Brownian motion interesting question

Question

I found this interesting question on the internet, but unfortunately I could not solve it.

What is probability that Brownian motion (starting at origin) has value 1 before having value -2?

Answer 1

Take $B$ to be a Brownian motion. Let

$$\tau = \inf\{t\geq 0: B_t\geq 1, B_t\leq -2\}.$$

Then $\tau$ is a stopping time. $B$ is a martingale, so $B^\tau$ - $B$ stopped at time $\tau$, is also a martingale. $B^\tau$ is uniformly integrable, since it is bounded, so we may apply the Optional Stopping Theorem to $B^\tau$ at time $\tau$ to get

$$ 0 = \mathbb{E}[B^\tau_\tau] = \mathbb{E}[B_\tau] = 1\mathbb{P}[B \text{ hits 1 before -2}]+(-2)\mathbb{P}[B \text{ hits -2 before 1}]. $$

Since precisely one of these events occurs, we may solve for $\mathbb{P}[B \text{ hits 1 before -2}]$, giving the probability as $\frac23$.