Expected value of the larger of two dice

Question

You roll two fair six-sided dice and keep the larger of the two numbers. What is the expected value of that larger number?

Accepted Answer

Build the expected maximum from its tail probabilities, E[	ext{max}] = \sum{v \ge 1} P(	ext{max} \ge v). The larger of two dice is at least v unless both dice come up below v, and each die is below v with probability \frac{v-1}{6}, so P(	ext{max} \ge v) = 1 - \left(\frac{v-1}{6}ight)^2. Adding the six terms for v = 1 to 6: E[	ext{max}] = 1 + 	frac{35}{36} + 	frac{32}{36} + 	frac{27}{36} + 	frac{20}{36} + 	frac{11}{36} = \frac{161}{36} \approx 4.472. Even with just two dice the average of the larger jumps from 3.5 up past 4.4. Takeaway. For the larger of two m-sided dice, each tail …

Expected value of the larger of two dice

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