How long the coin-flip gambler lasts

Question

A gambler starts with \3. On each round she bets \1 on a fair coin flip: heads she gains \1, tails she loses \1. She stops as soon as she is either broke (\0) or has reached her target of \8. On average, how many rounds does she play before stopping?

Accepted Answer

Her bankroll is a symmetric random walk on the integers 0, 1, \ldots, 8 that stops when it hits either edge. Let Di be the expected number of rounds remaining when she has i, with D0 = D8 = 0 (already stopped). From any interior amount i, one round passes and she moves to i+1 or i-1 with equal chance: Di = 1 + 	frac12 D{i-1} + 	frac12 D{i+1}. This recurrence has the neat closed-form solution Di = i\,(N - i), \qquad N = 8. You can verify it satisfies the recurrence: i(N-i) has constant second difference -2, and 1 + 	frac12[(i-1)(N-i+1) + (i+1)(N-i-1)] = 1 + i(N-i) - 1 = i(N-i). It also give…

How long the coin-flip gambler lasts

Your answer

Discussion

Learn the concepts

Related questions