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 pn(k), the probability that the highest score is k.
There are kn 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 kn − (k − 1)n cases out of 6n.
In other words, pn(k), the probability that the highest individual score is k, is (kn − (k − 1)n)/6n.
The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k · pn(k).
|Hence E(n)||= [6(6n−5n) + 5(5n−4n) + 4(4n−3n) + 3(3n−2n) + 2(2n−1n) + 1(1n−0n)]/6n.|
|= 6 − (1n + 2n + 3n + 4n + 5n)/6n.|
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.