Dice equal percentage

Question

There are 2x 20 sided dice.

According to anydice.com, "output 2d20" has a probability percentage mean of 5.0% (The chance of getting a 21) on 2x 20 sided dice.

Say, I wanted the probability to be evenly distributed among 5 categories, each having 20% chance equaling 100%.

Example:

A - ~20% probability
B - ~20% probability
C - ~20% probability
D - ~20% probability
E - ~20% probability

Which sequencing numbers would each category have to reach an evenly (or closely even) distributed probability of rolling two 20 sided dice? What is the mathematic formula for coming up with this distributed probability of rolling?

Answer 1

If you must throw two such dice, then the probability distribution of their sum has $P(S=s) = \dfrac{20 - |s-21|}{400}$ for $2 \le s \le 40$,

and so the following five patterns of sums are equally probable:

• $2,3,4,5,6,7,8,9,10,11,12,13$ or $39$
• $14,15,16,17,18$ or $36$
• $19,20,21,22$ or $37$
• $23,24,25,26,27$
• $28,29,30,31,32,33,34,35$ or $38$ or $40$

though there are many other partitions which also achieve this

Answer 2

Easier to remember than @Henry's proposed partition

The following five patterns of sums of two fair d20's are equally probable:

• $1,6,11,16,\dots,36$
• $2,7,12,17,\dots,37$
• $3,8,13,18,\dots,38$
• $4,9,14,19,\dots,39$
• $5,10,15,20,\dots,40$

Simply look at the last digit of the sum. Sums whose last digit is a $1$ or $6$ correspond to the first category while sums whose last digit is a $2$ or $7$ correspond to the second category etc...

Rolling two of a same number are equally distributed among these as well: $(3,3),(8,8),(13,13),(18,18)$ all belonging to the first category: $(1,1),(6,6),(11,11),(16,16)$ all belonging to the second category, etc...

That these are equally represented follows from the observation that each die is equally likely to have any possible remainder modulo 5.