Q: 12 men leave their hats with the hat check. If the hats are randomly returned, what is the probability that nobody gets the correct hat?

A: Suppose we are handing out hats to n people. First we start with all the possible outcomes. Then we subtract off those that assign the right hat to a given person, for each of the n people. But this double-counts each outcome that assigned 2 hats correctly, so we have to add those outcomes back in. But now we've counted one net copy of each outcome with 3 correct hats in our set, so we have to subtract those off again. But now we've taken away each 4-correct-hat outcome once too often, and have to put it back in, and so forth ... until we add or subtract the outcome that involves all n people getting the correct hats.

Putting it all in probabilities, the measure of the original set is 1. For a given subset of k people, the probability that they all get their correct hats is (n-k)!/n!, while there are (n choose k) such subsets of k people altogether. But then

(n choose k)*(n-k)!/n! = (n!/((n-k)!*k!))*(n-k)!/n! = 1/k!

is the total probability measure we get by counting each subset of k people once each. So we end up generating the finite series

1 - 1/1! + 1/2! - 1/3! +- ... +/- 1/n!

which is of course just the first n+1 terms of the Taylor series expansion for f(x) = e^x centered at 0 and evaluated at -1, which converges to 1/e quite fast. You can compute the exact probability for any n >= 1 simply by rounding n!/e to the nearest whole number, then dividing again by n!. Moreover I think you will find you are always rounding down for odd n and rounding up for even n.

For example,

12! / e = 176214840.95798...

which is within 0.05 (absolute error, not relative) of the correct intermediate result, 176214841.

Another fact is that if you set the probability of no matching hats equal to m/n!, then m is an integer divisible by (n-1). That's because the number of ways you can give hat #2 to person #1 is exactly the same as the number of ways you can give that person hat #3, likewise for hat #4, and so forth.

-- David Karr (karr@cs.cornell.edu)