# How many fair dice exist?

We know a coin is a fair die with a 50-50 probability for two alternatives. Similarly, all five Platonic solids are fair dice. That makes six solids that can be fair dice, but can there be more? One example could be a two tetrahedra pasted together along one face. The resulting solid is not platonic since two vertices have three faces meeting at them while three of them have four faces meeting at them. However, this too can be a regular die as far as I can tell since all faces are identical. The question is, how many solids can exist that can be used as fair dice?

• If you join two tetrahedra as you described, don’t you have three vertices with four faces? I don’t know that it’s possible to have only two faces at a vertex of any solid. – WGroleau Dec 27 '17 at 15:55
• I have seen a D10 land on a point and remain standing, not once, but twice, in the 30 years I have been using them as randomizers. I think somewhere I have a photo of one. One of the things that should be considered in a fair die, beyond extra miniscule faces (coin edge, die tip) is the fact that the moment of a die will change depending on its attitude to the imparting force, which biases the die to certain faces. – Pete Mancini Dec 27 '17 at 17:31
• What geometric property defines a fair die? I would start by assuming it it related to the angles between the normals of adjacent faces and the area of each of those faces. But that's not yet a workable definition, and looking at the answers so far it seems non-trivial to define it. An alternative approach would be to start with the notion of a random rotation, which should "select" one face. But that hinges on a definition of the centre of the die. – MSalters Dec 29 '17 at 14:00
• @M.Winter - the video is actually by one of the authors of the paper linked in the answer by J. Loreaux and he mentions the paper in his video. – Rohit Pandey Jan 1 '18 at 7:53

Infinitely many. For example, I think that by taking the intersection of $n$ identical spheres with adequate radius and centres on the vertices of a regular $n$-sided polygon, you create a fair die with $n$ outcomes. An example with 3 sides would look similar to the following:

[Chkhartishvili, Levan & Suryamurthy, Gokul. (2015). Volume of intersection of six spheres: A special case of practical interest. Nano Studies. 11.]

If we restrict ourselves to polyhedra, one could build a generalisation of your two-tetrahedra example: simply build two identical regular pyramids with $n$-sided bases and paste them together to make the two bases coincide. This gives a fair die with $2n$ outcomes.

Finally, while all answers (including this one) focus on solids with sides of equal surface, this need not be the case. There can be additional sides with arbitrary surface as long as the die can never land and stay on that side, for example when the projection of the centre of mass falls outside the convex hull of the side. For example a pencil sharpened on both sides: (Some symmetry of these "impossible sides" is required so that they do not alter the probabilities of the other sides.)

• And this cojoined pyramid solution has been used extensively by the industry to make dice. I have a 24 and 36 sided dice that are made like that. +1 for the infinite demonstration. – Mindwin Dec 27 '17 at 12:32
• Interestingly the last one that you noted is what I immediately thought of. A long regular n-gon prism with short pyramids on the ends such that it can never stable rest on any side of the pyramids is a trivially obvious die with n possible outcomes. I feel it a shame that it is relegated to almost a footnote in your answer since I feel it is much easier to picture how it works than your primary example. – Chris Dec 27 '17 at 15:49
• In fact, when I was a child, forty or fifty years ago, I remember reading a suggestion that if you lose te dice from your game, just poke different numbers of holes in the sides of a pencil. – WGroleau Dec 27 '17 at 15:52
• Example of prism dice: cdn.shopify.com/s/files/1/1483/3510/products/…. Most have triangular faces to reduce circumference, but it'd be easy enough to make them all with rectangular faces (like the d4 in the picture). – Bobson Dec 27 '17 at 22:34
• I was about to post the 'pencil' example before seeing this answer. :( +1 – tomasz Dec 28 '17 at 5:33

This problem was addressed in a 1989 paper of Diaconis and Keller entitled "Fair Dice" which appeared in the American Mathematical Monthly Vol. 96, No. 4, pp. 337-339.

There, they define a die to be "fair by symmetry" if it is a convex polyhedron whose symmetry group acts transitively on the faces. They determine all such polyhedra.

The polyhedra which are fair by symmetry are duals of the polyhedra symmetric with respect to their vertices. Each symmetry group of a fair polyhedron is represented by a regular solid or the dual of a semiregular solid. Thus in addition to the five regular solids there are thirteen individual polyhedra [the duals of the Archimedean solids] and two infinite classes [the duals of the prisms and antiprisms] among the fair polyhedra.

In other words, the Platonic solids, the Catalan solids, and the bipyramids and trapezohedra.

• The paper ends with: “The problem of characterizing all fair dice, not just those which are fair by symmetry or by continuity, is still unsolved.” – ShreevatsaR Dec 27 '17 at 8:06
• If I may add (somehow late) that some solids (like Pyritohdera and Tetartoid) are also fair dices. They combinatorially equivalent to the dodecahedron however (so they are mostly of aesthetical interest). – ARG Apr 14 '20 at 15:35

Just to add a type that I'm surprised no-one else has mentioned...

This class of shapes is called a "trapezohedron" or "deltohedron", and can be extended to any even number of faces. Typically, it is used for numbers that are not multiples of four, with a dipyramid used instead for those, so that a face faces up when it is resting on a flat surface.

A cube, and arguably a tetrahedron, is an example of a trapezohedron. An octahedron is a dipyramid.

All the answers here are very informative and interesting. I came across another family of shapes which will act as a fair dice called the tetartoid. It is a distorted version of a Dodecahedron. It relaxes the criterion for all faces to be regular polygons, but does require that all faces be equivalent. I thought these weren't directly mentioned in any of the answers, so worth adding.

• That is interesting indeed, but do you have any reason to believe it would be a fair die? – Wildcard Dec 29 '17 at 6:41
• Since all faces are identical, what would make one face more likely than any other to be the one it lands on when tossed randomly? – Rohit Pandey Dec 29 '17 at 18:23
• The cited article calls it the tetartoid, not the "tetroid". – John Bentin Dec 30 '17 at 23:41
• @JohnBentin - thanks, fixed. Tetroid is actually the name of a Math game, apparently. – Rohit Pandey Dec 31 '17 at 0:04
• @RohitPandey The (chiral) tetrahedral symmetry group consists of twelve rotations: eight (by ±1/3 turn) around the main diagonals of a cube, three (by 1/2 turn) around the coordinate axes, and the identity or null rotation. – Anton Sherwood Apr 5 '18 at 2:46

Assuming the criterion is that the chance of resting on any face is the same there are an infinite number, but I don't know how to define them. Think of two regular $n-$gon prisms, one long like a needle and one flat like a pancake. Each will land on the rectangular faces equally, but the first will almost never land on an end and the second will almost always land on an end. There is some length ratio where the chance of landing on an end matches the chance of landing on one lateral face. This will be a fair die. A similar approach applies to regular antiprisms. I haven't seen anybody suggest a way to calculate what the aspect ratio should be to make the die fair, but there has to be such a ratio. The same argument applies to regular $n-$gon pyramids and to many convex polyhedra. Just roll it and if one face is too probable as the base make it smaller, if it is not probable enough make it larger.

• It may not be fully justified, but if we borrow Boltzmann statistics from physics, we see that the probabilities will depend on the "temperature" of the die roll - unless the die is highly symmetric and in particular all faces have the same energy = distance to the centre of gravity. But such a prism will not be fair for $n\ne 4$ ... – Hagen von Eitzen Dec 27 '17 at 7:20
• @HagenvonEitzen: I like the physics analogy but do not think it applies here. If you roll a die fast it just rolls a long time, losing energy each time it flips over a face. Eventually it slows down and settles on a face. My intuition is that the speed it stops from is a small range. To get a "random" throw you need to throw it fast enough that it samples all the faces. – Ross Millikan Dec 27 '17 at 15:49

There is a brilliant explanation to the question here. For the sake of the completeness of the answer, I will give a small list of solids that can be used as fair dice:

• The five Platonic solids

• Triakis tetrahedron

• A fusion 8-sided die created by the author, among many others.

See here also, a reference from MathWorld and here with a lot of other referencing pages.

But even the fair coin is a theoretical abstraction, as is every other shape that has been suggested. Between any two adjacent obviously stable positions there is an "unstable" (or at least metastable) landing position. In any practical implementation of the die, there is always a miniscule but finite probability that the object will land on this edge, aided and abetted by the roughness of the surface on which it lands (and/or of the die itself), and stay there. [To see this, consider a graph of the die's potential energy plotted against a rotation around the edge.] However, we can still insist that all the stable positions have equal probabilities - but they don't quite add up to 100%.

• positive > finite – hkBst Dec 27 '17 at 13:00
• They do add up to 100% (except on a coin with nonzero thickness) - the set of initial positions that result in "landing" on a metastable position is one-dimensional, meaning the die will almost never land on them. – Random832 Dec 27 '17 at 16:14
• You can "sharpen" the edge of a coin; that will reduce the probability of landing on it to zero while still keeping the symmetry. – Tgr Dec 28 '17 at 23:32
• @Random832, and if it does land on its edge, it's just a coincidence, right? ;) – Wildcard Dec 29 '17 at 6:43
• I've often wondered how thick a cylindrical coin needs to be such that it has a probability of 1/3 of landing on the edge (and of course a probability of 1/3 of landing on either of the flat sides). I think the thickness has to equal the diameter, but I don't know how to prove (or disprove) it. – PM 2Ring Dec 29 '17 at 8:15

This list claims to be complete, though it offers no proof. Note that most of those with the highest symmetries are special cases of others shown with lower symmetry; for example, in the top row, the regular tetrahedron (symmetry group Td) is a special case of the isosceles tetrahedron (D2d), which in turn is a special case of the scalene tetrahedron (D2).

The obvious ones are the Platonic and Catalan solids; the latter are the duals of Archimedean (uniform but not regular) solids, in which we can include the infinite families of dipyramids (duals of prisms) and trapezohedra (duals of antiprisms).

More broadly, you get a fair die from the intersection of half-spaces defined by any plane (not passing through the origin!) and all of its images under one of the point groups of symmetries. (In the more regular dice, some of the images coincide.) Because the ‘parent’ plane can be anywhere relative to the axes or mirrors of the symmetry group, you get an infinite family by sliding it around, but for practical use you'd want to place it so that the dihedral angles are roughly equal.

There are designs for 5-, 7-, 9-, 11-, 13-, 15-, 17- and 19-sided dice here. As for the dice with 7 or more sides, that web page states of each design "This design is based on spacing points as equally as possible on a sphere and then cutting planar slices perpendicular to those directions".

• The fairness of such dice (unlike isohedral dice) depends on assumptions about how they're thrown. See also Hagen von Eitzen's reply to Ross Millikan, above. – Anton Sherwood Dec 30 '17 at 22:48
• @AntonSherwood - I think any dice needs to be thrown with a random rotation for all faces to be equally likely. If I drop a coin with heads facing up, no rotation then I suspect the probabilities might no longer be 50-50. – Rohit Pandey Jan 1 '18 at 6:13
• You could repeat numbers on several faces of regular polyhedera to get certain numbers of outcomes. For example, with six faces on a cube, you could have two each of 1, 2, and 3 to get a fair three sided die. A tetrahederon could have two faces each of 1 and 2. A dodecahederon could yield 2, 3, 4, 6, or 12 outcomes, etc. – richard1941 Jan 3 '18 at 4:15