发信人: pcasnik (pcasnik), 信区: BrainTeaser
标 题: Re: 一个老题:醉汉过桥
发信站: BBS 未名空间站 (Wed Dec 3 15:36:11 2008)
presumably, as long as he gets to one end of the bridge, it is considered that he has walked over the bridge?
then it is a typical random-walk with two barriers. if you know optional stopping theorem, the easiest way to get the answer is to construct the following martingale:
x(t)*[x(t)-100]-t.
note x(0)=17, easy to get the average time to reach either end of the bridge is 17*83=1411 seconds.
for reference, try google "gambler's ruin".
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※ 修改:·pcasnik 於 Dec 3 15:37:59 2008 修改本文·[FROM: 75.34.]
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