Expected larger of two ten-sided dice

Question

You roll two fair ten-sided dice (faces 1 through 10) and keep the larger number. What is the expected value of the larger face?

Accepted Answer

The expected maximum equals \sum{v \ge 1} P(	ext{max} \ge v). A tidy shortcut collapses this sum. Writing the maximum of two m-sided dice, E[	ext{max}] = \sum{v=1}^{m}\left[1 - \left(	frac{v-1}{m}ight)^2ight] = m - \frac{1}{m^2}\sum{j=0}^{m-1} j^2. For m = 10, the sum of squares 0^2 + 1^2 + \cdots + 9^2 = 285, so E[	ext{max}] = 10 - \frac{285}{100} = 10 - 2.85 = 7.15. The single-die average is 5.5; keeping the larger of two lifts it to 7.15. Takeaway. For the larger of two m-sided dice the answer is m - \frac{1}{m^2}\sum{j=0}^{m-1} j^2, a one-line formula once you know the sum of squa…

Expected larger of two ten-sided dice

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