Creating a binomial die with constraints

Question

I had an idea for making a custom 20-sided die that would have a binomial distribution of numbers printed on it, but I'm not sure exactly how to pick the numbers to put on the die itself.

So we have a few constraints:

• We only have 20 spaces for the numbers to go
• The numbers must be integers between 1 and 20 inclusive
• The average result of the die needs to be the same as a regular D20
• The distribution should be binomial (as defined by some parameters)

(and of course the solution could be further generalised to other dice types by tweaking the parameters)

To get the average result the same I figure we ought to be adding pairs of numbers at the same time equating to 21, so that's easy.

Having a limited number of spaces would probably be best solved by some iterative algorithm that based on the desired binomial curve and the current distribution would suggest the next pair of numbers to add to improve the curve / distort it the least.

So what would be a good way of going about figuring out which numbers should be added in each step? Or is there some better way of designing something like that?

Answer 1

Assuming you want a symmetric distribution, you're not left with a lot of choices.

• To get a mean of 10.5 means that you have to center it halfway between two integers, which implies ten faces above the mean and ten below.
• You probably want a single face at each of the extreme results, since lots of players love the "crit". This leaves nine other faces on each side of the mean.
• Allocating the rest 2/3/4 leaves you with a triangular distribution: one 7, two 8s, three 9s, four 10s, four 11s, three 12s, two 13s, one 14.
• Or, you could reduce the range by one on each side. A quick glance at Pascal's triangle suggests one 8, three 9s, six 10s, six 11s, three 12s, one 13 is a reasonable approximation of the 5-row.

You may also consider using the middle of 3d20, which doesn't quite have a binomial shape, but doesn't require custom dice, spans the same range, and is relatively easy to evaluate. Or, the classic sum of 3d6.