Make a market on the higher of two dice

Question

Two fair six-sided dice will be rolled. The contract settles at the maximum of the two faces. The interviewer will buy or sell against your quote before the roll.

Make a market: give a bid and an ask, and justify where fair sits.

Accepted Answer

Fair value. The maximum equals k when at least one die shows k and neither shows more than k. Counting ordered pairs, P(\max = k) = 	frac{2k - 1}{36}: that is 	frac{1}{36}, 	frac{3}{36}, 	frac{5}{36}, 	frac{7}{36}, 	frac{9}{36}, 	frac{11}{36} for k = 1, \dots, 6. So E[\max] = \frac{1(1) + 2(3) + 3(5) + 4(7) + 5(9) + 6(11)}{36} = \frac{161}{36} \approx 4.47. This is the key trap: taking the max of two draws pulls the expectation well above the single-die mean of 3.5. Candidates who blurt "3.5" or "4" are already offside. Dispersion. E[\max^2] = 	frac{791}{36} \approx 21.97, so \mathrm{…

Make a market on the higher of two dice

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