Renumbering tetrahedral dice but retaining the probabilities Let us have two regular tetrahedrons, with numbers 1,2,3,4 written on their sides.
We "throw" this object and play with it, like with the regular cube, and the side on the bottom counts as our point. It's easy to see that with 25% chance, we throw 1,2,3, or 4.
Now we should make two tetrahedrons, which are different from these two, and write positive integers on their side, it is 25% to throw either of them, and if we throw with both of them, we have the same chance for the sum of our throws, like with our regular tetrahedrons.
It is quite hard to understand, but here is the case:
We have these several cases:


*

*1 case -> We throw 1 with both of them, when the sum is 2.   

*2 cases -> We throw 1,2 or 2,1 with them, when the sum is 3.

*3 cases -> We throw 1,3 or 2,2 or 3,1 with them, when the sum is 4.

*4 cases -> We throw 1,4 or 2,3 or 3,2 or 4,1 with them, when sum is 5.

*3 cases -> We throw 2,4 or 3,3 or 4,2  with them, when the sum is 6.

*2 cases -> We throw 3,4 or 4,3 with them, then the sum is 7.

*1 case -> We throw 4 with both of them, then the sum is 8.


Concluded: We have a total of 16 cases, and we have 1/16 chance for 2, 2/16 chance for 3,... and finally 1/16 chance again for 8.
I don't really have any idea how to solve this task, so what exactly should I do to get two tetrahedrons, which have the same cases if I throw with them, but they are not like these?
 A: Using positive integers and having the ability to throw 2 in only one way $\Rightarrow$ there must be one 1 on each die.
For getting 3 two ways, assuming we want some different numbers, we can try putting the two 2s on one die (the first); then the second die cannot have a 2. 
To get 4 three ways, then, we need three 3s; these can't all be on the second die because that would give too many ways of making 5. So put two 3s on the second die and one on the first, which gives use four ways to make 5, and two (of three required) ways to make 6. 
Add the last option for making 6 by putting a 5 on the second die, and it all works out for 7s and 8s:
$$
(1,2,2,3) \\
(1,3,3,5)
$$
A: Suppose you have your initial set of dice, but when you roll them, instead of just adding the numbers on the bottom faces, you add any number to the die on the left and subtract the same number from the die on the right? Does your total change? Instead of adding and subtracting "any number" after you roll the dice, why not do it before you roll them?
