How many fair dice of this kind exist? 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?
 A: 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.
A: 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.
A: 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: $\left[\matrix{0-n\\ 6-n\\ 12-n\\ 18-n\\ 24-n\\ 30-n}\right]$
