This is a very tricky problem. We want dice that generate outcomes of two fair dice but no 7s, and every other outcome having the same relative probabilities. Note the generating function (x + x^2 + x^3 + x^4 + x^5 + x^6)^2 - 6x^7 has no nontrivial factors, so we must get creative. My first solution was to consider the generating function x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + 5x^7 + 4x^8 + 3x^9 + 2x^10 + x^11 = (x^6 + x^5 + x^4 + x^3 + x^2 + x) (x^5 + x^4 + x^3 + x^2 + x). So we can do it with a standard 6-sided die and a 20-sided die with 4 each of 1,2,3,4,5. If you roll a 7 or more, add 1. My solution is completely different from anything mentioned in the article. I actually like my solution much more, as it uses two readily available Platonic solids. You don't even need to renumber! For the D20, you can divide the face number by 4 and ceiling, or mod 5 and add 1.
There's a more obvious interpretation here. After you've rolled one die, there is exactly one other roll for the second die that could produce a 7. If you remove that possibility, you have a d5, still numbered 1-6 but with one of those digits missing.
But this messes with the distribution (you now have two ways of rolling a 2: 1,1 and 2,5) and makes outcomes like '1' possible (1, 6), which shouldn't be
a normal D6 and a D10 with 1-5 twice. If total is 7, use the D6 plus 6 instead.
The roll table for that is exactly that of 2 D6 with the 7s removed. The 7s shift to the last column.
Wait, let me see if I got this right... I use a Red die and a Blue Die. Blue die is considered "secondary" (in the examples below, the second number is always taken from the blue die), and both dice are thrown together.
