I think this problem was first proposed by J.B. Kelly in the American Mathematical Monthly in 1950. The answer is no. Let aj and bj be the probabilities that the number j turns up on each of the two dice, and let Pj 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.