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