## Four Hats

#### Problem

A teacher shows three clever boys a pile of four hats: two blue and two red. They are all blind-folded, each given a hat to wear at random, and lined-up in a line so that they are all facing towards a wall. When the blind-folds are removed, the first boy can only see the wall, the second boy can see the first boy, and the third boy can see the first two boys. None of the boys can see the colour of his own hat.

The first boy to correctly shout out the colour of his own hat will have no homework that evening. However, if he guesses incorrectly, he will have to complete the other boy's homework.

Which of the boys, 1 to 3, is most likely to correctly deduce by logic alone the colour of his hat?

#### Solution

There are exactly six ways in which the four hats can be allocated:

Option | Boy 1 | Boy 2 | Boy 3 | Not Used |
---|---|---|---|---|

1 | B | B | R | R |

2 | B | R | B | R |

3 | B | R | R | B |

4 | R | B | B | R |

5 | R | B | R | B |

6 | R | R | B | B |

If boy 3 can see two blue hats (option 1) or two red hats (option 6) then he will know the colour of his own hat and quickly shout out the colour. Therefore the probability that boy 3 correctly deduces the colour of his own hat is $\frac{2}{6} = \frac{1}{3}$.

If boy 3 sees one blue and one red hat, he will not be able to determine the colour of his own hat, so if there is a long silence then boy 2 can deduce that boy 3 is unable to see two red hats or two blue hats. He can then further deduce that he is wearing the "opposite" colour to boy 1. Therefore the probability that boy 2 correctly deduces the colour of his own hat is $\frac{4}{6} = \frac{2}{3}$.

Boy 1 will never be able to guess the colour of his hat, so it is the second boy that is most likely to escape homework this evening.

Of course, if boy 3 could see two red hats or two blue hats and was feeling particularly mean he could deliberately say nothing, knowing that boy 2 would (wrongly) deduce the colour of his own hat.