This question is under the topic of Brownian Motion. The question:

I don't understand how the P{down 2 before up 1} can translate into 1/3. What's the logic behind that?

Thanks a lot!

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Call $u(x)$ the probability that the path hits $-1$ before $2$, starting from $x$ in $[-1,2]$. Then $u(-1)=1$ and $u(2)=0$. Furthermore, for every $x$ in $(-1,2)$, consider some positive $z$ such that $-1\lt x-z$ and $x+z\lt 2$. To hit $\{-1,2\}$ starting from $x$, one must hit $\{x-z,x+z\}$ first. By symmetry, one hits $x-z$ or $x+z$ first with equal probabilities. Once at $x\pm z$, one must hit $\{-1,2\}$, and the point where one hits first does not depend on the way the path reaches $x\pm z$ (this is not obvious, but is called the strong Markov property).
All this yields $u(x)=\frac12(u(x+z)+u(x-z))$ for every suitable $x$ and $z$. Thus, the function $x\mapsto u(x)$ is affine. Since one knows $u$ at the boundary points, this yields $u(x)=\frac13(2-x)$, in particular the quantity you are after is $u(1)=\frac13$.