Fair but irregular polyhedral dice
I occasionally get essentially this question from non-mathematicians, some of whom are very pleased with their appeal to the Intermediate Value Theorem. But I always reply that the whole discussion is absolute nonsense: the only possible notion of a fair die is an isohedral one, because for any other die, it depends how you throw it.
What do we mean by saying that a die is equally likely to land on any of its faces: what is the random element? In theory, if we could accurately measure the speed and spin of the die as it is released as well as coefficients of friction and restitution etc, we could work out in advance which face the die will land on, so in this sense there is no randomness. The randomness only arises when you choose how to throw the die: the thrower selects from a continuum of "possible throws", and we assume that he samples from some probability distribution on this continuum. Morever, there is an action of the rotation group of the die on this continuum, and what we mean by saying that an isohedral die is fair is that we assume the probability distribution the thrower uses is invariant under this action. There is no sensible further assumption that we can make about the probability distribution which will ensure that two faces of the die which are not in the same orbit under the rotation group will occur equally often. However "obvious" it is that an octagonal coin is very unlikely to land on its edge, and a pencil is very unlikely to land on its end, there's no meaningful way to find a happy medium between these. So there is no meaningful mathematics to be done here. (Feel free to substitute "pure" for "meaningful" in the last sentence.) If you wanted to make a machine to throw a die in a standard way, then you'd have no chance of making a fair die, because with a perfect machine the die would always land on the same face.
Incidentally, the above discussion suggests that even isohedrality is not enough for a fair die - we require there to be only one orbit of faces under the rotation group of the die, not the full symetry group.
Depending on rules and technique, a person throwing a die can reasonably control the amount of angular momentum, the total kinetic energy when it first lands and the angle of its trajectory.
I want to suppose that the collisions of the die with the throwing surface have a reasonably high coefficient of restitution, enough that the die will undergo a good number of hits on the surface before it comes to rest. (One could imagine an alternate model, where the main randomization is by shaking the die before throwing, and the die stops where it lands ---- but that's not hte situation I want to discuss).
I think there's a reasonable range of polyhedral die shapes and energies where if the bouncing were completely elastic, the system would be ergodic --- the possible positions and motions of the die, up to translations in the plane of the throwing surface, would be visited a.s. in proportion to their measure among all states of the same energy. If the surface of the die were smooth, but just marked off into different areas of contact, this would not generally be true: KAM theory (small divisors and invariant tori) very often makes it non-ergodic. If the die can act like a top, it's not ergodic at that energy level. But I think of the rolling die more like a particle bumping into numerous obstacles, and systems like that are often ergodic.
The rolling of a real die is not perfectly elastic, and kinetic energy is gradually lost.
Here's a hypothesis that should guarantee fairness in the limiting case where energy is lost very slowly: Let's suppose we have a rule to partition phase space into sets $A_i$ associaed with the different possible outcomes. We want
The intersection of $A_i$ with each energy level E has volume $V(E)$ independent of $i$.
The dynamics is ergodic in each component of each energy level that intersects more than one $A_i$.
The labeling by $A_i$ depends reasonably on energy level --- for every pair of nearby energy levels $E$ and $F$, for most $x$ at energy level $E$ and most $y$ in energy level $F$ such that $d(x,y) < \epsilon$, $x$ and $y$ are in the same state. Also, the labeling shouldn't tunnel to a different connected components of the phase space of energy $<\le E$: the ratios of measure of $A_i$ intersected with each connected component of an energy level should stay the same.
With these hypotheses, with sufficiently slow loss of energy, the final state should be uniformly distributed.
If the dynamics at enough energy levels also has reasonably high entropy and is mixing (I think both are likely to be true for reasonable die shapes), then the uniformization should occur fairly well at realistic rates of energy loss.
The big difficulty though is condition (1). I think that pretty often, even with a symmetric die, the phase space becomes disconnected well before the die comes to rest.
For a standard cubical die, what are the components of phase space just above where it settles on a face? I think it can roll on 4 sides and not have enough energy to switch which four sides. At the same energy level, it might be spinning slowly on a vertical axis on one face. If so, that would make 9 components, 6 of which are already committed to one face. These are the kinds of things one would need to understand to show an asymmetrical die is fair. With more complicated dice, the fragmentation into components looks much trickier.
The volume of phase space must be equitably apportioned at each transition where the phase space disconnects, until the final outcome is determined, otherwise one could influence the outcome by the energy of the throw.
Connely's suggestion of a twisted deltoidal Icositetrahedron might work, but it might fail the test (1) as energy levels are decreased. Even though each face is the same, I'm not convinced that the fragmentation into continents, before the die has settled on one face, would be fair, since it presumably depends on bigger neighborhoods that are not exactly the same.
If one understood in detail how the components of energy levels separate and if there aren't too many of them, then one should in principle be able to engineer a die to be fair with the help of the Brouwer fixed point theorem (multi-dimensional intermediate value theorem). It would appear to be quite a challenge, though, for all but the simplest examples when there is only a small number of symmetry classes of faces.
Proving 2 seems like a significant challenge. My guess could be wrong, it might usually be non-ergodic. It's worth thinking about in it's own right.
Further thoughts:
Actual dice are made with rounded edges and rounded corners. How this rounding is done seems significant.
If the projected image of a die along a certain axis is almost round, then at low energy levels it should roll more easily about those axes than about axes where the projection is bumpy, other things being equal. This suggests larger components of the phase space for these kinds of rolls, when the phase space becomes disconnected. Also, it depending on the details of the shape, it seems likely there are ergodic components associated with rolling at energy levels above where the phase space becomes disconnected --- this is similar to the stability of a wheel rolling on somwhat bumpy terrain, which is explained by KAM theory.
It seems interesting to try to design shapes that appear fair, but are not, by exploiting this kind of behavior: creating rolling modes or rocking modes that, as the die settles down, tend to funnel the behavior into preferred outcomes.
Joe,
Here is convex polytope that could be a fair die. Consider the rhombicuboctahedron which is one of the Archimedean solids. There are three rings of square faces, each of which divides the polyhedron in half. Take one of these rings and rotate the top half by 45 degrees. This is a famous "fake" Archimedean solid. The symmetry group of this polytope is not transitive on all its vertices. There are two orbits of vertices of 8 and 16 vertices each. Take the dual of this polytope. Each face is a deltoid/kite. All 24 faces are congruent and at the same distance from the centroid, but it is not isohedral. I think that all the second moments are equal as well. Is this a fair die?
Bob C.