Brownian motion reaching minus two before one

Question

A standard Brownian motion starts at 0. What is the probability that it reaches $-2$ before it reaches $1$ ?

Short answer

$1/3$ . Brownian motion is a martingale, so by optional stopping $0=-2p+1\cdot(1-p)$ , giving $p=\tfrac13$ . In general the chance of hitting $-a$ before $b$ is $\tfrac{b}{a+b}$ .

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Lessons behind this question

Sample Spaces, Events, and the Probability Axioms
Gambler's Ruin and First Passage Times
Simple Random Walks — Symmetric and Asymmetric

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