I think this problem was first proposed by J.B. Kelly in the American Mathematical Monthly in 1950. The answer is no. Let a_j and b_j be the probabilities that the number j turns up on each of the two dice, and let P_j be the probability of getting the sum j on the pair of dice. So

P₂ = a₁b₁ = a₆b₆ = P₁₂.

Now, to maintain this balance, if the first die favors 1 over 6, then the second must favor 6 over 1, and vice versa, so that

(a₁ – a₆)(b₁ – b₆) ≤ 0.

But this means that

P₂ + P₁₂ = a₁b₁ + a₆b₆ ≤ a₁b₆ + a₆b₁ ≤ P₇.

The probabilities of rolling a 2, a 7, and a 12 cannot be equal.