# 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?

• Well, the answer to your question depends on the base area of each pyramid, doesn't it? Jan 10, 2017 at 8:59
• I guess the solid you are describing is a cuboctahedron, isn't it?
– AugSB
Jan 10, 2017 at 9:01
• I think this is more suited for physics stackexchange. This is because here one assume that the physical model is trivial or given which is not true in this case.
– skyking
Jan 10, 2017 at 10:24
• I'm voting to close this question as off-topic because this is essentially a maths question. Jan 11, 2017 at 6:29
• Jan 11, 2017 at 15:34

## 1 Answer

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.

• This and that model is seriously flawed. We know from experience that flipping a dollar coin almost always lands on head or tail, but this model predicts it to land on the edge 7% of the tosses (the edge area is 7% of the coin area) - how often do that happen in reality?
– skyking
Jan 10, 2017 at 10:40
• @skyking The way coins are thrown varies a lot between people, and none of those ways are mathematically ideal. Furthermore, a coin is never a perfect cylinder (due to milling, inscriptions, wear, etc.) and the irregularities cause the coin to tip towards head or tail if it lands on the edge first. Since this is a mathematical model I can only talk in the ideal sense, without the variations introduced in reality.
– Parcly Taxel
Jan 10, 2017 at 10:52
• Do you really think that irregularities of the edge is the only thing that makes the coin end up on either side? I don't think that's a reasonable explaination. These factors do not help your model, even with a perfect cylinder you would probably not observe 7% edge. Besides your model wouldn't be affected by these factors anyway.
– skyking
Jan 10, 2017 at 11:55
• @skyking I forgot to mention inertia. That was something I had considered while writing the other dice answer, and it was agreed there that it introduces too many variables to be adequately represented in a model. Also affecting the coin's result are the ground and any energy losses due to sound or friction.
– Parcly Taxel
Jan 10, 2017 at 11:59
• If I understand correctly, this is not really a model for rolling a die, but rather for releasing it above a flat surface that absorbs all its kinetic energy (after randomly orienting it). Alternatively this could be seen as a model in which the cuboctahedron is projected on a sphere centered on its centre of mass and which is then dropped (or rolled). Jan 11, 2017 at 8:09