Expected value of the maximum of dice

Question

You roll n fair six-sided dice and record the largest value that appears. What is the expected value of this maximum? Work out the case n = 2 exactly, then give the general formula.

Accepted Answer

Let M be the maximum of n i.i.d. fair dice, each uniform on \{1,\dots,6\}. CDF of the max. All dice are independent, so P(M \le m) = \left(\frac{m}{6}\right)^{n}, \qquad m = 0,1,\dots,6. Direct approach. The probability the max equals exactly m is P(M = m) = \left(\frac{m}{6}\right)^{n} - \left(\frac{m-1}{6}\right)^{n}. For n = 2 this is \dfrac{m^2 - (m-1)^2}{36} = \dfrac{2m-1}{36}, so E[M] = \sum{m=1}^{6} m\cdot\frac{2m-1}{36} = \frac{1}{36}\left(2\sum{m=1}^{6} m^2 - \sum{m=1}^{6} m\right) = \frac{2(91) - 21}{36} = \frac{161}{36} \approx 4.472. Tail-sum approach (general n). Using E[M] = \su…

Expected value of the maximum of dice

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