A frog on stones 0 to 20 aiming for the far bank

Question

A frog sits on stone 9 in a line of stepping stones numbered 0 to 20. Each hop it jumps to an adjacent stone: toward stone 20 with probability 	frac12, toward stone 0 with probability 	frac12. Stone 20 is the far bank (safe) and stone 0 is open water (lost); the frog stops on reaching either. What is the probability the frog reaches stone 20 before stone 0?

Accepted Answer

Let pk be the probability of reaching 20 before 0 from stone k, with p0 = 0 and p{20} = 1. Each hop is a fair step, so pk = 	frac12 p{k-1} + 	frac12 p{k+1}, which makes pk linear in k: pk = k/20. Thus p9 = \frac{9}{20} = 0.45. Takeaway. For a fair walk between two ends, distance is destiny: the chance of reaching the far end first is k/N, here 9/20.

A frog on stones 0 to 20 aiming for the far bank

Your answer

Discussion

Learn the concepts

Related questions