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7 points

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.

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4 points

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.

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3 points

I always divide by two and round up for d3

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4 points

Honestly you only need a d20 and a d6. D4? Divide by 5. D8? D20/5 x d20/10. D12? D6xd10/2

MATH BABY

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6 points

D8? D20/5 x d20/10

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?

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2 points

You could do something like ((d6-1)*20+d20)/15.

But that’s an awful lot of work just to avoid having a d8.

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1 point

Really it should be just using a d/20 itself divided into 5 parts. For instance, 1-4, 5-8, 9-12, etc.

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1 point

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

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1 point
*

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.

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1 point
*

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.

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2 points

I have a “D3” that’s just a D6 with two of each.

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2 points

You already have d10 and d100 (d00? What do we call the other one?), so there’s precedent for duplicating shapes.

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3 points
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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.

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2 points

what do we call the other one?

A golf ball?

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2 points

d% is what I usually see

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1 point

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.

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