Expected maximum of two four-sided dice

Question

You roll n fair four-sided dice (faces 1 to 4) and record the largest value that appears. What is the expected value of this maximum? Work out n = 2 exactly, then give the general formula.

Accepted Answer

Let M be the maximum of n i.i.d. fair four-sided dice. CDF of the max. P(M \le m) = \left(\frac{m}{4}\right)^{n}, \qquad m = 0, 1, \dots, 4. Tail-sum formula. Using E[M] = \sum{k=1}^{4} P(M \ge k) and P(M \ge k) = 1 - \left(\frac{k-1}{4}\right)^n, E[M] = \sum{k=1}^{4}\left[1 - \left(\frac{k-1}{4}\right)^{n}\right] = 4 - \frac{0^n + 1^n + 2^n + 3^n}{4^{n}}. Case n = 2: E[M] = 4 - \frac{0 + 1 + 4 + 9}{16} = 4 - \frac{14}{16} = 4 - 0.875 = 3.125 = \frac{25}{8}. Direct check: P(M = m) = \frac{m^2 - (m-1)^2}{16} = \frac{2m - 1}{16}, so E[M] = \frac{1\cdot 1 + 2\cdot 3 + 3\cdot 5 + 4\cdot 7}{16} = …

Expected maximum of two four-sided dice

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