Wikipedia:Reference desk/Archives/Science/2022 January 22

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January 22

Cuboctahedral dice (D14? D6/8?)

I asked this question in Mathematics, looking for a theoretical treatment, but didn't come up with a definitive answer.

A cuboctahedron is a solid shape with six square and eight equilateral triangular faces. It could conceivably be used as a 14-sided die. By symmetry, the probability of rolling any one the the square faces is equal, likewise the probability of rolling any one of the triangular faces. It seems likely, however, that those probabilities differ from each other, so such a die would be more likely to land on a square (or maybe a triangular?) face. Has anyone ever run this as an experiment and found out the probabilities? --Verbarson ^talk_edits 23:44, 22 January 2022 (UTC)

As a first guess, I'd look at the difference in area of each face: a square vs equilateral triangle with the same edge-length. I wonder if the distance from the centroid of mass to the different kinds of faces matters, as far as a roll "settling down" to lower energy? Would be an easy experiment. DMacks (talk) 00:42, 23 January 2022 (UTC)

You could cut smooth wood with carpenter tools, use potters clay to fill and glue the mold together, rent (probably) or buy (definitely) a kiln (or say Hey school, I just met you boy, and this is crazy, but here's my number, so call me maybe to schedule a bloc of cuboctahedron baking time), bake it, remove the ashes, shake it in a large box which you've padded enough to ensure it stays an uncracked cuboctahedron then drop it on a mattress and see which shape you got. Then repeat many times till the x-sigma "real probability of square" is down to plus or minus 0.05% from the ratio you threw so far or whatever's enough to satisfy you. Of course it'd take more throws to prove fair to very high accuracy than to prove not fair in a very biased die to the same accuracy. There's likely a calculator online that tells you where the sigmas are for any ratio of square throws to triangle ones. Sagittarian Milky Way (talk) 01:02, 23 January 2022 (UTC)

Or if you happen to be more skilled with computers than with hand tools, create a software model (some maths department might have one already) and print a bunch of copies using a 3-D printer. GeorgiaDC (talk) 02:25, 24 January 2022 (UTC)

Tangential to your query, the article Dice has tables of dice with various numbers of sides that are actually manufactured. The only entry for a 14-sided die is a Heptagonal trapezohedron, a very different form from the cuboctahedron you are considering: this suggests to me that the latter would not be easy to modify and make "fair", otherwise it would be in use. {The poster formerly known as 87.81.230.195} 90.193.128.231 (talk) 04:02, 23 January 2022 (UTC)

That table restricts itself to "uniform fair dice", where symmetry forces the probability of each of n faces to be 1/n. I want to know about an explicitly unfair shape for dice. How unfair is it? Is the probability of each square face greater than that of each triangular face, but is it by less than 8⁄6 so any roll is more likely to land on a triangular face (which is my guess)? --Verbarson ^talk_edits 22:31, 23 January 2022 (UTC)

I'd guess the centre of gravity of the die is lower when it sits on a square face, and that would then be the preferred final state. Another way of looking at it would be how much does the CG have to lift to rotate around one edge. Greglocock (talk) 05:27, 23 January 2022 (UTC)

I haven't proven it, but my gut feeling is that it's more likely to fall on one of the square faces, but the exact ratio may depend on the elastic properties and friction of the die and the surface on which is bounces. With more damping, I expect the square faces will be less favoured. So if you want to know, that's something to look into. PiusImpavidus (talk) 11:56, 23 January 2022 (UTC)

The only feasible way to find out for sure is the experimental way (which may turn out to produce different results depending on the details of the experimental set-up).  --Lambiam 12:58, 23 January 2022 (UTC)

What are the elastic properties and friction of craps dice following the rules of Vegas craps and what are cheaper ways of getting close without having to buy a casino-grade craps table and commission casino-quality cuboctahedral dice from whoever makes those? Sagittarian Milky Way (talk) 16:29, 23 January 2022 (UTC)

This is just a rough sketch. First, define what you mean by landing on a face: do you require friction to take away all kinetic energy, or is it good enough as long as any bouncing doesn't change which face is 'down'--in which case you should define what that means as well. Of course, without friction and with enough energy, the die might bounce around changing face forever, but maybe on the lower end, the amount of kinetic energy required to turn over a triangular face is different from that for a square face, and you can still make a quantitative comparison. Second, analyze how likely it is for a die to contact the floor at a vertex, along an edge, or flat on a face (some of these might be negligible depending on how you simplify your approach). For any such contact scenario, further analyze where the die would go from there depending on its momentum. Finally, tabulate the probabilities for 2 categories of outcomes--(1) die lands in a way with such energy that a triangular face is 'down', either at rest or oscillating along that face's edges and vertices, and (2) similarly for a square face. The probabilities might not add up to 1 because of possible existence of other scenarios, but at least you can compare the two. GeorgiaDC (talk) 20:47, 23 January 2022 (UTC)

Even with a sophisticated physical model that fixes parameters for the shape of the die (size, edge rounding), its mass and Reynolds number, the air and surface friction, and the elastic moduli of the die and surface materials, an unknown factor of paramount importance remains: the distribution of the initial state in the state space of states the die can be in (height, orientation, and linear and angular momentum). For the directions of the orientation and the momenta we can use a uniform distribution over the unit sphere, but there remain three scalars (height and the magnitudes of the momenta) with no plausible candidates for their distribution. I conjecture that if we use one-point distributions and let their centres grow without bounds, the distribution of outcomes of the die throws approaches a limit.  --Lambiam 13:21, 24 January 2022 (UTC)

Thinking a bit more about the effect of the initial state, I realized that as the release height grows – unless the other scalars grow exponentially with it (which in reality would cause the die to melt) – the velocity of the die on first impact will approach more and more terminal velocity in a vertical direction. On one hand, one might expect it to assume an orientation that minimizes drag; but, on the other hand, its wake will be turbulent and I'm not sure if that results in an irregular but strong rotational force. In any case, the limit behaviour as release height goes to infinity may not be "typical" for dice loosely thrown out of a player's hand.  --Lambiam 21:38, 24 January 2022 (UTC)

I figured there needs to be a lot of simplifications to even a theoretical model. For example, there would be no dropping from height, only imparting angular momentum to the die before allowing floor contact with negligible drop. Constant gravity. Perfect elastic collisions. No friction. I might even assume that any initial state is already floor contact along an edge of a face (triangular or square) at some angle. Then I might calculate how much angular momentum is necessary to overturn that face versus oscillating under a restorative gravity force. Finally, tabulating the results for all such edges, angles, and angular momenta, and see by how much the likelihood differs between settling on a triangle vs settling on a square. If one's more ambitious, can add linear momentum and static friction (only for rolling calculations, no damping/dissipation effects) as well. Although by this point, it's probably easier to gather empirical data instead by 3-D printing a die and rolling it under some setup. GeorgiaDC (talk) 05:29, 25 January 2022 (UTC)

Here is a thought that should be "unsettling" (LOL). With perfect elastic collisions and no friction, there is no energy loss, so the die keeps bouncing forever. As to the suggested simplification for the orientation at initial contact with the floor, the most likely case is with just one corner (vertex) touching the floor; for idealized dice with unrounded edges and corners, when in a uniformly random orientation, the other cases (a whole edge or even face) almost never occur.  --Lambiam 08:02, 25 January 2022 (UTC)

I think not necessarily. If initial energy isn't sufficient, the die can rock back and forth, revolve or even bounce about the same face. Cases of "down" face changing forever do not have to be included for comparison (triangle vs square). And say, only corners make contact at first, ignoring edges or faces. There is only a finite number of them, so should be able to analyse in some form. Notice also that in cases where there's no bouncing but only rolling, after an initial corner contact, I get the feeling that what follows is probably an edge contact. GeorgiaDC (talk) 10:12, 25 January 2022 (UTC)

As could be expected, the problem has been considered before.^[1] Letting p stand for the probability of the die settling on a square face, an experiment with a cardboard cuboctahedron in a run of 100 trials reportedly gave a value p ≈ 0.8 with σ ≈ 0.04. The proposer gives a mathematical solution based on the assumption that at some point the die stops tumbling or sliding but is then still rolling, turning along its edges from face to face. Each turn is from a square to a triangular face or vice versa, and the two require different kinetic energy levels; when insufficient for a next turn, the die settles. As the die tumbles, it loses energy in a somewhat unpredictable way; assuming a uniform distribution of the kinetic energy from 0 to a level in which the die must turn anyway, the proposer obtains p = 0.76239..., close enough to the experimental result. In an exact formula,

$p={\frac {~\,\,9+6{\sqrt {2}}-4{\sqrt {3}}-3{\sqrt {6}}}{10+6{\sqrt {2}}-4{\sqrt {3}}-3{\sqrt {6}}}}\approx 0.76239\,.$

(The fraction of a cuboctahedron's surface area contributed by its square faces is only ${\tfrac {1}{2}}(3-{\sqrt {3}})\approx 0.63397\,.$)  --Lambiam 10:01, 25 January 2022 (UTC)