Expected larger of two four-sided dice

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You roll two fair four-sided dice (faces 1,2,3,4) and keep the larger number. What is the expected value of the larger face?

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Use the tail-sum form of the expected maximum, E[	ext{max}] = \sum{v \ge 1} P(	ext{max} \ge v). The larger of the two is at least v unless both land below v, and each die is below v with probability \frac{v-1}{4}, so P(	ext{max} \ge v) = 1 - \left(\frac{v-1}{4}ight)^2. Summing over v = 1 to 4: E[	ext{max}] = 1 + 	frac{15}{16} + 	frac{12}{16} + 	frac{7}{16} = \frac{50}{16} = \frac{25}{8} = 3.125. The single-die average here is 2.5, and keeping the larger of two nudges it up to 3.125. Takeaway. Shrinking the number of faces to four does not change the recipe: sum 1 - ((v-1)/4)^2 over t…

Expected larger of two four-sided dice

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