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?
2026-03-25 12:22:36.1774441356
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How many fair dice exist?
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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.
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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.
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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.
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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%.
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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.
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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.
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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.
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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".
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:
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.)