riddle of the day

Two dice can carry different labels and still leave exactly the same fingerprint when their results are added.

The Dice Signature

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Chance and uncertainty

Riddle statement

When two ordinary dice are rolled, the sums from 2 to 12 occur, among the 36 possible outcomes, with frequencies

1,2,3,4,5,6,5,4,3,2,1.

You may make two other fair six-sided dice. Every face must carry a positive integer, and labels may be repeated.

Is there a pair different from two ordinary dice that produces exactly the same distribution of sums? If so, find it and determine whether any other pair exists.

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Solution

Answer: yes. Up to swapping the dice, the unique nonstandard pair is

$ 1,2,2,3,3,4 \qquad\text{and}\qquad 1,3,4,5,6,8. $

These are known as the Sicherman dice.

Encoding the sums. For a die whose faces are \(a_1,\ldots,a_6\), define

$ P(x)=x^{a_1}+\cdots+x^{a_6}. $

If the two dice correspond to \(P(x)\) and \(Q(x)\), the coefficient of \(x^s\) in \(P(x)Q(x)\) is exactly the number of the 36 outcomes whose sum is \(s\).

For an ordinary die,

$ D(x)=x+x^2+\cdots+x^6 =x(x+1)(x^2+x+1)(x^2-x+1). $

We therefore seek two polynomials \(A(x)\) and \(B(x)\) satisfying

$ A(x)B(x)=D(x)^2. $

Constructing the pair. Distribute the factors as follows:

$ A(x)=x(x+1)(x^2+x+1) =x+2x^2+2x^3+x^4, $
$ B(x)=x(x+1)(x^2+x+1)(x^2-x+1)^2 =x+x^3+x^4+x^5+x^6+x^8. $

The coefficients specify how many faces carry each label. Thus the dice are

$ 1,2,2,3,3,4 \qquad\text{and}\qquad 1,3,4,5,6,8. $

Since \(A(x)B(x)=D(x)^2\), their sum frequencies are exactly the same as those of two ordinary dice.

Why no other pair exists. The smallest sum is 2 and occurs only once. Each die must therefore have exactly one face labeled 1, so

$ A(x)=xP(x), \qquad B(x)=xQ(x). $

The three factors

$ x+1,\qquad x^2+x+1,\qquad x^2-x+1 $

are irreducible in \(\mathbb Z[x]\). Unique factorization means that \(P\) and \(Q\) must be obtained by distributing between them the two copies of each factor occurring in \(D(x)^2\).

Each die has six faces, so

$ P(1)=Q(1)=6. $

At \(x=1\), the three factors have values \(2\), \(3\), and \(1\). Each polynomial must therefore receive one copy of \(x+1\) and one copy of \(x^2+x+1\).

Only the two copies of \(x^2-x+1\) remain:

  • giving one to each die recovers the ordinary pair;
  • giving both to the same die produces the Sicherman pair;
  • swapping which die receives both copies does not create a new pair.

There are no other possibilities.

The essential idea is to replace a search through labels with a factorization of the entire sum distribution.