Question 2.31: At a party N men throw their hats into the center of a room.......

Letting X denote the number of men that select their own hats, we can best compute E[X] by noting that

where

Now, because the ith man is equally likely to select any of the N hats, it follows that

$P\{X_{i}=1\}$ = P{ith man selects his own hat} = $\frac{1}{N}$

and so

Hence, from Equation (2.11)

$E[a_{1}X_{1}+a_{2}X_{2}+\cdot\cdot\cdot+a_{n}X_{n}]=a_{1}E[X_{1}]+a_{2}E[X_{2}]+\cdot\cdot\cdot+a_{n}E[X_{n}]$       (2.11)

we obtain

Hence, no matter how many people are at the party, on the average exactly one of the men will select his own hat.