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
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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
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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 one
method for finding the solution for Sicherman Dice. Using the same
polynomial factoring method, I found a set of dice that mimic two
eight-sided dice, and two unusual dice that mimic 3 6-sided dice. Can you
find them, or use this method to find something else interesting? Write me.
R William Gosper investigated a different order 12 polynomial, and found a
way 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 to
factors with 4, 6, 8, 12, or 20 terms. In addition to dice with
4/6/8/12/20
sides, gamers also use 10-sided dice.
http://www.8westhall.freeserve.co.uk/polydice.htm

8-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 three
tetrahedral 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
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There are actually 3 solutions for a pair of 8 sided Sicherman dice.

1,2,3,3,4,4,5,6 and 1,2,5,5,6,6,9,10
1,2,2,3,3,4,4,5 and 1,3,5,5,7,7,9,11
1,2,2,3,5,6,6,7 and 1,3,3,5,5,7,7,9

Brian Trial