How many flips until the gambler stops?

Question

You start with \20 and repeatedly bet \1 on a fair coin flip, stopping when you reach \100 or go broke. You already know the probability of reaching \100 is 	frac15. What is the expected number of bets you make before stopping?

Accepted Answer

Let Di be the expected number of bets still to come starting from bankroll i, with D0 = D{100} = 0 (you have already stopped at either boundary). Conditioning on the next flip: you make one bet, then land at i-1 or i+1 with equal probability: Di = 1 + 	frac12 D{i-1} + 	frac12 D{i+1}. Rearranging, D{i+1} - 2Di + D{i-1} = -2: the second difference is constant, which means Di is a downward parabola in i. The parabola vanishing at i = 0 and i = N with the right curvature is Di = i\,(N - i). You can verify the curvature: for f(i) = i(N-i), the second difference f(i+1) - 2f(i) + f(i-1) = -2, matc…

How many flips until the gambler stops?

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