Roll three dice, keep the highest, fair price?

Question

You roll three fair dice and are paid the largest of the three faces, in dollars. Solve in your head. What is the fair price of this game? Aim: under 60 seconds.

Accepted Answer

Use the tail-sum formula with P(\max \le k) = (k/6)^3: E[\max] = \sum{k=0}^{5}\left(1 - \frac{k^3}{216}ight) = 6 - \frac{0 + 1 + 8 + 27 + 64 + 125}{216} = 6 - \frac{225}{216}. The cubes 1, 8, 27, 64, 125 sum to 225, so E[\max] = \frac{1296 - 225}{216} = \frac{1071}{216} \approx 4.96. Adding a third die pulls the expected maximum up from the two-dice value of 	frac{161}{36} \approx 4.47 toward 6, exactly as intuition suggests: more draws, higher expected best. Takeaway. For the maximum of n independent dice, the CDF factorizes as (k/6)^n, so the tail sum turns the expectation into 6 - \frac…

Roll three dice, keep the highest, fair price?

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