= Solution
Again assume independent die scores. Their sum has the discrete <convolution> of the two score distributions, so
$$
\boxed{P(X=2)=p_1q_1,\qquad
P(X=7)=\sum_{i=1}^6p_iq_{7-i},\qquad
P(X=12)=p_6q_6.}
$$
All terms are nonnegative. The <arithmetic-geometric mean inequality> applied to the two extreme-score contributions gives
$$
P(X=7)\geq p_1q_6+p_6q_1
\geq2\sqrt{p_1q_6p_6q_1}
=2\sqrt{P(X=2)P(X=12)}.
$$
If all eleven probabilities were equal, normalization would make each $1/11$, but the inequality would require $1/11\geq2/11$. This contradiction proves \b[the sum cannot be uniform on $\{2,\ldots,12\}$], whatever the individual die probabilities.
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