 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ia/paper-2/9f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Again assume independent die scores. Their sum has the discrete convolution of the two score distributions, so

P (X = 2) = p_{1} q_{1}, P (X = 7) = i = 1 \sum 6 p_{i} q_{7 - i}, P (X = 12) = p_{6} q_{6} .

All terms are nonnegative. The arithmetic-geometric mean inequality applied to the two extreme-score contributions gives

P (X = 7) \geq p_{1} q_{6} + p_{6} q_{1} \geq 2 p_{1} q_{6} p_{6} q_{1} = 2 P (X = 2) P (X = 12) .

If all eleven probabilities were equal, normalization would make each

1/11

, but the inequality would require

1/11 \geq 2/11

. This contradiction proves the sum cannot be uniform on ${2, \dots, 12}$ , whatever the individual die probabilities.

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