First off, assuming this problem - or one similar enough to directly apply - hasn't already got an official theorem or conjecture proving it one way or the other, I'd be content with simply being pointed in the direction of whatever maths would help me understand and learn about the problem myself. In light of this, I should say that the highest level of maths I studied have been a high school statistics class and intro to trig.

Anyway, so the question proper. What first got me thinking about this was playing old D&D-based CRPGs like Baldur's Gate and KotOR where things like weapon damage are listed as "2-8" or "3-10". But these number ranges actually correspond to different combinations of dice (two 4-sided dice, and one 8-sided dice adding 2 to the result, respectively; often referred to as 2d4 and 1d8+2 in RPG circles, for those who don't know), and I'm wondering if it's possible to derive a single combination of dice from any given (valid) pair of numbers.

Further, I'd be interested to know what sorts of constraints there might be on being able to derive things this way. Like is it only possible with specific "sizes" of dice, or so long as all the dice in a given pool are the same size, etc. Which is kind of why, if someone hasn't already written a paper on the subject, I'd be content with simply being directed to the sorts of math that would help me solve these problems myself, rather than simply being given an answer (though I certainly wouldn't decline one if it is offered).


1 Answer 1


In general you can't figure out which specific number of dice of which size are implied by the interval alone, at least not uniquely.

Consider d3+d4 vs d6+1 -- both are on the interval 2-7.

To make things harder, let's restrict consideration only to the "usual" dice used in tabletop roleplaying games (d4,d6,d8,d10,d12,d20,d100), and further, restrict it to cases where you only use a single die type in one roll - though perhaps that latter restriction was already your intent with the phrase 'dice pool'.

Consider that the range (max-min) on any die is the number of faces on the die, less one.

So d4 has a range of 3, 2d4 has a range of 6, and so on.

So the range of $m\ \!\text{d}\ \!n$ is $m\times (n-1)$.

If we see an interval like 5-20, that's a range of 15. Now $15=5\times 3 = 5 \times (4-1) = 3 \times (6-1)$

Consequently, we can clearly write it as either 3d6 + mods or as 5d4 + mods. Specifically, either as 3d6+2 or 5d4. Consequently, in general, just expressing the interval is not sufficient to uniquely define the dice required to produce it, even under those restricted conditions.

It does make some difference - the variance of 3d6+2 (105/12) is different from the variance of 5d4 (75/12); and you're about 5 times more likely to get a "20" using the first method than the second.

You can figure out a way to get the range of the desired interval if you can find at least one such factorization of the range into two numbers one of which is one less than the number of faces on an available die.

e.g. if you have an interval with a range of 12 you have 1$\times$12 or 2$\times$6 or 3$\times$4 or 4$\times$3 or 6$\times$2 or 12$\times$1, implying (ignoring the +mod term for the moment) you need a d13 or 2d7 or 3d5 or 4d4 or 6d3 or 12d2 - not all of which are necessarily available to you.

Note that if you allow a d2 - e.g. a coin labelled 1 and 2 or a d6 numbered 1,1,1,2,2,2 - and you allow negative modifiers, then you can always produce any interval of outcomes, because a d2 has a range of 1. If you have an interval from $a$ to $b$, then $(b-a) \text{d}2 + (2a-b)$ will work e.g. values on 110-173 can be done with 63d2+47.

More generally, what intervals can be produced depends on which dice you are prepared to count as 'legitimate'. Then you're in the arena of partition problems with restrictions on part size and change-making problems, and is related to the Frobenius coin problem.

As an example, combinations of d3 and d4 can cover every range on an interval except 1.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.