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