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?

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: $\left[\matrix{0-n\\ 6-n\\ 12-n\\ 18-n\\ 24-n\\ 30-n}\right]$
