`Hi Ed-The dice thing is cool. It's a problem I remember solving as a kid, without any machinery, just keep going until you find it, but I also ran across the generating-function idea earlier this year and was very happy to see there was a more mechanical way of going about it, something that I could throw out to a Combinatorics class.Your problems seem a little loose though - a solution to the second might be to have a regular 6-sided die along with a 36-sided die that just carries the distribution of the sum for two dice. I don't imagine this is what you were looking for - what did you have in mind? I did find something this morning that I think has a unique solution: find two 8-sided dice which mimic three four-sided dice. The #s on the 8-siders should be positive, and the max # should be the same on each.Dave Molnar----------------------------------------------------------------------------This is pretty well known, likely being the first think one learns about generating functions. But nevertheless, as you discovered, is a source of endless fun.For equivalents for two "standard" 8-sided dice, I found three sets (assuming two 8-sided with values at least 1):(1,3,3,5,5,7,7,9)+(1,2,2,3,5,6,6,7)(1,2,5,5,6,6,9,10)+(1,2,3,3,4,4,5,6)(1,3,5,5,7,7,9,11)+(1,2,2,3,3,4,4,5)For three 6-sided dice, there are too many possible combinations, so I mixed things up a bit:(1,2,4,5)+(1,3,4,5,6,8)+(1,2,2,3,3,3,4,4,5)Nick Baxter-----------------------------------------------------------------------------x2+ 2x3+ 3x4+ 4x5+ 5x6+ 6x7+ 5x8+ 4x9+ 3x10+ 2x11+ x12 = (x1+ x2+ x3+ x4+x5+ x6)2 =  (x1+ x2+ x2+ x3+ x3+ x4)(x1+ x3+ x4+ x5+ x6+ x8).  That's onemethod for finding the solution for Sicherman Dice. Using the samepolynomial factoring method, I found a set of dice that mimic twoeight-sided dice, and two unusual dice that mimic 3 6-sided dice. Can youfind them, or use this method to find something else interesting? Write me.R William Gosper investigated a different order 12 polynomial, and found away to divide a square into 10 acute isosceles triangles.With two-eight-sided dice, we have:    (x^1 + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8)^2 =    (x^1)^2 * (x^1 + 1)^2 * (x^2 + 1)^2 * (x^4 + 1)^2.If we assume that "dice" must be regular polyhedra, then we are limited tofactors with 4, 6, 8, 12, or 20 terms.  In addition to dice with4/6/8/12/20sides, gamers also use 10-sided dice.http://www.8westhall.freeserve.co.uk/polydice.htm8-sided solution:    {5, 4, 4, 3, 3, 2, 2, 1} and {11, 9, 7, 7, 5, 5, 3, 1}    Die 1:  (x^1) * (x^1 + 1)^2 * (x^2 + 1) =            (x^5 + x^4 + x^4 + x^3 + x^3 + x^2 + x^2 + x^1)    Die 2:  (x^1) * (x^4 + 1)^2 * (x^2 + 1) =            (x^11 + x^9 + x^7 + x^7 + x^5 + x^5 + x^3 + x^1)If we allow 0s, then we can form this set of three tetrahedral dice:    {7, 5, 3, 1} and {6, 5, 2, 1} and {3, 2, 1, 0}    Die 1:  (x^1) * (x^4 + 1) * (x^2 + 1) =            (x^7 + x^5 + x^3 + x^1)    Die 2:  (x^1) * (x^4 + 1) * (x^1 + 1) =            (x^6 + x^5 + x^2 + x^1)    Die 3:  (x^2 + 1) * (x^1 + 1) =            (x^3 + x^2 + x^1 + x^0)We can also redistribute the (x^1) factors to get other sets of threetetrahedral dice:    {6, 4, 2, 0} and {6, 5, 2, 1} and {4, 3, 2, 1}    {7, 5, 3, 1} and {5, 4, 1, 0} and {4, 3, 2, 1}    {8, 6, 4, 2} and {5, 4, 1, 0} and {3, 2, 1, 0}    {6, 4, 2, 0} and {7, 6, 3, 2} and {3, 2, 1, 0}    {6, 4, 2, 0} and {5, 4, 1, 0} and {5, 4, 3, 2}... Still working on the three 6-sided dice.-- Susan Hoover---------------------------------------------------------------------There are actually 3 solutions for a pair of 8 sided Sicherman dice.1,2,3,3,4,4,5,6  and 1,2,5,5,6,6,9,101,2,2,3,3,4,4,5 and 1,3,5,5,7,7,9,111,2,2,3,5,6,6,7 and 1,3,3,5,5,7,7,9Brian Trial`