Problem: we have n people sitting around in a circle. I tap one of them, and then go on a drunken/random walk going in either direction with equal probablility, tapping people as I go on my merrily drunken way. For each person besides the first, what is the probability he/she will be the last one I tap?

Consider the probability of not hitting the guy to A’s right. Clearly it is equal to the probability that he is the last one hit. Now, for any dude in the middle X, the probability that he is the last one to be hit is equal to the sum of: “we go and hit the guy to X’s left and then never hit X” and “we go and hit the guy to X’s right and then never hit X.” Notice that because you have to hit one of them first at some point, the sum of these two is just: “we go and tap a person adjacent to X and then never hit X.” But note this is exactly equal to the probability of not hitting the guy to A’s right. Thus, the probability of being tapped last is equal for everyone besides A, and is 1/(n-1).

I thank Boris Alexeev for giving me this problem.

-YZ