Make me a market on the max of three dice

Question

Three fair six-sided dice are rolled. A contract pays \max(X, Y, Z) in dollars. Compute the fair value exactly and make a two-sided market.

Accepted Answer

For the max of three independent dice, P(\max \le k) = \left(	frac{k}{6}ight)^3, so P(\max = k) = \frac{k^3 - (k-1)^3}{216}. The numerators for k = 1,\dots,6 are 1, 7, 19, 37, 61, 91, which sum to 216 as they must. Then E[\max] = \sum{k=1}^{6} k\cdot\frac{k^3 - (k-1)^3}{216} = \frac{1 + 14 + 57 + 148 + 305 + 546}{216} = \frac{1071}{216} \approx 4.96. Compare with the max of two dice, which is 161/36 \approx 4.47: adding a third die pulls the expected maximum up, since it is one more chance to roll high, and with more dice the max drifts toward 6. The market. Fair \approx 4.96. Quote for ex…

Make me a market on the max of three dice

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