How many fair dice of this kind exist?

Question

I am not talking about the shape of the dice here, I am talking about another type. You will see what I mean soon.

For example, when there are 1 dice, a normal dice is a fair dice, because the probability of getting each number is the same by $ \frac{1}{6} $

When 2 normal dice are thrown, you can get numbers from 2 to 12, but the probability of getting each of them are different. Think of Monopoly, the probability of getting 7 is $ \frac{6}{36} $ while the probability of getting 2 is $ \frac{1}{36} $ the goal is to make the probability of getting each number the same.

I got an example:

Dice 1: $ [1,2,3,4,5,6] $

Dice 2: $ [0,6,12,18,24,30] $

When these two dice are thrown and add the two number, the possible outcome can be every number from 1 to 36 and the probability of each is

So the question is, How many different pairs of dice of this kind are possible so that the possible outcome can be every number from 1 to 36 and the probability of each is $ \frac{1}{36} $

The number on the dice can be negative too.

Example:

Dice 1: $ [-1,1,11,13,23,25] $

Dice 2: $ [2,3,6,7,10,11] $

Same number, different order is considered the same.

$$ [-1,1,11,13,23,25] [2,3,6,7,10,11] $$

$$ [1,11,13,23,25,-1] [2,3,6,7,10,11] $$

$$ [2,3,6,7,10,11] [-1,1,11,13,23,25] $$

They are all the same

So how many are there?

Is there a way to find it without trying one by one?

Answer 1

If you accept dice with negative numbers, then the answer is, obviously, infinity. You can add $N$ to all numbers on one dice, and subtract $N$ from all numbers on another.

If you require them to be all non-negative, then it boils down to factorizing the polynomial $1+X+X^2+\ldots+X^{35}$ into the product of two polynomials of degree $6$ with all their coefficients being $0$ or $1$. Which, I guess, could be done by hand, but probably not worth it.

Correction: not of degree $6$, but having exactly $6$ non-zero (and hence equal to $1$) coefficients.

Answer 2

Let $n$ be any integer.

Dice 1: $\left[\matrix{1+n\\ 2+n\\ 3+n\\ 4+n\\ 5+n\\ 6+n}\right]$ $\qquad $Dice 2: $\left[\matrix{0-n\\ 6-n\\ 12-n\\ 18-n\\ 24-n\\ 30-n}\right]$

Answer 3

Add $1$ to each number in one of the dice:

$A: (0,1,2,3,4,5),(0,6,12,18,24,30)$

$B: (0,1,2,6,7,8),(0,3,12,15,24,27)$

$C: (0,1;2,9,10,11),(0,3,6,18,21,24)$

$D: (0,1,2,18,19,20),(0,3,6,9,12,15)$

$E: (0,1,4,5,8,9),(0,2,12,14,24,26)$

$F: (0,1,6,7,12,13),(0,2,4,18,20,22)$

$G: (0,1,12,13,24,25),(0,2,4,6,8,10)$

36 is the product of four prime numbers: 2, 2, 3, 3. Arrange them in any of six ways, for example $a=2, b=3,c =3, d=2$.
$a=2$ so I let $A=\{0,1\}$.
( If $a$ were 3 I would let $A=\{0,1,2\}$. )
$b=3$ so $B=\{0,a,2a\}=\{0,2,4\}$.
$c=3$ so $C=\{0,ab,2ab\}=\{0,6,12\}$.
$d=2$ so $D=\{0,abc\}=\{0,18\}$
Form one die from $A+B$ and the other from $C+D$; or else $A+C$ and $B+D$.
Here, $A+C=\{0,1,6,7,12,13\}$ and $B+D=\{0,2,4,18,20,22\}$.
Finally add 1 to either die so they go from 1 to 36 instead of 0 to 35.

Answer 4

As stated in MigMit's answer, we need two polynomials, each having the coefficient $1$ for exactly six terms coefficient $0$ for all other terms, whose product is $1 + x + x^2 + \cdots + x^{35}.$

Note that $$ x^{36} - 1 = (x - 1)(x^{35} + x^{34} + \cdots + x^3 + x^2 + x + 1),$$ and since the roots of $x^{36} - 1$ over complex numbers are all the $36$th roots of unity, the roots of $x^{35} + x^{34} + \cdots + x + 1$ are all the $36$th roots of unity except $1$ itself.

That is, $$ x^{35} + x^{34} + \cdots + x^3 + x^2 + x + 1 = \prod_{n=1}^{17}(x - e^{i\pi n/18}). $$

Multiplying together the various factors of $(x - e^{i\pi n/18})$ to obtain polynomials with integer coefficients, \begin{multline} x^{35} + x^{34} + \cdots + x^3 + x^2 + x + 1 = \\ (x + 1)(x^2 + 1) (x^2 - x + 1)(x^2 + x + 1) (x^4 - x^2 + 1)\\ (x^6 - x^3 + 1) (x^6 + x^3 + 1) (x^{12} - x^6 + 1). \end{multline}

To eliminate the negative coefficients we can multiply as follows: \begin{align} (x + 1)(x^2 - x + 1) &= x^3 + 1,\\ (x^2 + 1)(x^4 - x^2 + 1) &= x^6 + 1,\\ (x^2 - x + 1)(x^2 + x + 1) &= x^4 + x^2 + 1,\\ (x^4 - x^2 + 1)(x^4 + x^2 + 1) &= x^8 + x^4 + 1,\\ (x^6 - x^3 + 1)(x^6 + x^3 + 1) &= x^{12} + x^6 + 1,\\ (x^{12} - x^6 + 1)(x^{12} + x^6 + 1) &= x^{24} + x^{12} + 1,\\ (x + 1)(x^2 - x + 1)(x^6 - x^3 + 1) &= x^9 + 1,\\ (x^2 + 1)(x^4 - x^2 + 1)(x^{12} - x^6 + 1) &= x^{18} + 1. \end{align}

From this we get the following factorizations in which all coefficients are $1$ or $0$: \begin{align} x^{35} + x^{34} + \cdots + x + 1 &= (x^2 + x + 1) (x^3 + 1) (x^6 + 1) (x^{24} + x^{12} + 1)\\ &= (x + 1)(x^2 + 1) (x^8 + x^4 + 1) (x^{24} + x^{12} + 1)\\ &= (x^2 + x + 1) (x^6 + x^3 + 1) (x^9 + 1)(x^{18} + 1) \\ &= (x + 1) (x^4 + x^2 + 1) (x^{12} + x^6 + 1) (x^{18} + 1)\\ &= (x + 1)(x^4 + x^2 + 1) (x^6 + 1) (x^{24} + x^{12} + 1). \end{align}

In each case we can multiply either polynomial with two non-zero terms by either polynomial with three non-zero terms to get a polynomial with six non-zero terms. That is, \begin{align} x^{35} + x^{34} + \cdots + x + 1 &= (x^5 + x^4 + x^3 + x^2 + x + 1) (x^{30} + x^{24} + x^{18} + x^{12} + x^6 + 1) \\ &= (x^8 + x^7 + x^6 + x^2 + x + 1) (x^{27} + x^{24} + x^{15} + x^{12} + x^3 + 1) \\ &= (x^9 + x^8 + x^5 + x^4 + x + 1) (x^{26} + x^{24} + x^{14} + x^{12} + x^2 + 1) \\ &= (x^{10} + x^8 + x^6 + x^4 + x^2 + 1) (x^{25} + x^{24} + x^{13} + x^{12} + x + 1) \\ &= (x^{11} + x^{10} + x^9 + x^2 + x + 1) (x^{24} + x^{21} + x^{18} + x^6 + x^3 + 1) \\ &= (x^{13} + x^{12} + x^7 + x^6 + x + 1) (x^{22} + x^{20} + x^{18} + x^4 + x^2 + 1)\\ &= (x^{15} + x^{12} + x^9 + x^6 + x^3 + 1) (x^{20} + x^{19} + x^{18} + x^2 + x + 1). \end{align}

Although each of the five factorizations into four polynomials is capable of producing two factorizations into polynomials of six non-zero terms, some of them are duplicates of each other, and only seven are distinct.