Conference rusure::math

    re .0
    
    Are the two dice labelled the same as each other? That is, it would
    seem that if we took two normally labelled dice and on one we increased
    the label on each face by r and on the other we decreased the label on
    each face by r then the sum for any combination will be unchanged. If
    faces are restricted to non-negative integers (i.e. that which can be
    represented by putting whole dots on the face) then r must be equal to
    1.
    
    If the two dice must be labelled the same, nothing springs to mind.

    Let's specify that each face of each dice shows at least 1.
    
    A

    Try {1,2,2,3,3,4} and {1,3,4,5,6,8}. I believe this and the
    conventional labelling are the only solutions with all numbers greater
    than or equal to 1. Now all I need to do is prove it!
    
    Andy.

.3 works for me - very cute - and a quick (& possibly buggy) computer search
reveals no other solutions. 

I love problems that twist the absolutely familiar, and I wonder why I haven't
seen these dice in gimmick shops... 

(Not to pick nits, but there is a "begging the issue" answer to the problem;
all "ordinary" dice have the spots arranged so that opposite faces sum to 7,
so technically a rearrangement of the faces that breaks this would be a
solution - but .3 is clearly the intended solution.)

    .3 is the solution.
    
    One approach is to use polynomials. Let ax^b denote the fact that a
    dice rolls b with probability a. So a regular dice can be represented
    as:
    	r = (x+x^2+x^3+x^4+x^5+x^6)/6
    
	So our two dice can be coded as f and g such that:
    
    	fg=r^2
    
    We also know that f(1)=g(1)=1, and f(0)=g(0)=0.
    
    We can use the uniqueness of factorization of Z[x], and it takes just a
    moment to show that the only answer is the one identified in .3.
    
    If we junk the requirement that each face shows a +ve number, then we
    can modify the two dice in .3 so that the sum of opposite faces is 7.
    One of the dice is then:
    
    	{2,3,3,4,4,5}
    
    which is used in figure wargaming (WRG rules, UK) and is known as an
    "average dice". I find it neat that the other dice:
    
        {0,2,3,4,5,7}
    
    "cancels out" the average dice exactly.
    
    What about nD6? What about 2Dm?
    
    Andrew

Using the polynomial metaphor its easy to come up with non-cube solutions as
well; for instance, a (0,3,3,6) tetrahedron and two (1,2,3) toblerones [named
after the chocolate bar - what's the official name of that shape? - except with
rounded edges so they never land on end] would give the same distribution. Or,
if you don't like zero as a face value and can construct an equiprobable
nonahedron, you could use (1,4,4,7) and (1,2,2,3,3,3,4,4,5)...

The only tetrahedral dice pair giving the same sum probabilities as
{(1,2,3,4),(1,2,3,4)} is {(1,2,2,3),(1,3,3,5)}.

There are three octahedral dice pairs giving the same sum probabilities as
{(1,2,3,4,5,6,7,8),(1,2,3,4,5,6,7,8)}, and they are:

	{(1,2,2,3,3,4,4,5),(1,3,5,5,7,7,9,11)}
	{(1,2,3,3,4,4,5,6),(1,2,5,5,6,6,9,10)}
	{(1,2,2,3,5,6,7,7),(1,3,3,5,5,7,7,9)}

Somebody wanna do deca-, dodeca-, icosahedrons? There seem to be 1, 5, and 2
possibilities, respectively, but its getting late...
							Richie