# Can a die be made unfair without changing the center of mass or the exterior?

This question came from trying to design some 3d printed dnd dice. I think it's possible to make a die unfair without changing the center of mass or affecting the exterior shape*, but I'm not sure exactly how to do this, or what the result would be. That being said, I have some theories:

• A cubic die would be the most influenced by one of two strategies:

1. A heavy rod connecting the centers of two opposite faces.
2. A disk (or ring) equidistant between two faces.
• If the internal weights look the same from when viewed from any side (after rotation/mirroring in the viewing plane), the die is fair. Some examples:

• A sphere at the center
• Weights with xy, yz and xz symmetry
• A rod connecting opposite corners, such as (-1,-1,-1) and (1,1,1)

So, is it possible?

*As a side note, only the vertices are necessary for a convex polyhedron to correctly collide with a flat surface. Some really cool 3d printed dice consist only of vertices connected to the center.

• Bouncy rubber on one side, and double sticky tape on the other? Dec 11, 2015 at 5:09
• What exactly do you mean by "is it possible"? To make an isotropic dice? You give examples yourself. Do you want to know whether it will actually behave unfair? (And what's your definition of unfair?) Dec 11, 2015 at 10:37
• Yes, one can, for example, write the same number on every side. Dec 11, 2015 at 15:34
• @linuxick doesn't that change the exterior shape? Dec 11, 2015 at 19:04

TL;DR: Yes.

If you distribute the (interior) mass so there is a large anisotropy in the moments of inertia about the different axes of the die, then it will get a "preferred axis" of rotation (along the axis with the lowest moment of inertia). If you then launch the dice so they initially rotate about this axis, they are likely to maintain their rotation about this axis - meaning that two numbers will be less likely than the other four. Note that this will not affect the mean of the value rolled, because (on most dice) the sum of values on opposite sides is constant (7 for a regular D6).

You may have see the piece of equipment ("balancing caliper") used in casinos to confirm fairness of dice: this test for an offset in the center of mass, but doesn't checks for uniformity of the moment of inertia. In principle, if you detected lateral forces (torque) on the support while rotating about the corners, you could have that additional information - but I don't know of any equipment that does it. Of course every time you have your tires balanced, this is the kind of thing that is done all the time: they calculate not only the offset in the center of mass, but the second moment of mass distribution to make sure the tire will behave at all velocities.

And finally, you might be interested in this article describing a 20,000 roll comparison of D20 dice - neither of which turned out to be fair.

• I think the "preferred axis" is the one with the highest moment of inertia, due to dissipative forces in the material; see the story of Explorer 1. You'd also need the object to be fairly non-rigid if you wanted this effect to work over the time of flight of a die; maybe you could fill a toroidal channel in the die with mercury or something. Dec 11, 2015 at 16:15
• Even a fair die that is launched so that it initially rotates around one of its principal axes will maintain its rotation around that axis (though it will be more susceptible to slight errors, so it may be harder to throw in an unfair way). With a fair throw a die with a non-trivial tensor of inertia might move in a strange way, but while in flight, if we disregard aerodynamic effects, any orientation should be equally likely (not during a single throw, but over many throws). I would think the unfairness would have to come from interaction with the air or with the table. Dec 12, 2015 at 2:30

I am not sure I understand what you ask for, but here's a trial answer:

You want something that "looks" like a regular dice, i.e. $$\rho = 0$$ for $$x,y,z<0$$ or $$x,y,z>1$$. You want the same center of mass as a regular (homogeneous, isotropic) dice this fixes $$\int \mathrm{d}V\,\vec{r} \rho$$. But a dice is thrown, i.e. its fairness depends on its rotational properties, in technical terms its tensor of inertia, which is given by the second moments of $\rho$ without looking stuff up $$I \sim \int\mathrm{d}V\, \vec{r}\otimes \vec{r} \rho$$. So yes it is possible to at least affect the throw statistics with the given constraints. Some thought should go into how the dice is thrown to determine what you have to do to achieve a wanted result. But what you could do is start from the final position you want to avoid (probably 6 facing down) and adjust the distribution of mass such that this becomes unstable against small rotations. Such a reasoning leads to a higher density close to the center at the 6 side and far from the density at the 1 side

An intuitive solution might be to imagine the die balanced on a corner, attempting to spin. In the rod case, there would be torque to lay the rod flat, making sides parallel to the rod more likely. In the weighted disk case, there is torque to rotate along the disk's axis. Depending on the corner, one of the two sides parallel to the disk becomes more likely.

Simulating every combination of orientation and momentum for a settling die*, I'd expect most to result in the same face up as before, but some to be similar enough to the ideal case to result with a different face up.

*defined as having reached the part of the roll that every corner that touches the ground that will continue to touch the ground until the die comes to rest.

It would be possible: you could use ultra-thin iron filings (ferrofluid) mixed with the plastic on one side and balance the die out with another non-magnetic metal fluid on all other sides (right under the plastic so that it's not visible), and then stick a big, powerful magnet under the table. Make sure the magnet is not big enough to make the die stick to the table, but just to influence the die.

There is a similar method in Ocean's 13. They use a much more compicated method with weird-ass zippo-lighters and stuff.

I hope this helps.