We have 2 players, A and B, competing. The probability that A wins a match is p, making the probability that B wins a match (1-p) = q.
The game is won by player A as soon as he gets one more win than B, whereas B wins when he gets ahead x won matches.
What is the probability P(B wins) of A losing the game (i.e. B getting ahead x won matches before A gets ahead one)?
The difficulty for me is the unbalance in the game and the fact that there is a seemingly infinite amount of sequences that need to be considered. My experience with such games is too little to find an elegant solution.
Any help is appreciated.
Let $A_n$ (resp. $B_n$) be the number of matches won by $A$ (resp. $B$) by the $n$th match. Then, $$ X_n = A_n - B_n $$ is a random walk with transition probabilities
\begin{align} X_{n+1} = \left\{\begin{array}{ccc} X_n + 1 & \mbox{with prob.} & p;\\ X_n - 1 & \mbox{with prob.} & 1-p, \end{array}\right. \end{align} and $X_0=0$. You are interested in the probability that $X_n$ hits $-x$ before hitting $1$, and the probability of $X_n$ hitting $1$ before $-x$. I believe this problem is equivalent to the Gambler's ruin problem. The expressions for these probabilities and their derivation can be found in Feller's book "Introduction to Probability Theory and its Applications : Volume 1".