Probability of Brownian Motion hitting -2 before 1?

Question

Why is the probability of Brownian Motion hitting -2 before 1 is equal to 1/3? This is an interview question asked for Quant roles.

I found a similar question was previously asked: Brownian motion interesting question. However, the given solution does not answer my question.

Answer 1

We can compute the probability that $a$ is hit before $-b$ using the optional stopping theorem for martingales. Observe that $$E(B(\tau)) = E(B(0)) = 0$$ where $\tau = \inf\{t \geq 0 : B(t) \in \{a, −b\}\}$ where $a,b>0.$ Thus we have that $$a p - b(1-p) = 0\implies p =\frac{b}{a+b}$$ Your answer is thus $1-p = \frac{a}{a+b}.$ Take $a=1$ and $b=2$ and observe that $1-p=\frac{1}{3}.$

Answer 2

Denote $p$ 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 hit zero.

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

$$p=1/2 ( 1/2 + 1/2p) \Rightarrow 4p-p=1 \Rightarrow p=1/3. $$