A pair of dice, when rolled, produces sums of 2 to 12, with varying probabilities. Can the dice be reweighted (each face assigned a probability other than 1/6) so that all 11 sums occur with the same frequency?
If so how, if not how close can the difference between the least and most likely sum be made?
I agree with those who say it is probably not possible. There are 6 probabilities to be assigned to each die, but they are not independent (since they must total to 1). There are really only 5 free variables on each die, for a total of ten variables. But we have 11 different results that must all equal 1/6. 11 equations with 10 unknowns is generally not solvable.
The same applies to two n-sided dice. This leads to 2n-1 equations with only 2n-2 variables.
And increasing the number of dice doesn't help either. If we have m n-sided dice, then there are m(n-1) variables, which is still one less than the number of different totals, which mn - m + 1.
degrees of freedom
