Gambler's Ruin

Question

You start with 30 and bet 1 on fair coin flips (win 1 on heads, lose 1 on tails), stopping when you reach 100 or go broke. What is the probability you reach 100?

Accepted Answer

For a fair game your bankroll X_t is a martingale: mathbbE[X_t+1 mid X_t] = X_t. The game stops at T when X_T in 0, 100. By the optional stopping theorem (valid here, bounded process, finite expected stopping time): 30 = mathbbE[X_T] = 100 cdot p + 0 cdot (1-p) implies p = frac30100 = 0.3. In general, from k aiming for N: p = k/N. Approach 2, first principles Let p_k be the probability of reaching N from k. Then p_k = frac12 p_k+1 + frac12 p_k-1 with p_0 = 0, p_N = 1. The recurrence says p is linear in k, so p_k = k/N, same answer without invoking martingales, but the martingale argument generalizes much further (biased games, expected durations). Trading relevance: this is the cleanest model of risk-of-ruin, and the reason position sizing matters even in zero-edge games. With any negative edge the formula becomes exponential and ruin becomes near-certain, see the Kelly questions.

Gambler's Ruin

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