N=\tfrac12-\tfrac12(1-2P)^{N}.$$
A: @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 $P_N$ ?
