>To Ted Alper and others wondering about the dice and equal probablities. Here >are the dice: > Die one: 1,2,3,4,5,6 > Die two: 0,0,0,6,6,6 > Die three: 0,0,6,6,12,12 > And die four is an octahedron with sides 0,0,6,6,12,12,18,18
>Then depending on which 2 die you roll you have 1 to 24 and can combine more >to get higher numbers.
OK -- I guess what you mean is you roll die one always and (a) pair it with die two to get 1 -- 12 with equal probability (b) pair it with die three to get 1 -- 18 with equal probability (c) pair it with die four to get 1 -- 24 with equal probability
When you say "combine more to get higher numbers" you won't have equal probabilities any more if you use any of dice two three and four together.
(Of course there are other ways to construct/label dice to get equiprobable outcomes.)