Alternative Dice

Here is an fun proof I saw during a stay at MSRI (Mathematical Sciences Research Institute) – unfortunately, I can not remember the reference (or none was given). If you happen to know the originator of this work, please let me know and I will attribute it appropriately. In any event, it is a “fun” problem, so I typed up the following summary:

Prop: There exists precisely one pair of numerically non-symmetric six-sided dice with no blank sides such that the sum-roll probability distribution is equivalent to normal symmetric six-sided dice. (Where by sum-roll it is meant the probability of two dice rolling a sum of x, etc.)

Proof: Let a die be represented by a Polynomial in the following fashion. Powers represent the number on a side of a die, and the coefficient of a specific power represents the number of sides with that number of dots. For example, a normal symmetric die has the following representation:

P(x) = x + x^2 + x^3 + x^4 + x^5 + x^6

And a four sided die with three sevens and one four would be:

P(x) = x^4 + 3*x^7

Notice that with this representation, P(1) is the number of sides on the die. Also notice that this representation encapsulates the probability of a specific roll, i.e., the probability of a die P(x) rolling a “n” is:

\dfrac{\text{coefficient of }x^n}{P(1)}

Finally, the action of rolling two dice is equivalent to forming the product of the polynomial representations. For example, when two normal symmetric six-sided dice are rolled, the probability of rolling a one (in sum) is zero, the probability of rolling a two is 1/36, the probability of rolling a seven is 1/6, etc. Notice that if we form the product

(x + x^2 + x^3 + x^4 + x^5 + x^6)^2

Then the coefficient of x^1 is zero, the coefficient of x^2 is one, and the coefficient of x^7 is six (0/36, 1/36, and 6/36).

Now, to answer the original question, we find two polynomials, T(x) and S(x) such that:

1) T(x)*S(x) = (x + x^2 + x^3 + x^4 + x^5 + x^6)^2
2) T(1) = S(1) = 6
3) T(x) \neq S(x)
4) T(0) = S(0) = 0

Condition #1 gives us the same sum-roll distribution as normal symmetric dice.
Condition #2 gives us two six-sided dice.
Condition #3 assures us of the non-symmetry of T(x) and S(x).
Condition #4 is equivalent to saying that the polynomials have a zero coefficient for the x^0 term, which is the same as saying that the dice have no blank sides.

So, to proceed, we simply need to factor the polynomial in condition #1. Here it is:

x^2 * (x^2 + 1)^2 * (x^2 - x + 1)^2 * (x^2 + x + 1)^2

Which I’ll write as:

a(x)^2 * b(x)^2 * c(x)^2 * d(x)^2


a(x) = x
b(x) = x^2 + 1
c(x) = x^2 - x + 1
d(x) = x^2 + x + 1

Notice that a(1)=1 so multiplying by a(x) will not change the number of sides on the die. Also notice that if we put both a(x) terms in one of T(x) or S(x) then the other die would have a blank side (since the product of any of the remaining terms b(x), c(x) or d(x) would incur a monomial). Thus, both T(x) and S(x) must have a(x) as a factor.

Now, notice that b(1)=2 and d(1)=3, whose product is six. Also notice that c(1)=1. Thus, if we want T(x) and S(x) to represent six-sided dice then they must both have b(x) and d(x) as factors, since multiplying by c(x) will not change the number of sides. So far we have
the following:

T(x) = a(x) * b(x) * d(x) * ??
S(x) = a(x) * b(x) * d(x) * ??

The only factor we have left is the c(x)^2. If we put a c(x) in both of T(x) and S(x) then we will have created symmetric dice since T(x) will be equal to S(x) (in violation of condition #3). The only other option is to put the entire c(x)^2 term into either T(x) or S(x) By symmetry, it is irrelevant which polynomial gets the factor, so we’ll try:

T(x) = a(x) * b(x) * c(x)^2 * d(x)
S(x) = a(x) * b(x) * d(x)

Now let’s check our work! Condition #1 is satisfied since T(x)*S(x) = a(x)^2 * b(x)^2 * c(x)^2 * d(x)^2. How about condition #2? Well, T(1) = 6 and S(1) = 6 so that looks good. Clearly, T(x) != S(x) so condition #3 is ok. Finally, neither term has a monic, so T(0)=S(0)=0 and neither die has a blank side.

So, now you have all the information you need to find these dice. I didn’t want to spoil it by writing out the actual sides, so I’ll leave you the final step.

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Alternative Dice — 1 Comment

  1. I saw this pair of dice around 2006. I don’t remember a proof, but stumbled across yours as I was inspired to try to find the dice today.

    As stated, a(x)^2 * b(x)^2 * c(x)^2 * d(x)^2 would yield a 14th degree polynomial, which would imply that one could roll a 14 on a normal pair of dice.

    It seems that your b(x) is incorrect. If b(x)=x+1, things work out for a normal pair of dice.

    The error was easy to miss since when x=1, x+1 = x^2 +1 = 2.

    The proof using polynomials is clever. Definitely a different way to look at things. Yay math!

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