Yep, that works =) That's a nice conceptual shortcut, actually. The way he did it you don't even have to look for a pattern by getting your hands dirty, you just have to evaluate the anchor case and use the logic stated.
I'll explain what he did. Let's say the bug is on its nth move. If the bug got to the starting point on its (n-1)th move (lol @ notation) then it has a zero probability of getting to the starting point on its nth move. So you take the probability that it DIDN'T get to the starting point on the (n-1)th move, which is 1 - a_n-1, which implies that the bug is on one of the two vertices that's not the starting point. In this situation, the bug has two places to move to, and each are equally likely, so you take half the probability, to make it (1 - a_n-1) / 2.
I probably should have used LaTeX on that hehe... I may get around to prettifying it later.
By the way, why doesn't Nick visit these forums?
(i.e. without the iago proxy ;-))