回复：another way to look at Quant面试题：喝醉了（2.5星）

This is a first hitting time problem of the symmetric random walk.

Assume the poor drunk guy starts from n steps far away from left door(n<=99), define M(0)=0, and M(t+1)=M(t)+1 if he walks to the right or M(t+1)=M(t)-1 if he walks to the left, let tau(stopping time) be the steps he takes to reach either -n or (100-n), then the problem here is to find E[tau].

since M(t) here is martingale, it's easy to show that E[tau]=n(100-n), here we have n=1, therefore E[tau]=99.

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• 回复：回复：another way to look at Quant面试题：喝醉了（2.5星） -tianxing- ♂ (8 bytes) () 10/24/2009 postreply 23:49:10