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Comment by Veedrac | Hacker News Reader

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.

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My simple solution was:

- Use a normal d5 and d6.

- On a roll, if the sum on the dice face is >= 7, add 1 to the sum.

If I roll 2+3, the result is 5. If I roll 3+4, the sum is 7, so add 1 to get the result of 8.

The gives you this distribution, and is simple in practice.

      1  2  3  4  5
    1 2  3  4  5  6
    2 3  4  5  6  8
    3 4  5  6  8  9
    4 5  6  8  9  10
    5 6  8  9  10 11
    6 8  9  10 11 12

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.

      1  2  3  4  5    <- D10
    1 2  3  4  5  6
    2 3  4  5  6  >  8
    3 4  5  6  >  8  9
    4 5  6  >  8  9  10
    5 6  >  8  9  10 11
    6 >  8  9  10 11 12

Oh I see, I misunderstood. Yep I like your way best so far although I didn't see the problem with rerolling

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.

- I roll 4+2 - result is 6

- I roll 4+6 - result is 10

- I roll 1+6 - result is 1

- I roll 3+4 - result is 3

- I roll 4+3 - result is 4

Correct?

Wouldn’t that change the probabilities though?

This is what I suspect, too, and I am too lazy to try this out on Anydice anyway, so I asked to be sure I am not missing something.

Excellent point!

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