Probability question: names in hats

[Chas]

Pro tip:  It is not about probabilities, it is about combinatorics.

For 2 players (A, B), there is 1 legal order, and 0 ways for B to end up with 'B'.  P= 0.

For 3 players (A, B, C), there are 3 legal orders  and 1 way for C to end up with 'C'.  P= 1/3.

<Recomputing>

Farting around with the combinations, I come up with n! combinations of which n!/2 are legal, and of those (n-2)! end with the last guy stuck.

So the probability P is (n-2)! / (n!/2)  -> 2/n(n-1).

For 10 players, that comes out to P = 1/45.