Let the hats of monkeys A, B and C be a, b and c respectively.
This table shows the ways in which the monkeys can select hats.

A B C
a b c
a c b
b a c
b c a
c a b
c b a

In only two of the six ways do none of the monkeys have the same hat as when they arrived, hence the required probability is

$\frac{2}{6}=\frac{1}{3}$.

This problem is taken from the UKMT Mathematical Challenges.