Lowest of Three Dice

Question

Roll three fair dice. What is the expected value of the lowest of the three results?

Accepted Answer

The minimum is at least k exactly when all three dice are, so P(min ge k) = left(frac7-k6right)^3. The tail-sum formula turns this directly into the expectation: mathbbE[min] = sum_k=1^6 left(frac7-k6right)^3 = frac6^3 + 5^3 + 4^3 + 3^3 + 2^3 + 1^3216 = frac441216 = frac4924 approx 2.04. The numerator is the sum-of-cubes identity left(frac6 cdot 72right)^2 = 21^2 = 441, a tidy arithmetic check, and full enumeration of all 216 outcomes confirms it. Symmetry shortcut: the relabeling x mapsto 7 - x swaps min and max, so mathbbE[max] = 7 - frac4924 = frac11924 approx 4.96 comes free. And since the three order statistics must sum to the plain total 3 times 3.5 = 10.5, the middle die averages 10.5 - 7 = 3.5, the median order statistic sits exactly at the mean here. Trading relevance: this is the simplest worst-of payoff. Worst-of structures (lowest of several underlyings) are systematically cheaper than intuition suggests because the minimum concentrates downward fast, three dice already drag fair value to 2.04, well below the 2.5 many people guess. Tail-sum plus the 7-x reflection should both be reflexes for order-statistic pricing.

Lowest of Three Dice

Problem

Your answer

Hints

More Expected Value to train on