Pairs of Dice

If you roll two six-sided die and add up their values, you get a certain probability distribution. Here it is for reference (rescaled):

\[\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline X & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline P(X) & 1 & 2 & 3 & 4 & 5 & 6 & 5 & 4 & 3 & 2 & 1 \\ \hline \end{array}\]

Our goal is two find all other pairs of die so that it has the same probability distribution.

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Now I will impose the following conditions:

• sides must be a positive whole number
• the die must be six sided

This problem is almost asking for the generating function approach. So here it is!

Each dice is represented by the polynomial \((x + x^2 + x^3 + x^4 + x^5 + x^6)\). Recall that the exponent gives the value and its coefficient the probability. Now for two die it is:

\[(x + x^2 + x^3 + x^4 + x^5 + x^6)^2 = x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + 6x^7 + 5x^8 + 4x^9 + 3x^{10} + 2x^{11} + x^{12}\]

(the reason this works is because polynomial multiplication is the same as convolution of their coefficients.)

So in essence we want another factorization of \(f(x)^2\), where \(f(x) = (x + x^2 + x^3 + x^4 + x^5 + x^6)\). In particular, we want a factorization of \(f\) into two polynomials, ie, \(f = P_1 P_2\) where both \(P_1\) and \(P_2\) satisfy

• \(P(0) = 0\) (no \(x^0\) terms)
• \(P(1) = 6\) (if \(P = \sum a_i x^i\), \(\sum a_i = 6\) since it’s six sided)
• all coefficients are non-negative integers.

Note that \(f(x) = x(1 + x + x^2 + \cdots + x^5) = x \frac{x^6 - 1}{x-1}\). Now using the some roots of unity stuff,

\[f(x) = x(x+1)(x^2 - x + 1)(x^2 + x + 1)\]

(A faster way to get these are the cyclotomic polynomials)

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Therefore

\[f(x)^2 = x^2(x+1)^2(x^2+x+1)^2(x^2-x+1)^2\]

Writing \(P = x Q\) (each of the \(P\)’s need an \(x\), otherwise they would have a constant term!), we need \(Q_1 Q_2 = (x+1)^2(x^2+x+1)^2(x^2-x+1)^2\) with \(Q(1) = 6\).

Now evaluating each factor at \(x = 1\):

\[(1+1) = 2, \quad (1+1+1) = 3, \quad (1-1+1) = 1\]

Now \(Q(1) = 6\) implies each die must receive a factor of \((x+1)\) and a factor of \((x^2+x+1)\). The only freedom lies in which way we divvy up \((x^2-x+1)^2\).

Case 1: Equal split

\[Q_1 = Q_2 = (x+1)(x^2+x+1)(x^2-x+1) = 1+x+x^2+x^3+x^4+x^5\]

This is our original solution.

Case 2:

\(Q_1 = (x+1)(x^2+x+1) = x^3+2x^2+2x+1 \implies P_1 = x^4+2x^3+2x^2+x\) \(Q_2 = (x^2+x+1)(x^3+1)(x^2-x+1) = x^7+x^5+x^4+x^3+x^2+1 \implies P_2 = x^8+x^6+x^5+x^4+x^3+x\)

This corresponds to dice with faces \(\{1, 2, 2, 3, 3, 4\}\) and \(\{1, 3, 4, 5, 6, 8\}\).

Note that swapping the factor gives the same answer.

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Since we covered all possible cases, we have proved that there is exactly one non-trivial solution!

These are known as Sicherman dice, first discovered by George Sicherman in 1978. Solving discrete problems with generating functions always does feel a bit magical!

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I wanted to write this since I eventually want to write something motivating the convolution theorem. And generating functions really help with this, I feel. So I would like to rewrite this to introduce and motivate generating functions a bit more in the future.