Dice, (De)Convolution and Generating Functions " The blog at the bottom of the sea

I stumbled across a very cool YouTube video today that talks about how to look for alternate ways of labeling sides of two dice to give you the same random number distribution as if you added two standard dice together. https://www.youtube.com/watch?v=xVRRykIyll0

Making customized dice give desired probability distributions is an interesting topic (like for game design!), but the math that was used in the video was also pretty neat. I’ll walk you through what they did and then we can talk about why it works, from two different points of view.

(I’m going to refer to a single die as "a dice", apologies in advance!)

The Problem Statement

If you roll a single dice, the probability of getting each possible value is the same: 16.666…%.

However, if you roll two dice and add them together, the probability of getting each possible value is not equal. You can only get a two by rolling a 1 and a 1, but you can roll a four by rolling a 1 and a 3, a 2 and a 2, or a 3 and a 1. There is only one way to roll a two, but there are three ways to roll a four. So, rolling a four is three times more likely than rolling a two. The probability for each possible value is described by the binomial probability distribution below.

Dice Sum # Of Ways Probability (Rounded) Dice Values 2 13%113 26%12, 214 38%13, 22, 315 411%14, 23, 32, 416 514%15, 24, 33, 42, 517 617%16, 25, 34, 43, 52, 618 514%26, 35, 44, 53, 629 411%36, 45, 54, 6310 38%46, 55, 6411 26%56, 6512 13%66

The problem statement is this: Is there a different way to label two dice such that when you add them together, you get the exact same probability distribution as when using standard dice?

It turns out there is exactly one other way to label the dice, and they are called "Sicherman Dice", named after the person who invented them, in 1978.

Dice A: 1, 2, 2, 3, 3, 4

Dice B: 1, 3, 4, 5, 6, 8

How To Find These Dice

The video explains that you can turn a dice into a polynomial summing up terms of the form . The value is what number is on the dice face, and the value says how many faces have that value.

For instance, a standard dice which has faces 1,2,3,4,5,6 would have this polynomial below, which states that it has one of each number 1 through 6:

If we changed the 2 face to be a 4, we’d have the polynomial below, which says we have faces 1,3,4,4,5,6:

These polynomials are valuable because if you have two of them, you can multiply them together to get a "virtual dice" that is the same as if we added the dice together into a single dice.

If we multiply two standard dice together:

We get the polynomial:

If you add up how many faces there are, that would be a 36 sided dice, and if you look at the number of faces of each value, you can see they follow the same binomial distribution that we showed in the problem statement section, when adding two dice together.

We aren’t done yet though, we need to see if there’s an alternate way to break this 36 sided dice polynomial back into two 6 sided dice, that aren’t just the standard 6 sided dice.

We need to factor that large polynomial into two smaller ones. These smaller polynomials have requirements (constraints) as well, to be valid:

Each needs to be non negative. That is, you can’t have a negative number of face values on a dice. That’d be weird.

The labels need to be positive. 0 is not a valid face value. This means that so that is 0.

The dice should have 6 faces. So if you sum up for all , it should give you 6.

The second bullet point can be described as because it makes all the power terms go away and leaves only the value of .

The third bullet point can be described as because that will make all power terms become 1, and so it will just be the values summed up.

The first step is to decompose the large polynomial into irreducible (prime) factors. In the video, instead of factoring the large polynomial, he factors the standard dice polynomial instead, and duplicates the factors, to get the list below. This video is a good intro to factoring polynomials if you need it: https://www.youtube.com/watch?v=KUMhpKGwpCY

To make new dice, we are going to assign each of those factors above to one of the dice. The factors assigned to a dice are going to be multiplied together to get the final polynomial for the dice.

We start with the second constraint that . The only way that can be true is if each dice has the factor. So, we assign one of those to the first dice, and the other one to the second dice.

Dice A:

Dice B:

Let’s move onto the second constraint that . If we plug 1 in for , the factor we already have for each dice evaluates to 1,and here are the values we get for each of the factors remaining:

Since we are multiplying factors together, and we need both of the dice to evaluate to 6, that means we assign to each dice to multiply by 2, and also assign to each dice to multiply by 3, so that we are multiplying each dice by...