[jay_shark]

Here is a rigorous solution to the expected number of hat matches .

Let X denote the number of matches we can compute E(x) by writing X= X1 + X2 + X3+ ...+ Xn

where Xi = { 1 if the ith person selects his own hat , or 0 otherwise }

For each i , the ith person is equally likely to select any of the N hats .

E(Xi]= P(Xi=1)= 1/N

E(X) = E(X1) + E(X2) + ... + E(Xn) = 1/N*N = 1

Hence on average , exactly one person selects his own hat .