# Solution to puzzle 50: Highest score

Suppose n fair 6-sided dice are rolled simultaneously. What is the expected value of the score on the highest valued die?

We seek the expected value of the highest individual score when n dice are thrown. We first find p_{n}(k), the probability that the highest score is k.

There are k^{n} ways in which n dice can each show k or less.

For the highest score to *equal* k, we must subtract those cases for which each die shows less than k; these number (k − 1)^{n}.

So, k is the highest score in k^{n} − (k − 1)^{n} cases out of 6^{n}.

In other words, p_{n}(k), the probability that the highest individual score is k, is (k^{n} − (k − 1)^{n})/6^{n}.

The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k · p_{n}(k).

Hence E(n) | = [6(6^{n}−5^{n}) + 5(5^{n}−4^{n}) + 4(4^{n}−3^{n}) + 3(3^{n}−2^{n}) + 2(2^{n}−1^{n}) + 1(1^{n}−0^{n})]/6^{n}. |

| = 6 − (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n})/6^{n}. |

The table below shows E(n), correct to four decimal places, for various values of n, from 1 to 50. As expected intuitively, E(n) approaches 6, as n increases.

Expected values
n | E(n) |
---|

1 | 3.5 |

2 | 4.4722 |

3 | 4.9583 |

4 | 5.2446 |

5 | 5.4309 |

6 | 5.5603 |

7 | 5.6541 |

8 | 5.7244 |

9 | 5.7782 |

10 | 5.8202 |

20 | 5.9736 |

30 | 5.9958 |

40 | 5.9993 |

50 | 5.9999 |

## Further reading

- Sums of Powers

Source: Inspired by A Collection of Dice Problems, by Matthew M. Conroy. See problem number 3.

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