# Fair d5 with as few faces as possible

Challenge: design a polyhedral die that will always give one of five outcomes, each with equal probability.

While achieving that, minimize the total number of faces of the die. Think of the die as convex and made of a uniform density material.

More specifically, design a polyhedron that has 5 symmetric faces that are stable, where the die won't tip over if sitting on those faces, and any number of unstable faces, where the die will tip over. There should be no stable, asymmetric faces. Minimize the number of unstable faces.

By "symmetric", I mean that there should exist symmetries of the polyhedron mapping each of the five stable faces to each other.

By a "stable" face, I mean that the center of mass of the polyhedron, projected onto the plane of the face, lies within the face. If placed face-down on a table, the center of mass should lie above a stable face, and off to the side of an unstable face. Avoid having the center of mass exactly above the boundary of the face.

As a starting point, this double pencil construction by Julian Rosen achieves 15 total faces (5 stable, 10 unstable). Can you do better?

Edit: A die of 10 faces, as posted by @Bass, is the best solution I know. Can you prove it is optimal, or find a better one?

Bonus question: Can you find a construction with less than $$3n$$ faces for arbitrary $$n$$?

• Are you aware of a provable optimum? Do you suspect there to be a provable optimum? Mar 18 at 3:59
• Asking to "minimize the number of unstable faces" is, technically, open-ended unless either the expected solution has a minima of zero (leaving no room for a better answer), or it is known that a provable optimum exists (what optimization is all about). Otherwise, making the objective of a puzzle to find a "best" solution is de facto open-ended. Having said that, if you have in mind a possibly best solution, making the objective "Can you equal (or do better than) <x>?" is permissible, as any/all solutions that meet that criteria remain valid irrespective of other/later answers.
– Rubio
Mar 18 at 6:50
• I don't think you've defined "outcomes" sufficiently. It sounds like you want each side to be a distinct outcome; if not, a D10 with 5 labels each assigned to 2 faces obviously solves the problem. Mar 18 at 16:50
• @Rubio given that the accepted answer on that meta explicitly says that both "optimization questions" and "questions about whether something is provably optimizable" are still on topic, this question seems on topic to me. Furthermore, I don't think changing the objective from finding the optimum to surpassing some threshold changes the whether the question may be considered off topic. A more optimum answer which comes after another sufficient answer is still a better answer in spirit, if not technically. Mar 18 at 17:13
• @Vaelus There's more to the meta answer than those sub-headings. Optimization questions are ones which "should have a provably best answer[....] Answers should come with justification of why they are optimal[....]"—in other words, optimization questions are about finding and proving the optimal solution. The alternative, whether something is provably optimizable, goes hand in hand with this, but only seeks "answers [...] of the form 'yes' or 'no', with justification". Both cases are about provable optimizations, specifically distinguished from "the best someone has posted thus far."
– Rubio
Mar 18 at 22:17

By "symmetric", I mean that there should exist symmetries of the polyhedron mapping each of the five stable faces to each other.

Sets of independent symmetry elements

combine multiplicatively. For example, the combined operation of a reflection and a threefold (120°) rotation produces up to five distinct images of any given point, for a total of 6 = 2 * 3 (for the reflection and rotation, respectively) symmetry-related points.

The symmetry of a cube contains several such combinations that relate all of the faces to each other.

But

5 is prime, so an orbit with that multiplicity cannot arise without a symmetry element of exactly order 5. The only non-decomposable symmetry elements having finite order greater than 2 are rotations, so the die must have fivefold rotational symmetry.

In order, then, for the die to have a number of faces that is not

a multiple of 5

, one or more faces would need to be

fixed by the five-fold axis. That is, the operation of the fivefold axis on such a face must yield back that face.

This is only possible if

the face is perpendicular to the axis, with the axis passing through its center.

But no matter how small you make such a face,

the center of gravity of the die must lie on the fivefold axis, above the center of that face,

so it would be an additional stable face, which we cannot have. We have established, then, that

the number of faces is indeed a multiple of 5.

Because the 5 stable faces are all related to each other by

a fivefold rotation

they cannot by themselves

form a closed 3D figure

.

10 is the next smallest multiple of 5,

and another answer demonstrates a solution with that many faces, so it is optimal.

• Note: this assumes mass distribution having the same symmetry as the die's faces, but it does not require uniform mass distribution throughout the die. Mar 19 at 16:14
• @SextusEmpiricus, the die must be symmetric because that's stipulated in the puzzle: "there exist symmetries of the polyhedron that [...]." If you chop the point off of the Bass solution in such a way that the die still satisfies the required symmetry constraint, you thereby create an additional stable face, as this answer already describes. Mar 20 at 14:08
• you are right, it is more specific that the symmetry is related to the entire polyhedron. I misread it as only the stable faces requiring a symmetry. Mar 20 at 15:23

Surely

2n

will suffice for all n > 2.

Consider a symmetrical cone with an n-sided regular polygon as its base.

If we take a "stubby" one (of a shortish height) and a "pointy" one (with great height), and join them at their identical bases, the centre of mass of the combined piece is dominated by the pointy cone. Therefore we can make the faces belonging to the stubby cone unstable by arbitrarily increasing the height of the pointy one.

Something like this:

• So for the specific n=5, this will give 5 stable symmetric faces, and 5 unstable faces, right? Mar 18 at 11:11
• @justhalf Yes, exactly.
– Bass
Mar 18 at 13:40

I think 6 sides would suffice. Make it fairly tall compared to the pentagon at the base, and "roll" it by spinning it like a top. If, by some miracle of fate, it manages to balance on the pentagon base (and if designed correctly I don't think there's very much odds of that), it's a do-over, or at the DM's discretion the WORST roll the player could have made, with the least benefit to them. I can see a DM crafting a fairly large one of these for a special event in their D&D session.

• This brings the probability of landing in the undesired side arbitraily close to zero, which is a nice outside-the-box solution. Not what I was going for, but I like it! Mar 20 at 3:42

If the die is not required to be convex, and faces are not required to be simple polygons, I think one could fabricate a nine-faced die by taking a five-sided pyramid and attaching a very long and slender tetrahedral "spike" to the middle of the base. If thrown onto a flat surface, there would be ten ways the shape could have at least three points touch the surface, but if the shape was slender enough, five of those would be unstable.

As to whether such a design would fit the intended constraints, that's debatable, but I think it's the closest thing to a solution with less than the "obvious and boring" solutions with ten faces.

• Yo momma was obvious and boring. That out of the way, the tetrahedral spike would have a mass distribution that would make the dice ever so slightly unfair.
– Bass
Mar 20 at 7:58
• @Bass: If the main die and axial spike were both symmetric around the same axis, under what circumstances would the mass behavior of the die be observably radially non-symmetric? From what I can tell, all division of mass into three or more three or more equal point masses that are evenly spaced around a circle will have the same moments of inertia as each other, about all possible axes. Mar 20 at 14:52