Expected absolute difference of two dice

Question

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

Accepted Answer

For a gap of d \ge 1, the ordered pairs are (a, a+d) and (a+d, a), and a can run from 1 to 6 - d, so there are 2(6 - d) of them. Weight each gap by its count: E[X - Y] = \frac{1}{36}\sum{d=1}^{5} d \cdot 2(6 - d) = \frac{2}{36}\big(1\cdot5 + 2\cdot4 + 3\cdot3 + 4\cdot2 + 5\cdot1\big). The bracket is 5 + 8 + 9 + 8 + 5 = 35, so E[X - Y] = \frac{2 \cdot 35}{36} = \frac{70}{36} = \frac{35}{18} \approx 1.94. A quick sanity check: the gap is 0 only on the six doubles, and the largest gap 5 happens just twice, so a mean near 2 feels right. Takeaway. When a payoff depends on the gap between two dice,…

Expected absolute difference of two dice

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