I'm imagining a simple random walk on $\mathbb{Z}$ with three independent particles (maybe add laziness so they don't jump over each other). Suppose the particles are initially placed at, say, $-10$, $0$ and $10$. The distance between any two particles evolves like a random walk, so the expected hitting time for those two specific particles is infinite. Now, if we require only that any two of the three particles hit, is the expected hitting time still infinite?



Hitting time for two out of three random walk particles
The answer is the same for random walks and for Brownian motion.


If you project a $3$-dimensional Brownian motion perpendicular to $x=y=z$, you get a $2$-dimensional Brownian motion. The projection of the set where $x\lt y\lt z$ is a wedge of angle $\frac{\pi}{3}$. Your question is whether the first exit time from a $\frac{\pi}{3}$ wedge has finite expected value. This question and more is answered in Spitzer (1958) [Some theorems concerning $2$-dimensional Brownian motion.](http://www.ams.org/journals/tran/1958-087-01/S0002-9947-1958-0104296-5/S0002-9947-1958-0104296-5.pdf) *Trans. Amer. Math. Society* 87 pp. 187-197.


In that paper, theorem $2$ on page $192$ is that in a wedge of angle $\beta$, the $\delta$ power of the first exit time has finite expected value iff $2\delta \beta \lt \pi$. In your case, $\frac{2 \pi}{3} \lt \pi$ so the expected value is finite.


I believe you can also check this by the reflection principle, since the probability that you remain in a wedge of angle $\frac{\pi}{3}$ is a finite alternating sum of $6$ densities, $3$ positive and $3$ negative. You should be able to estimate the value in terms of derivatives of the normal density. However, Spitzer's work is worth checking out.


