Random Walk: Drunk Man

A drunk man is at the 17th meter of a 100-meter-long bridge. He has a 50% probability of staggering forward or backward one meter each step. What is the probability that he will make it to the end of the bridge (the 100th meter) before the beginning (the 0th meter)? What is the expected number of steps he takes to reach either the beginning or the end of the bridge?

Answer

Here we are dealing with a simple symmetric random walk with a starting point at \(x_0 = 17\) and with two absorbing barriers at \(0\) and \(100\). From this we infer that \(a=17\) and \(b=100-17=83\). Hence,

  1. Hitting probability.

    \[ P(\text{hit }100\text{ before }0) = \frac{a}{\,a+b\,} = \frac{17}{100}. \]

  2. Expected absorption time.

    \[ E[\text{steps}] = a\cdot b = 17\times83 = 1411. \]
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