Expected largest of three four-sided dice

Question

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

Accepted Answer

The expected maximum equals \sum{v \ge 1} P(	ext{max} \ge v), and for three dice each die falls below v with probability \frac{v-1}{4}, so E[	ext{max}] = \sum{v=1}^{4}\left[1 - \left(	frac{v-1}{4}ight)^3ight] = 4 - \frac{1}{4^3}\sum{j=0}^{3} j^3. The cubes sum to 0 + 1 + 8 + 27 = 36, so E[	ext{max}] = 4 - \frac{36}{64} = 4 - 0.5625 = 3.4375 = \frac{55}{16}. Compared with the two-dice answer of 3.125, adding a third die pushes the expected largest even closer to the ceiling of 4. Takeaway. Raising the die count from two to three swaps the square for a cube: E[	ext{max}] = m - \frac{1}…

Expected largest of three four-sided dice

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