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# Rolling an Archimedean Die

Suppose we construct a die in the shape of a cuboctahedron:

Anyone have any idea how to calculate the theoretical probability that when we roll it, it will land on a triangle as opposed to a square?

Image credit: http://www.korthalsaltes.com/

Note by Geoff Pilling
11 months, 2 weeks ago

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If it were just a function of relative surface area then the probability that it would land on a square face would be $$\dfrac{3 - \sqrt{3}}{2} \approx 0.634$$, but this of course does not take into account the mechanics of actually rolling a cuboctahedron. The square sides offer a more stable base so it would take more angular momentum to roll from this state onto a triangle, but as the square has one extra side there would be more directions in which to roll. A precise calculation would appear unlikely, so I would probably just go ahead and roll it 100 times and see how it compares to the $$0.634$$ value. (I don't have such a solid to experiment with but if I were to make a guess, I would anticipate a square probability of in excess of $$0.75$$.)

- 11 months, 2 weeks ago

That's an interesting experiment! I wonder what the results are like.

I've wondered if it is sufficient to project the edges onto a sphere, and then take the "sector surface area" as the relative probabilities (scale to a sum of 1). In a sense, I feel that accounts for 1) Landing on a vertex and spinning, 2) Landing on an edge, 3) Landing on a face.

(Though, this doesn't change the "actual surface area relative probabilities" by much)

Staff - 11 months, 2 weeks ago

Next time I get my hands on a cuboctahedron, I'll definitely give it a go! (At least find the experimental value)

- 11 months, 2 weeks ago

Depends on the thing's mass, bouncy-ness, angular momentum, and momentum. You can say the momentums are random within a certain range, but mass and bouncyness are still variables. Are you asking us to calculate it in terms of mass and bouncyness?

- 7 months ago