Rolling a 14-sided die Suppose we make a 14 sided die by taking a cube and slicing a triangular pyramid off each corner, such that there are now 6 square faces and 8 triangular faces, each with unit side length (in other words, a cuboctahedron).
Assuming that the die is an evenly weighted solid and rolled with plenty of energy, what is the probability of it landing on a square (and hence with a square face up)?
There are two values that I know of that might yield insight to this problem.
Centre of Mass: the die seems more likely to remain on a face which lowers its centre of mass.
Area: the die seems more likely to land on a face with greater area.
The square both has a larger area, and a lower centre of mass; indicating that surely it is more likely to be landed on. But exactly how much more likely - and could such a probability be calculated as a function of the centre of mass and/or area?
 A: This is how I model rolling a die without actually rolling it, based on this answer of mine:

Cast a ray from the die's centre of mass in a random direction. Whichever face the ray passes through is the face the die lands on.

In other words, the probability of a (convex) polyhedron landing on a given face is the solid angle subtended by that face at the centre of mass divided by $4\pi$ (the solid angle of a sphere).
In the case of the cuboctahedron the centre of mass is the geometric centre. We can work out the solid angle $\Omega$ subtended by a square face, noting that it forms a square pyramid with base side length 1 and height $\frac{\sqrt2}2$. Using the formula provided on Wikipedia:
$$\Omega=4\tan^{-1}\frac{1^2}{2\frac{\sqrt2}2\sqrt{4\left(\frac{\sqrt2}2\right)^2+1^2+1^2}}=4\tan^{-1}\frac1{\sqrt8}$$
The proportion of $4\pi$ this is then works out to
$$\Box=\frac{4\tan^{-1}\frac1{\sqrt8}}{4\pi}=\frac{\tan^{-1}\frac1{\sqrt8}}\pi=0.108173\dots$$
This is for one square face. To get the probability of landing on any square face, multiply by six:
$$6\Box=\frac{6\tan^{-1}\frac1{\sqrt8}}\pi=0.649040\dots$$
Thus the cuboctahedral die described has around a 65% chance of landing on square faces, or around an 11% chance of landing on a given square face.
