Alternative Dice

	Here is an interesting 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, ie, the probability of a die P(x)
rolling a "n" is:

	coeffecient(P(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 all we need to do is 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) != 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). And 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 

	where:

	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 incurr
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.
alleft.com

Alternative Dice
