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