Solution to the Plane Boarding Puzzle

This started as a simple response to Aunt Joan's proposed solution and turned into... um...

Ahem.
The clearest rigorous argument for the 1/2 result, I think, is like this. Think of it like a game: if the 100th guy gets his seat, we win; else we lose.
Also, change the rules and make the people more aggressive: so if you find your seat is taken, you don't move; you kick the offending fellow out to a random seat. This doesn't change which seats are occupied at any time--the thing which actually matters, since we're trying to figure out whether the 100th seat will be occupied when the 100th guy boards. But it makes it so that there's only one guy in the wrong seat at a given time--namely, the original confused guy.
So if the original confused guy gets kicked out and lands in his own seat, at that point everyone who has boarded has their assigned seat. So everyone else can board without confusion, and the 100th guy gets his seat--we win!
On the other hand, if the original confused guy gets kicked out and lands in the 100th guy's seat, he won't be kicked out again until the 100th guy comes to board and finds his seat taken. So we lose.
And in fact, this is the only way we can win or lose, because by the time we get to the 100th guy, either his seat, or the original guy's seat, will be taken by someone who got kicked there.
But, (the punchline), the win condition at each step (kick into first seat) is exactly as probable as the loss condition (kick into last seat)
So the total probability of winning must be the same as the total probability of losing--the situation is symmetric.
QED! :D