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## 'Monkey Puzzle' printed from http://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

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

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