Gambler's ruin with a fair coin

Question

You start with \20 and repeatedly bet \1 on a fair coin flip (win \1 or lose \1, each with probability 	frac12). You stop when you reach \100 or go broke. What is the probability you reach \100 before hitting \0?

Accepted Answer

Recurrence approach. Let hi be the probability of reaching N = 100 from bankroll i, with h0 = 0 and hN = 1. Conditioning on one flip, hi = 	frac12 h{i-1} + 	frac12 h{i+1}, which says h{i+1} - hi = hi - h{i-1}: the increments are constant, so h is linear in i. With the boundary values, hi = i/N, hence h{20} = \frac{20}{100} = \frac{1}{5}. Martingale shortcut. With fair bets, your bankroll St is a martingale, so the optional stopping theorem gives E[ST] = S0 = 20. At stopping, ST is 100 with probability p and 0 otherwise: 100p + 0(1-p) = 20 \implies p = \frac{1}{5}. You go broke with probabil…

Gambler's ruin with a fair coin

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