Ruin probability for a biased random walk

Question

A gambler starts with \3 and makes repeated bets, each winning \1 with probability p = 0.6 and losing \1 with probability q = 0.4, independently. She stops at \0 (broke) or \6 (target). What is the probability she reaches \6 before going broke?

Accepted Answer

Because p 
e q the walk drifts, so St is not a martingale. The right object is Mt = (q/p)^{St}. Check it is a martingale with ratio r = q/p: E\!\left[r^{S{t+1}} \mid Stight] = r^{St}\left(p \cdot r + q \cdot r^{-1}ight) = r^{St}\left(p \cdot 	frac{q}{p} + q \cdot 	frac{p}{q}ight) = r^{St}(q + p) = r^{St}. Let the walk start at i = 3 with absorbing barriers at 0 and N = 6, and let P be the probability of reaching N. Optional stopping gives E[r^{ST}] = r^{i}, and expanding over the two exit states yields the gambler's-ruin formula P = \frac{1 - (q/p)^{\,i}}{1 - (q/p)^{\,N}}. Here q/p =…

Ruin probability for a biased random walk

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