For what follows, we define a * loaded die *as a discrete probability distribution with six outcomes (labelled {1,…,6}), each of which has positive probability. To each such die, we associate a generating polynomial, given by

in which denotes the probability of the outcome *i*. If corresponds to another such die, we note that the product

has coefficients which reflect the probabilities of certain dice sums for *p* and *q *(and this is the utility of generating polynomials). We are now ready to ask the following question:

**Question: ***Does there exist a pair of loaded dice such that the probability of rolling any dice sum ({2,…12}) is equally likely?*

*Proof: *Equivalently, we question the existence of dice polynomials *p* and *q *such that

To see that the answer must be no, we observe that *pq* must then have precisely two real roots (counting with multiplicity). On the other hand, *p* (resp. *q*) has at least two real roots (counting multiplicity), since is a real polynomial of odd degree (hence admits a real root). Thus *pq* has at minimum four real roots, and we have a contradiction.

An extension of this argument shows the following:

**Theorem: ***There is no finite set of n>1 loaded dice such that the probability of any dice sum ({n,…6n}) is equally likely.*

We can extend the class of loaded dice to the class of * null-loaded dice* by allowing the probabilities of certain die rolls to equal zero. In this case, our theorem still holds: the proof holds verbatim provided that and remain non-zero, and if either is zero, our theorem holds trivially. Moreover, we may consider the class of null-loaded

*dice, and ask if there exists a collection of*

**k-sided***n>1*such dice (throughout which

*k*may vary, in general) such that the probability of any (attainable) dice sum is equally likely. If is fixed as our dice vary, we have the following:

**Theorem: ***There is no simulation of a fair (m(k-1)+1)-sided die via dice sums on a collection of m null-loaded k-sided dice.*

*Proof: *If, to the contrary, such a collection were to exist, we may write

Equating coefficients gives . Moreover, consideration of the coefficient gives us

It follows by the AM-GM inequality that

which presents a contradiction for .

*Note: when k is even, this result admits a proof analogous to that given in the ** case.* If we restrict the scope of our previous theorem to loaded *k*-sided dice, the following is immediate:

**Corollary: ***There is no simulation of a fair die via dice sums on a collection of m loaded k-sided dice.*

Matters become far more complicated when is allowed to vary with the die . In this case, the simulation of a fair die via dice sums becomes possible:

**Example: ** Let denote the fair 2-sided die with outcomes , and let denote the null-loaded 3-sided die with outcomes and respective probabilities . Then the dice sums for and have outcomes {2,3,4,5}, each of which is equally likely. *(In other words, we may simulate a fair 4-sided die via dice sums of two fair 2-sided dice (as long as one die has been renumbered {1,3}.)*

Underlying all of this lies the factorization of the polynomial

into scaled cyclotomic polynomials. To wit, if may be written as a product of polynomials with non-negative coefficients such that , then the null-loaded -sided dice (now numbered for convenience) associated to the polynomials have equally probable dice sums. In particular, it is an easy exercise to show that any fair *m*-sided die can be simulated as a null-loaded dice sum, provided that *m* is composite. In what follows, we’ll prove just one theorem concerning varied (concerning the factorization of over ).

**Theorem: ***Suppose that the dice sums of a collection of n null-loaded -sided dice simulate a fair m-sided die. If the probability of each outcome of is rational, then each non-null outcome of the die is equally likely.*

*Proof: *Let denote the generating polynomials for the dice (numbered ). Then

and Gauss’s lemma gives rational constants such that (with ). Specifically, is the constant term of , and we note that each coefficient of is a non-negative integer. Let be the coefficient of . If , then the coefficient of exceeds 1 (by non-negativity of the coefficients), a contradiction. Hence is 0 or 1, and our result is immediate.

Some food for thought –

**Exercise: **We say that a null-loaded *k*-sided die (labelled {0,…k-1}) is *primitive** *if the associated polynomial is irreducible in . Prove that a fair *m-*sided die arises as the dice sum of a collection of primitive dice if and only if *m* is a prime power.