There are commercially-available 5- (and 7-) sided dice, the best of which is designed by Lou Zocchi of GameScience: http://www.gamescience.com/3502-0013.jpg These dice of course aren't "fair by symmetry", but you can argue that there is some die of this combinatorial type that is "fair by continuity", and you just need to find the right height. But figuring out that height by physics is an awful idea. Instead, Zocchi worked with engineers and build a die-rolling robot, then cut and rolled dice with thicknesses in increments of 1mm, rolling each die some 10,000 times, if I recall correctly, and then interpolating the curve to find out the "fair" thickness. But because it's not "fair by symmetry", this extreme precision is a little unwarranted: if you're rolling your die onto 10mm-thick plexiglass, as the robot did, then the die really is fair. But if you're rolling it onto something that's bouncier or harder (or sticky or slanted or underwater or on the moon or...), then the fair thickness would be different. --Michael On Thu, Sep 20, 2012 at 11:20 AM, Warren Smith <warren.wds@gmail.com> wrote:
Neat idea! Synopsis for people who didn't click through: a set of four 12-sided dice, faces numbered from 1 to 48 with each number appearing once, such that if you roll all four of them, all four permutations of the dice are equally likely.
With three people, it looks to me like you can do it with six-sided dice. Here are the 11 solutions, according to a little Mma, though I'd be happy to have someone else confirm that I didn't mess things up:
{{1, 5, 9, 12, 14, 16}, {2, 6, 7, 11, 13, 18}, {3, 4, 8, 10, 15, 17}},...
--yes but, you could just roll ONE die with the 6 permutations written on its 6 faces. And for 4-item perms use an icosahedral die plus a coinflip to get the 24 possibilities... or a cube & tetrahedron... So I don't see the point. (Although perhaps this has some other application.)
--Here's a different problem: Devise a convex polyhedron with 5 faces, such that the probability it lands on each face, is 1/5. I have a solution in mind which in fact works for any number N>=4 of faces (here N=5), although I'm not sure whether we should accept my solution. The question inside the question is: "what is the right probabilistic model?" And the answer inside the answer is not so obvious to me.
--Warren D. Smith
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