Probability about hitting time of Brownian motion

Question

The question is as following:

Denote $T_a = inf\{t\geq 0|B_t =a\}$, how to compute $P(T_1<T_{-1}<T_2)$?

Answer 1

Sketch:

For this to happen, the first exit from $(-1,1)$ started at $0$ must be through $1$. Compute the probability of this happening.

Next, compute the probability that the first exit from $(-1,2)$ of a BM started at $1$ is through $-1$. (I will point out for you that the result here is not the same as the result for the first part.)

Now use the strong Markov property to infer that the desired probability is the product of these two.

Answer 2

Now I have figure out how to compute $P(T_{-1}<T_2)$.

Let $T=min(T_{-1},T_2)$, which is a new bounded stopping time. By bounded stopping time theorem, we have $E(B_{T})=E(B_0)=-1P(T_{-1}<T_2)+2P(T_{-1}>T_2)=2-3P(T_{-1}<T_2)$, then $P(T_{-1}<T_2)=\frac{2}{3}$