5 ms·
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. N
by octonion 5y ago
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.
- Veedrac 5y agoThere'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 agoExcellent point!
- Pamar 5y agoWait, 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?
- nihakue 5y agoBut 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 agoa 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 agoOh I see, I misunderstood. Yep I like your way best so far although I didn't see the problem with rerolling
- quacker 5y agoMy 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