math puzzle

all sides of both dice 0, 1-12 have a 0% chance of showing up!

first die: 0,0,0,6,6,6
second die: 1,2,3,4,5,6

existance:
use standrad die and a die that has 3 0's and 3 6's. Each sum's probability is obviously 1/6*1/2=1/12

Uniqueness:
Say die 1 and die 2 has a1 and a2 number of values on their sides.

a1=1 =>a2=12 no solutions
a1=2 =>a2=6 =>a2 is standard die. a1 has to have 0's and 6's in order to form numbers 1 and 12. Clearly only 1 solution for a1=1 or 6(the one above)
3<=a1<6 =~> 4>=a2<6 =>solutions fail because they can't be equally distributed on the dice (a1 or a2 does not divide 6), creating different probabilities.
This only leaves a1=a2=6 which clearly is no solution.

Got too long for being obvious, but wth

Given a die, we assign a polynomial P(x)=ax^0+bx^1+cx^2+dx^3+ex^4+fx^5+gx^6 to that die.
The coefficient of x^j is defined to be the number of faces of the die with a j written on them, so the sum of the coefficients of P is 6.

Now, if P is the polynomial associated to one die and Q to another, then P times Q represents all the possible sums that come from rolling the two dice and adding the results. In short, to solve the problem we need to find P and Q such that P times Q has all coefficients equal.

We basically want to solve
P*Q=3*(x+x^2+...+x^12) (we get the 3 by noticing that (P*Q)(1)=P(1)*Q(1)=6*6)

Now, (x+x^2+...+x^12)=(x^13-x)/(x-1)=(x)(x^6-1)(x^6+1)/(x-1)=(x)(1+x+x^2+...+x^5)(x^6+1)

Thus 3*(x+x^2+...+x^12)=(x+x^2+x^3+...+x^6)(3x^6+3x^0).

We conclude that one die should be the standard die with 1-6 on its faces, while the second die should have 3 0s and 3 6s.

First die: 0 2 4 6 8 10
Second die: 1 1 1 2 2 2

Didn't give it much thoughts at all, so probably wrong.

What if I leave 1 die alone, so its faces have 1-6.

I label three faces of the other die 0, and the other three faces 6.

The probability of obtaining a 0 or 6 on the new die is equally likely. The probability of obtaining any number is equally likely for the average die.

So you should be able to get any number b/w 1 and 12 with equal odds.