布朗运动

设 $p$ 为布朗运动先到达 $-2$ 再到达 $1$ 的概率。

由对称性可知，从 0 出发，先到达 $-1$ 而不是 $1$ 的概率是 $\frac{1}{2}$。

到达 $-1$ 之后，以概率 $\frac{1}{2}$ 会先到达 $-2$ 而不是回到 0；如果先回到 0，那么问题又重新开始，此时成功概率仍是 $p$。因此

$$p = \frac{1}{2}\left(\frac{1}{2} + \frac{1}{2}p\right)$$

解得

$$4p-p=1 \quad\Rightarrow\quad \boxed{p=\frac{1}{3}}$$

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Original Explanation

Denote $p$ as the probability you hit -2 before you hit 1.

From symmetry, the probability that the Brownian motion hits $-1$ before it hits $1$ is $1/2$.

When you are at $-1$, with probability $1/2$ you hit $-2$ before hitting zero.

Now if you are at zero, then the probability is once again $p$ since the problem has restarted. Therefore,

$$p = \frac{1}{2}(\frac{1}{2} + \frac{1}{2}p) \rightarrow 4p-p = 1 \rightarrow \boxed{p = \frac{1}{3}}$$