When you roll a standard pair of 6-sided dice, there are 11 possible outcomes, with a certain probability of occurrence for each possible result (the chance of getting a total of 8 is 5/36, for example). This is called the probability distribution.

Now, it turns out that you can take 2 6-sided dice and put different positive integers (whole numbers) on them such that the probability distribution is the same. That is, you can write a completely different set of numbers on the 2 dice but still have the same likelihood of rolling any total as if the dice were regular. It turns out that the only way to do this is to put 1, 2, 2, 3, 3, 4 on one die and 1, 3, 4, 5, 6, 8 on the other (clearly I'm allowing repeats of numbers, and if you think about it, repeats must occur).

The two problems I have relate to pairs of dice with more than 6 sides. The first problem can be done by trial and error, but the second one most certainly cannot be done that way, and would require an advanced method of solving.

Find the *2* possible relabelings for a pair of 8-sided dice.
Find the only possible relabeling for a pair of 35-sided dice.

Hint 1

Use generating functions. (Sub-hint: practice on the 6-sided case, where you already know the answer.)

Hint 2

This hint solves the problem for the 6-sided case (sometimes called Crazy Dice or Sicherman Dice) by using generating functions. Once you understand this case, you should be able to work the 8-sided and 35-sided cases the same way. It's helpful if you have a computer algebra program like Mathematica to do the heavy lifting for you, especially in the 35-sided case, but even by hand they are not too hard.

To write this problem in terms of generating functions we use a variable x and make the following observation: The coefficient of xⁿ in

is the number of way of throwing n using 2 regular dice. This is true because we interpret the first factor as the outcome of the first die (the exponent is how many dots came up) and the second factor as the outcome of the second die. For example, you can throw 8 in 5 ways because the factors that lead to x⁸ are

making 8 with plain dice

How about the Crazy Dice? They have polynomials too, which are

and the fact that they have the same frequency distribution as regular dice implies that this product of polynomials must be exactly the same as the product for the regular dice, and if fact if you multiply them out (or get Mathematica to do it for you) you'll see that they really are identical.

"So what?" you ask. So what is that we know a lot about polynomials, in particular about factoring them, so by converting our problem into a polynomial problem we can use this knowledge and solve it. Imagine that we didn't already know the other solution for 6; then we would want to find polynomials so that

solving crazy dice generating function

where the a and the b are the number of spots on each face, and we want to find them. A valid solution must have all a and b greater than or equal 1.

Now we use the polynomial properties. We can completely factor the left-hand side, because

factoring crazy dice polynomial

so the left-hand side of (*) is

We don't know what the terms on the right-hand side are, but we do know that if we factor them completely we must get exactly the same factorization, although the terms may be grouped differently. This means we can work backward to discover the terms on the right: we consider different ways of grouping the factors to make two polynomials

and see which ones represent valid 6-sided dice. Not every combination is valid; for example, we cannot give all factors to the first polynomial, and none to the second (leaving the second as a constant 1) because each polynomial must have 6 powers of x to represent the 6 faces. In fact we can eliminate most of the possibilities as follows:

The requirement that each polynomial represent a 6-sided die can be written numerically by saying that the value at x=1 of each polynomial in (**) must be 6. Looking at the complete factorization, we notice that the values of the factors at x=1 are 1, 2, 1, and 3. If one polynomial in (**) got both copies of (x + 1) and of (x² + x + 1), then at x=1 that polynomial would be divisible by 4 or by 9, and so could not be equal to 6. Therefore each polynomial in (**) gets one copy of these factors.
Each face must have at least one spot, so each polynomial in (**) must have an x factor (otherwise the polynomial would have a constant term, that is x⁰, that would represent 0 spots). So each polynomial in (**) must get one copy of the factor x.

The only uncertainy remaining is the two (x² - x + 1) factors. There's no obvious reason why we have to give one copy to each polynomial, or if we could give both copies to the same polynomial. In fact both choices work; giving one to each produces the regular dice, and (as we can verify by multiplying out the polynomials) giving both copies to one polynomial produces two valid sets of 6 spots, that are the Crazy Dice:

which multiplies out to

so the spots are

    1,2,2,3,3,4 and 1,3,4,5,6,8

That's the complete solution for 6-sided dice, now use the same ideas to solve the 8-sided and 35-sided cases. (The next hint gives the needed factorizations.)

Hint 3

For the 8-sided and 35-sided cases we have the complete factorizations

factorization for 8 and 35 sides

Click here for the complete solution.