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醉汉过桥

Drunk Man

专题
Probability / 概率
难度
L4

题目详情

一个醉汉站在 100 米桥的第 17 米处。每一步他以 1/21/2 概率前进 1 米，以 1/21/2 概率后退 1 米。到达 0 或 100 即停止。

  1. 他先到达 100（远端）而不是 0 的概率是多少？
  2. 到达 0 或 100 之一的期望步数是多少？

A drunk man stands at the 17th meter of a 100-meter bridge. Each step, he staggers forward or backward 1 meter with probability 1/21/2. He stops upon reaching meter 0 or meter 100.

  1. What is the probability he reaches meter 100 (the far end) before meter 0?
  2. What is the expected number of steps until he reaches either 0 or 100?
解析

把坐标平移到以 17 米处为 0，左右边界分别为 −17-17 与 +83+83。

  1. 无偏随机游走命中右边界的概率为
1717+83=0.17.\frac{17}{17+83}=0.17.
  1. 无偏随机游走从 0 出发命中 ±(α,β)\pm(\alpha,\beta) 的期望时间为 αβ\alpha\beta。这里 α=83,β=17\alpha=83,\beta=17，因此
E[N]=83×17=1411.\mathbb{E}[N]=83\times 17=1411.

Original Explanation

Shift coordinates so his position is 00, with boundaries at +83+83 and −17-17. This is a symmetric random walk starting at 00. Let pαp_\alpha be the probability of hitting +83+83 first; from the gambler’s ruin result, pα=1717+83=0.17.p_\alpha = \frac{17}{17 + 83} = 0.17. So the chance of reaching the far end (100m) first is 0.170.17.

For the expected time E[N]E[N] to hit either boundary ±(α,β)\pm(\alpha,\beta) in a symmetric walk, the known formula is E[N]=αβE[N] = \alpha\beta. Here α=83\alpha=83, β=17\beta=17, so E[N]=83×17=1411.E[N] = 83 \times 17 = 1411.