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## 'Weekly Problem 47 - 2012' printed from http://nrich.maths.org/

Let the hats of B, H and L be b, h and l respectively. Draw a table showing the ways in which the monkeys can select hats.

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

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

Alternatively, there are $3\times2\times1=6$ possible ways to choose the three hats.

There are two hats that B could choose.

If B chose h, then L would have to choose b and H would have to choose l.

If B chose l, then H would have to choose b and L would have to choose h.

So once B has chosen his hat the other two are fixed. So there are just the two possible alternatives out of six ways.

So the probability is $\frac{2}{6}=\frac{1}{3}$.

*This problem is taken from the UKMT Mathematical Challenges.*

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