# Conditional Gambler's Ruin

I've learned about the most canonical gambler's ruin problems, but what if winning or losing on a previous turn changes the probability of winning or losing on the following turn? Say each turn I either win or lose. Every turn after winning, I win with probability p, and every turn after losing, I win with probability q. How would I compute the long term proportion of turns won. Say I want to approximate the number of turns I won after n turns.

You can study this using a Markov chain with two states (1=post-win, 2 = post-loss), and transition matrix $$\pmatrix{ p & 1-p\cr q & 1-q\cr}$$ The equilibrium probabilities are $q/(q+1-p)$ and $(1-p)/(q+1-p)$. The long term proportion of wins is then $q/(q+1-p)$.
• Besides the asymptotic result, how would you address the question for finite $n$? – megas Dec 14 '14 at 21:26
• If $T$ is the transition matrix, $$T^n = \dfrac{1 - (p-q)^n}{1+q-p} T + \dfrac{(p-q)^n - p + q}{1 + q - p} I$$ – Robert Israel Dec 14 '14 at 21:48