Which barrier does the random walk hit first?

Question

A simple symmetric random walk starts at $0$ and takes $\pm 1$ steps with probability $\tfrac12$ each. It stops upon reaching $-a$ or $+b$ (with $a, b > 0$ integers).

What is the probability the walk exits at $+b$ rather than at $-a$ ?

Accepted Answer

Martingale route. For a symmetric walk, St is a martingale: E[S{t+1}\mid St] = 	frac12(St+1) + 	frac12(St-1) = St. The exit time T is finite (bounded interval), so optional stopping gives E[ST] = S0 = 0. The walk exits at either -a or +b. Let p = P(ST = b). Then E[ST] = (-a)(1 - p) + b\,p = 0 \implies p = \frac{a}{a + b}. So the probability of reaching the far barrier +b is \dfrac{a}{a+b}, and of reaching -a is \dfrac{b}{a+b}. The chance of exiting a given side is proportional to the distance to the opposite barrier: the closer barrier is easier to hit. Example. Barriers at -2 and +8: P(+8)…

Which barrier does the random walk hit first?

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