Roll two dice, keep the lower, fair price?

Question

You roll two fair dice and are paid the smaller of the two 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 in its "\ge" form, since the minimum is small and it is easier to ask when both dice clear a threshold: P(\min \ge k) = P(	ext{both} \ge k) = \left(\frac{7 - k}{6}ight)^2, because a single die is \ge k with probability \frac{6 - k + 1}{6} = \frac{7 - k}{6}. Summing k = 1 to 6: E[\min] = \sum{k=1}^{6}\left(\frac{7-k}{6}ight)^2 = \frac{6^2 + 5^2 + 4^2 + 3^2 + 2^2 + 1^2}{36} = \frac{91}{36} \approx 2.53. Check against the max: E[\min] + E[\max] = \frac{91}{36} + \frac{161}{36} = \frac{252}{36} = 7, which must hold because \min + \max = X + Y and E[X + Y] = 7. Takeaway. For …

Roll two dice, keep the lower, fair price?

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