Interesting math puzzle.

[Thudlow_Boink]

@MentalGuy

Here is a previous thread about this. The link to the puzzler is broken now, but it is the same problem.

I think the version in that thread presumes that the “crazy person” is the first passenger to board. But it turns out that that doesn’t affect the solution.

Proof by induction on n, the number of passengers who get on before you, that the answer is 50%:

If n = 1, that one passenger before you is the crazy person, and there’s a 50-50 chance he chooses your seat.

Now assume it’s true for n = k, and consider what happens when n = k+1.

If the first passenger to board is not the crazy passenger, there are k remaining passengers ahead of you, so by the inductive hypothesis there’s a 50% chance you’ll get your own seat.

If the first passenger to board is the crazy passenger, then the argument Xema gave in the other thread applies:

@Xema

How about an argument for 50%, based on symmetry:

There are two special seats: yours (call it Y), and the one that was assigned to the first passenger to board (call it F). If any passenger after the first happens to sit in F before Y is occupied, you will get to sit in Y. But if anyone grabs Y before F is occupied, you will not get to sit in Y. As there is nothing to make either of these more likely than the other, the chances are equal.

Does that work?