5 ms·

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 remov

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.

octonion 5y ago
Excellent point!
Pamar 5y ago
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?
- Aeolun 5y ago
  Wouldn’t that change the probabilities though?
  - Pamar 5y ago
    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.
- deleted 5y ago
  [deleted]
nihakue 5y ago
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
- bazzargh 5y ago
  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
  - nihakue 5y ago
    Oh I see, I misunderstood. Yep I like your way best so far although I didn't see the problem with rerolling
quacker 5y ago
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

octonion 5y ago

Excellent point!

Pamar 5y ago

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?

Aeolun 5y ago
Wouldn’t that change the probabilities though?
- Pamar 5y ago
  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.
deleted 5y ago
[deleted]

nihakue 5y ago

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

bazzargh 5y ago
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
- nihakue 5y ago
  Oh I see, I misunderstood. Yep I like your way best so far although I didn't see the problem with rerolling

quacker 5y ago

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