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'Monkey Puzzle' printed from https://nrich.maths.org/
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
$$\frac{2}{6}=\frac{1}{3}$$