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The drunkard near the cliff

Asked at DE Shaw, Optiver

A drunkard stands k steps from the edge of a cliff at position 0. Each step they stagger away from the cliff with probability p and toward it with probability 1 - p. There is no wall on the other side, the line runs to infinity. With p = 0.4 the pull toward the cliff wins, so a fall is certain.

Estimate the expected number of steps until they fall (hit 0) by Monte Carlo, then derive the closed form.

k = 5, p = 0.4   ->  ~25 steps
Show a hint

There is only one barrier, so the parabola formula for two barriers does not apply. Use the drift: on average, each step moves the drunkard (p)(1p)=2p1(p) - (1-p) = 2p - 1 toward safety, which is negative here.

Your answer

This one is open-ended. Work it through, then check your reasoning against the full solution.

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