Mean turns for a pawn to fall off a lane

Question

A pawn sits on square 4 of a lane numbered 0 to 8. Each turn it slides one square, up or down with equal probability 	frac12, and it drops off the lane when it reaches square 0 or square 8. What is the expected number of turns until the pawn drops off?

Accepted Answer

Let ek be the expected number of turns from square k, with e0 = e8 = 0. First-step analysis gives ek = 1 + 	frac12 e{k-1} + 	frac12 e{k+1}, with parabolic solution ek = k(8 - k). At k = 4 (the exact middle), e4 = 4 \cdot 4 = 16. Takeaway. Expected absorption time is k(N-k), largest right at the midpoint; the center of a length-8 lane takes 16 turns on average.

Mean turns for a pawn to fall off a lane

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