06 August 2007

the design of asymmetric dice

Mark Dominus says that "For some reason I've been trying to construct a die whose faces come up with probabilities 1/21, 2/21, 3/21, 4/21, 5/21, and 6/21 respectively." He has not been able to do this exactly, although it wouldn't be too hard to come up with an approximate solution given his claim that "the probability that the hexahedron will land on face F is not proportional to the area of F, but rather to the solid angle subtended by F from the hexahedron's center of gravity."

This sounds right -- if you throw a die in the air then it'll rotate around its center of gravity, and you expect that whichever face is pointing "down" when it hits a table for the first time is the face which will eventually settle as "down". (Note that because the faces of a general polyhedron are not necessarily parallel, I'm considering the "down" face, not the "up" face.) But I'm not entirely sure if it's true, because if the die comes close to landing on an edge it might take a funny bounce; I suspect the system is a little more chaotic than Mark's representation of it, and the only real way to solve the problem is experimentally. Unfortunately, the "experimental" part would require tossing a die a very large number of times, and would be incredibly boring. Stereotypically, one gets one's grad students to do that. But I'm a grad student and I wouldn't do it.

What makes Mark's problem difficult is the lack of symmetry; each face has to be different. Let's say, hypothetically, that I wanted to make a die which had two faces which come up with one frequency p and four with another frequency q; clearly 2p + 4q = 1. If q = 0 this is just a coin; if q = 1/4 it's some kind of four-sided stick; my guess is that we can smoothly vary q from 0 to 1/4 by taking a wide variety of rectangular boxes with two of the three dimensions the same.

Similarly, if we want a die whose sides come up with frequencies p, p, q, q, r, r we should be able to do it by taking rectangular boxes with all three dimensions different.

If we're willing to go away from the rectangular box, but have a shape which is still topologically a cube (by this I mean all six sides are quadrilaterals, meeting three at each vertex) I think we can get dice where four faces have the same frequency and each of the other two has a different frequency by using a truncated square pyramid; clearly the four trapezoidal sides each come up with the same frequency, by symmetry. There are two degrees of freedom in constructing such a truncated pyramid (say the ratio between the height and the side length of the base, and the ratio between the side length of the top and the side length of the base) so there's no obvious "dimensionality" reason why this wouldn't work. (A degenerate case of this has five of the six faces having the same frequency.)

And it's probably possible to get three faces with one frequency and three with another by "stretching" the three faces around a single corner, although I'm having trouble picturing how exactly one would do that.

But all of these constructions take into account the symmetry of the cube; I'm assuming that faces that "look the same" are landed on with the same frequency. Mark can't take advantage of that symmetry in his problem.

I also suspect that designing exotic dice like these is difficult, or perhaps impossible, because the probabilities might not be "stable"; they might depend on how hard one throws the die, which defeats the purpose of dice, which is to be a source of randomness that can't be controlled by the person throwing them. Wikipedia offers some oddly shaped dice, though, and seems to claim there exists a fair seven-sided die in the shape of a pentagonal prism, which suggests that stability isn't as much of an issue as I suspect.


Anonymous said...

"because if the die comes close to landing on an edge it might take a funny bounce; I suspect the system is a little more chaotic than Mark's representation of it, and the only real way to solve the problem is experimentally."

Wouldn't the edge&corner problem cancel for each side?
Taking a funny bounce is no different then spinning several more times in the air.
It hits and bounces in the air spinning; it doesn't make a difference whether it lands on its edge, side or corner.
The die either stops, bounces in the air or rolls on the table.

Even normal die land on their edges, sides and corners; so I don't see this as a difficulty.

My question would be: is spinning in the air the same as spinning/rolling on the table for the weighted die?

Anonymous said...

My intuition on the "funny bounce" problem:

Chaos can actually work to our advantage here. If the system is sufficiently ergodic, then the (small) region of phase space which leads to funny bounces will be more-or-less evenly spread over all of phase space. Most will land in one of the six "settling areas", while sometimes there will be yet another funny bounce, and so on.

The upshot is that (if the bouncing is indeed ergodic) the fraction of the volume of phase space in the basin of a particular attractor will be the same as the fraction of all the settling areas of that attractor's settling area.

Michael Lugo said...


I don't know that much of the requisite theory, so I can't be sure, but that certainly sounds plausible.

Anonymous said...

I'm pretty sure there's got to be more to dice than solid angle. If you assume that a penny is a perfect cylinder, the edge subtends about 7% of the solid angle through the center, and you don't get nearly that many coins landing on edge. There's some complicated physics going on here; if you place a penny on a table in a random orientation, it'll balance a fair fraction of the time, but if you drop it from a height, it'll hardly ever balance. The "settling area" depends on the kinetic energy.

Maybe you can model it that way: figure out some combination of the probability that it'll land with a particular face down and the probability that it'll land with kinetic energy small enough for it to stay there.

Anonymous said...

It seems like in the case of the truncated pyramid the larger square face will always have a probability greater than the trapezoidal faces of ending up face down, and the smaller square face will always have a smaller probabability. So this doesn't really give a general solution for a 6 sided die of 4 sides equal probability.

I have no math to back this up, but I suspect it fits in with the solid angle theory.

Anonymous said...

Instead of having grad students do the tossing one could have them build a robot to do the tossing and to tabulate the results. It might even be fun, if you're into that sort of thing.

Unknown said...

Isabel, as far as I can tell, Wikipedia doesn't say that the 7-sided pentagonal prism die is fair, only that it exists. (Even if Wikipedia did say it was fair, I wouldn't really trust it on that kind of detail.)

Steve, I think maybe you're not visualizing a sufficiently great variety of truncated square pyramids. For instance, consider the Washington Monument (without its cap); it seems clear to me that a die shaped like this would be much more likely to land on one of the trapezoidal sides than on either square endcap.

Isabel and Harald, it might be easier to get a grad student to figure out how to run the experiment using some sort of physics simulator.

Anonymous said...

You're absolutely right cwitty. I was to restricted in my thinking about the ratios of pyramids. I guess I was just thinking of a typical egyptian pyramid aspect ratio and truncating the top. thanks for clearing that up.

Thinking about it even more, If you chop enough off the top of any pyramid it won't even be stable on the trapezoidal side. It'll be like a big square coin. So I guess I just wasn't thinking enough.

I might have to make some out of wood and play with them. I actually made a small box about 4 inches on a side last year in this shape, but I don't really want to roll it around and bang it up.