Gabeland

First week of the new schedule! I already missed it. Anyways, take a basic six-sided die.
You know, the one that has the numbers from 1 through 6 on it. Roll it twice, and add up the total.
The result can be anywhere from 2 to 12, but it’s most likely to be 7. You can see this if you look at the number of possibilities for each:

2 – 11
3 – 12 21
4 – 13 22 31
5 – 14 23 32 41
6 – 15 24 33 42 51
7 – 16 25 34 43 52 61
8 – 26 35 44 53 62
9 – 36 45 54 63
10 – 46 55 64
11 – 56 65
12 – 66

What if I told you that there was another set of two dice that had the exact same distribution?
The first die has the numbers [1,2,2,3,3,4] on it, and the second die has the numbers [1,3,4,5,6,8].
Let’s just tally the results to check:

2 – 11
3 – 21 21
4 – 13 31 31
5 – 14 23 23 41
6 – 15 24 24 33 33
7 – 16 25 25 34 34 43
8 – 26 26 35 35 44
9 – 18 36 36 45
10 – 28 28 46
11 – 38 38
12 – 48

Yep, it’s the same! But why? Why should these two seemingly arbitrary dice work?

(d# is a die with that many sides, in case you didn’t know that terminology)
Well, you can split a d6 into two smaller dice, a d2 and a d3, and add those together. You could have a d2 with [0,1] and a d3 with [1,3,5], or you could have a d2 with [0,3] and a d3 with [1,2,3]. Adding all four of those dice together gives the same result as two d6. So what if we swap them around? Combining the [0,1] with the [1,2,3] makes [1,2,2,3,3,4]. And combining the [0,3] with the [1,3,5] makes [1,3,4,5,6,8].

But wait, there’s more! What if we combined them in different ways? You could get a d4 with [1,2,4,5] and a d9 with [1,2,3,3,4,5,5,6,7]. (I got this by subtracting one from the d9 sides and adding one to the d4 sides, since that doesn’t change the results.) Or you could have a coin which adds 1 if it’s heads, and a d18 with [2,3,4,4,5,5,6,6,6,7,7,7,8,8,9,9,10,11]. But that last one’s kind of ridiculous, because a d18 doesn’t actually exist. Right?

There’s also a different way you could look at this: Turn the dice into functions.
Consider this function: x + x² + x³ + x⁴ + x⁵ + x⁶. The exponents are the possible results, and the coefficients are the number of possibilities for each result.

If you square it, you get this,
x² + 2x³ + 3x⁴ + 4x⁵ + 5x⁶ + 6x⁷ + 5x⁸ + 4x⁹ + 3x¹⁰ + 2x¹¹ + x¹²
which is exactly the distribution for 2 dice.

However, the important thing about this is that now we can factor them. x + x² + x³ + x⁴ + x⁵ + x⁶ factors into x(x+1)(x²+x+1)(x²-x+1). What is that equivalent to in dice? A d1 with [1], a d2 with [0,1], a d3 with [0,1,2]… and another d1… that has one side with 0, one side with 2… and negative one side with 1? I’ve been trying to come up with a physical interpretation of that for a while, but I think there just is none.

But, regardless, now we can get the alternate dice a different way.
The dice need to be six-sided, so the sum of the coefficients has to be 6. How do we find that? Evaluate it at x=1. x+1 evaluates to 2, and x²+x+1 evaluates to 3, so we need those to be paired up. And each one needs an x, otherwise it’d just be equivalent to adding 1 to one of them and subtracting 1 from the other. That leaves only the x²-x+1 term. If we give one of them to each, then you get the original two d6. But if you give them both to one die, and none to the other, then it’s different.

x(x+1)(x²+x+1) = x + 2x² + 2x³ + x⁴
x(x+1)(x²+x+1)(x²-x+1)(x²-x+1) = x + x³ + x⁴ + x⁵ + x⁶ + x⁸

And there you have it! And, furthermore, I’ve now proven that this is the only other way to do it, because there’s no other way to factor x + x² + x³ + x⁴ + x⁵ + x⁶.