A near-broke gambler against a huge target

Question

A gambler has just \1 and plays a fair \1 coin-flip game, aiming for \100 before going broke at \0. Each round he gains or loses \1 with equal probability, stopping at \0 or \100. On average, how many rounds does he play before stopping?

Accepted Answer

The bankroll is a symmetric random walk on 0, 1, \ldots, 100 absorbed at both ends, starting at i = 1. The expected duration from position i is Di = i(N - i), \qquad N = 100. Starting at i = 1: D1 = 1 \cdot (100 - 1) = 99 	ext{ rounds.} This is the surprise: even though he is standing right next to the \0 cliff, and will almost certainly go broke (probability 99/100), the game still lasts 99 rounds on average. A fair walk wanders a long time before it is absorbed, and the rare runs toward \100 are extremely long, pulling the average way up. Takeaway. For a fair \pm 1 walk on 0..N the expecte…

A near-broke gambler against a huge target

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