Visits to 1000 before 0 for a walk started at 5

Question

A simple symmetric random walk on the integers starts at $5$ ; each step moves $+1$ or $-1$ with probability $\tfrac12$ . What is the expected number of times the walk is at $1000$ before it first reaches $0$ ?

Short answer

$10$ . By the gambler's ruin formula the walk reaches 1000 before 0 with probability $\tfrac5{1000}$ . From 1000, it comes back before hitting 0 with probability $\tfrac12+\tfrac12\cdot\tfrac{999}{1000}=1-\tfrac1{2000}$ , so the number of visits is geometric with mean 2000. The answer is $\tfrac{5}{1000}\cdot2000=10$ .

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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