show/hide this revision's text 4 added comments about "Markov Chain"

I just noticed Joseph's comment in his question about Markov chains. My observations about the correctness of trying to use Markov chains to describe the rolling of a die, fair or unfair:

If by state in the Markov chains, you mean just the "face" it is currently on or the "face" which is lower-most in attitude at a particular point in time, then it is inappropriate to use Markov chains because the likelihood of transitioning from die face $F_i$ to die face $F_j$ is not purely dependent upon the current state. If $F_j$ and $F_k$ are two "faces" adjacent to face $F_i$, then the likelihoods of transitioning $F_i \to F_j$ vs. $F_i \to F_k$ is not just dependent on the "current state" being $F_i$, but also dependent upon the velocity, position, and orientation of the die. The "faces" are necessary but not sufficient to encode state in such a way for Markov chains to be applicable: that the Bayesian requirement that "current state" at time $t$ is all that is needed to be known in order to be able to predict the likelihood of the state at time $t+1$ (if you talk about discrete time) or time $t+\varepsilon$ if you talk about continous time.

If by "state", you try to get around this factor that only current state be considered and not the history of how you came to currently be in that state, then you could try to add the vectors of position, velocity, and orientation as extra "states", which is valid in numerical simulation, because ultimately all reals are still encoded into limited precision "floating point" representations of reals. However, the transition table would be huge if you allowed even for 16-bit floating point representation.

I do not think that history-less "Markov chains" can be applied in this situation.

older answer components below
show/hide this revision's text 3 corrected spelling error "aon" -> "on", inserted steradians as appropriate

To answer Benoît Kloeckner's comment that <<[then] the solid angle under which each face is seen from the center of gravity alone would determine the probability of the dice landing on that face. But to determine all polyhedra for which this solid angle is constant is already a nice problem.>>

I don't believe that having similar solid angles is sufficient to determine the equal probabilities of the die landing on faces with similar solid angles.

Here is a construction for 2-d die (which can easily be converted into prismatic die, disregard if the die lands on "top" or "bottom" face, and look at the relative probabilities of landing on the prismatic faces)

Using polar coordinates $(r,\theta)$ , let's define a fair hexagonal die's profile as the closed path determined by the six vertices at

$(1,\frac{\pi}{3}), (1,\frac{2\pi}{3}), (1,\pi), (1,\frac{4\pi}{3}), (1,\frac{5\pi}{3}), (1,{2\pi})$

Now let us define an unfair hexagonal die's profile as the path defined by the polar coordinates

$(1,\frac{\pi}{3}), (1,\frac{2\pi}{3}), (100,\pi), (1,\frac{4\pi}{3}), (1,\frac{5\pi}{3}), (100,{2\pi})$

Now this die's center of mass (center of gravity) remains at $(0,0)$ since the material the die is composed of has uniformly homogeneous density. This unfair die also has each prismatic face subtending equal solid angles (and equal angles of $\pi/3$ for each edge in the $2$-dimensional case), however the this unfair die is highly biased towards landing on two faces to the detriment of the other four faces probabilities.

Thus Benoît Kloeckner's conjecture that

"the solid angle under which each face is seen from the center of gravity alone would determine the probability of the dice landing on that face"

is incorrect.

In fact, using this polar coordinate approach, it can be seen that using any three radii greater than $0$ in length yields a rotationally symmetric die profile with equiangular faces (edges which subtend equal angles , in $2$-d, prismatic faces which subtend equal angular regionssteradians of solid angle in $3$-d) and with center of mass still at $(0,0)$:

$(r_1,\pi/3), (r_2,2\pi/3), (r_3, \pi), (r_1,4\pi/3), (r_2,5\pi/3), (r_3,2\pi)$

but very few of these would be fair. Particularly the non-convex profiles, which also are equiangular, but make it possible to land on pairs of vertices/edges without landing aon on a specific face
show/hide this revision's text 2 corrected radii to 3 instead of 6, to keep center of mass at (0,0)

To answer Benoît Kloeckner's comment that <<[then] the solid angle under which each face is seen from the center of gravity alone would determine the probability of the dice landing on that face. But to determine all polyhedra for which this solid angle is constant is already a nice problem.>>

I don't believe that having similar solid angles is sufficient to determine the equal probabilities of the die landing on faces with similar solid angles.

Here is a construction for 2-d die (which can easily be converted into prismatic die, disregard if the die lands on "top" or "bottom" face, and look at the relative probabilities of landing on the prismatic faces)

Using polar coordinates $(r,\theta)$ , let's define a fair hexagonal die's profile as the closed path determined by the six vertices at

$(1,\frac{\pi}{3}), (1,\frac{2\pi}{3}), (1,\pi), (1,\frac{4\pi}{3}), (1,\frac{5\pi}{3}), (1,{2\pi})$

Now let us define an unfair hexagonal die's profile as the path defined by the polar coordinates

$(1,\frac{\pi}{3}), (1,\frac{2\pi}{3}), (100,\pi), (1,\frac{4\pi}{3}), (1,\frac{5\pi}{3}), (100,{2\pi})$

Now this die's center of mass (center of gravity) remains at $(0,0)$ since the material the die is composed of has uniformly homogeneous density. This unfair die also has each prismatic face subtending equal solid angles (and equal angles of $\pi/3$ for each edge in the $2$-dimensional case), however the this unfair die is highly biased towards landing on two faces to the detriment of the other four faces probabilities.

Thus Benoît Kloeckner's conjecture that

"the solid angle under which each face is seen from the center of gravity alone would determine the probability of the dice landing on that face"

is incorrect.

In fact, using this polar coordinate approach, it can be seen that using any six three radii greater than $0$ in length yields a rotationally symmetric die profile with equiangular faces (edges which subtend equal angles, prismatic faces which subtend equal angular regions) :and with center of mass still at $(0,0)$:

$(r_1,\pi/3), (r_2,2\pi/3), (r_3, \pi), (r_4,4\pi/3), r_1,4\pi/3), (r_5,5\pi/3), r_2,5\pi/3), (r_6,2\pi)$r_3,2\pi)$

but very few of these would be fair. Particularly the non-convex profiles, which also are equiangular, but make it possible to land on pairs of vertices/edges without landing aon a specific face
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