I am interested in determining a collection of geometric conditions that will guarantee that a convex polyhedron of $n$ faces is a fair die in the sense that, upon random rolling, it has an equal $1/n$ probability of landing on each of its faces. (Assume the polyhedron is composed of a homogeneous material; i.e., it is not "loaded.")

There has been study of what Grünbaum and Shephard call isohedral polyhedra, which always represent fair dice: "An isohedron is a convex polyhedron with symmetries acting transitively on its faces with respect to the center of gravity. Every isohedron has an even number of faces." It is clear such a polyhedral die is fair. Here is an example of the trapezoidal dodecahedron, an isohedron of 12 congruent faces, from an attractive web site on polyhedral dice:
          trapezoidal dodecahedron http://people.csail.mit.edu/%7Eorourke/MathOverflow/TrapDodeca.jpg
But a clever argument in a delightful paper by Persi Diaconis and Joseph Keller ("Fair Dice." Amer. Math. Monthly 96, 337-339, 1989) shows (essentially, by continuity) that there must be fair polyhedral dice that are not symmetric. For example, there is no reason to expect that equal face areas is a necessary condition for a polyhedral die to be fair. Nor is it reasonable to expect that the distance from each face to the center of gravity of the polyhedron is alone a determining condition. Rather it should depend on the dihedral angles between faces, the likelihood of one face rolling to the next—perhaps a Markov chain of transitions?

My question is:

Is there a collection of geometric conditions—broader than isohedral—that guarantee that a (perhaps asymetrical, perhaps unequal-face-areas) convex polyhedron represents a fair die?

Sufficient conditions welcomed; necessary and sufficient conditions may be too much to hope for! Speculations and literature leads appreciated!