Honestly I’m sure this is the best solution. I get that a d4 is the obvious choice for something that should have a 1/4 chance of happening but a d8 with 4 numbers twice would be the most appropriate.
The only downside I can see is that a d8 and a d8/4 would be easy to mix up at first glance.
Am I missing something here? Can this even generate 5 or 7?
D20/5 gives [1…4] and D20/10 [1…2], of course assuming whole numbers. Where to get the factors for 5? 5 can be factored only as 5x1 or 1x5 and the 5 cannot be found either in d20/5 or d20/10. Same is true for 7.
And I don’t see it happening either if we allow rational numbers. To get 5 we would get the following expressions
5= d120/5 x d220/10 = d120 x d220/50
or
250= d120 x d220
And two d20 multiplied together cannot give us 250.
You are right, in my mind the d20/2 was some sort of iterator over the d20/5, the correct math would be d20/5+(20/5*(d20/10-1)). To get 5 this expresion would be with a 1-5 in the first one and a 11-20 on the second, the first would be 1 (rounded up) , and the second one 4*(2-1), so 5. The idea is that you use the second one to decide how many batches of the full first batch you add to the first one. As if you were rolling a d100 with two d10 but in base 20/5 instead of base 10. It’s not actually base 20/5 but that’s the idea, one of the dice is the “tens” dice and the other is the “hundreds” dice.
But then do really need the d8? If we toss that in the bin we can go to the universal d60. This one dice will allow us to get
d2 (even/odd)
d3 (d60/20)
d4 (d60/15)
d5 (d60/12)
d6 (d60/10)
d10 (d60/6)
and d12, d15, d20, d30
To keep the same probabilities, you can only reduce and only to one that is a factor. E.g. d20 can be equivalent to d10, d5, d4 and d2.
Multiplying the rolls messes things up. As an example, for d12 as a d6xd2 you have double the chance to roll 2, 4, and 6 and no chance to roll 7, 9, and 11.
You could make the equation a little more complicated (6×(d2-1))+d6 to make it work.
Honestly I’m sure this is the best solution. I get that a d4 is the obvious choice for something that should have a 1/4 chance of happening but a d8 with 4 numbers twice would be the most appropriate.
The only downside I can see is that a d8 and a d8/4 would be easy to mix up at first glance.
Honestly you only need a d20 and a d6. D4? Divide by 5. D8? D20/5 x d20/10. D12? D6xd10/2
MATH BABY
Am I missing something here? Can this even generate 5 or 7?
D20/5 gives [1…4] and D20/10 [1…2], of course assuming whole numbers. Where to get the factors for 5? 5 can be factored only as 5x1 or 1x5 and the 5 cannot be found either in d20/5 or d20/10. Same is true for 7.
And I don’t see it happening either if we allow rational numbers. To get 5 we would get the following expressions
5= d120/5 x d220/10 = d120 x d220/50
or 250= d120 x d220
And two d20 multiplied together cannot give us 250.
Math baby?
You could do something like ((d6-1)*20+d20)/15.
But that’s an awful lot of work just to avoid having a d8.
Really it should be just using a d/20 itself divided into 5 parts. For instance, 1-4, 5-8, 9-12, etc.
You are right, in my mind the d20/2 was some sort of iterator over the d20/5, the correct math would be d20/5+(20/5*(d20/10-1)). To get 5 this expresion would be with a 1-5 in the first one and a 11-20 on the second, the first would be 1 (rounded up) , and the second one 4*(2-1), so 5. The idea is that you use the second one to decide how many batches of the full first batch you add to the first one. As if you were rolling a d100 with two d10 but in base 20/5 instead of base 10. It’s not actually base 20/5 but that’s the idea, one of the dice is the “tens” dice and the other is the “hundreds” dice.
… math baby
I hate math babies. Least favorite type of baby.
Math baby.
But then do really need the d8? If we toss that in the bin we can go to the universal d60. This one dice will allow us to get
d2 (even/odd)
d3 (d60/20)
d4 (d60/15)
d5 (d60/12)
d6 (d60/10)
d10 (d60/6)
and d12, d15, d20, d30
Base 60 is cool yo!
that dice would either be really big, or it would just be a ball that would take too long to stop rolling lol… I want it now.
To keep the same probabilities, you can only reduce and only to one that is a factor. E.g. d20 can be equivalent to d10, d5, d4 and d2.
Multiplying the rolls messes things up. As an example, for d12 as a d6xd2 you have double the chance to roll 2, 4, and 6 and no chance to roll 7, 9, and 11.
You could make the equation a little more complicated (6×(d2-1))+d6 to make it work.
You are absolutely right, I was thinking d6d2 as: the D2 rolls 1, it’s d6. The D2 rolls 2,its 6+d6. That’s not what my math said so my bad!
Edit: your equation is what I had in my mind, which is sorta what we do to roll d100.
You don’t even need a special die for this. Just roll a d8 and subtract 4 if it’s 5-8. Just like using a d6 as a d3.
I always divide by two and round up for d3
I have a “D3” that’s just a D6 with two of each.
You already have d10 and d100 (d00? What do we call the other one?), so there’s precedent for duplicating shapes.
But if you roll the d00 on accident, you can easily still treat it as a d10. If you roll the d8/4, you can’t.
d% is what I usually see
A golf ball?
I’ve heard it called a tens die or a percentile die. D100 is usually saved for the actual 100-sided die in my experience.