What kinds of fair dice?

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What kinds of fair dice?

#1  Postby lpetrich » Dec 20, 2012 5:26 am

Where a fair die is one with an equal probability of landing on each face. Real physical dice can only do an approximation of that, but one can find the shapes that give the best approximations. One wants a polyhedron shape with all faces interchangeable to within mirror imaging.

At first sight, one is stuck with the five Platonic solids: regular tetrahedron (d4), regular hexahedron or cube (d6), regular octahedron (d8), regular dodecahedron (d12), regular icosahedron (d20), and the degenerate case of two back-to-back polygons: a dihedron (d2).

In fact, the most familiar sorts of dice are cubes (d6) and coins and the like (d2).

-

But it turns out that one can make many more "isohedral" or "face-transitive" polyhedra by relaxing comparable conditions for vertices and edges.

I like to divide them into "axial" and "quasi-spherical" shapes. The axial ones have a main axis of symmetry, and n-fold symmetry under rotations around that axis. There are thus some infinite axial families. The quasi-spherical ones do not have a single main symmetry axis, but instead, evenly-spread ones.

Axial:

Original: dihedron: 2
Pyramids on originals' faces: bipyramid: 2n
Kite-shaped polygons extending from the shape's poles: trapezohedron: 2n

Their duals, the generalized prism and the antiprism, are sometimes used as dice, but often with some way of keeping them from landing on their end caps. Like high axial length to radius or rounded ends or pyramids. They thus land on their equatorial faces.
Prism: n
Antiprism: 2n

Quasi-spherical:

Original:
T: 4 ... C: 6 ... O: 8 ... D: 12 ... I: 20

Pyramids on faces:
T: 12 ... C, O: 24 ... D, I: 60

Split-face pyramids on faces:
T: 24 ... C/O: 48 ... D/I: 120

Rhombus edges:
T: 6 ... C/O: 12 ... D/I: 30

Faces to kites, faces to irregular pentagons:
T: 12 ... C/O: 24 .... D/I: 60

How many faces?
Axial: n, 2n
Quasi-spherical: 4, 6, 8, 12, 20, 24, 30, 48, 60, 120

Only prism dice can have an odd number of faces.


Some of these pages have pictures of lots of different kinds of dice many of them isohedral dice.

Fair Dice
DiceCollector.com - What shapes do dice have?
Alea Kybos' Dice Shapes
Math Games: Fair Dice
Isohedron -- from Wolfram MathWorld
A ROMAN GLASS GAMING DIE | CIRCA 2ND CENTURY A.D. | Christie's - an icosahedral die
Properties of Dice - mathematical proof

Isohedral figure -- Isogonal figure
face-transitive -- edge-transitive
Axial:
Bipyramid -- Prism (geometry)
Trapezohedron -- Antiprism
Quasi-spherical:
Platonic solid (both)
Catalan solid -- Archimedean solid

The remaining sort, Isotoxal figure or edge-transitive polyhedron, is special cases of the previous two.


I've posted at length because I've found this to be an interesting mathematical curiosity.
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Re: What kinds of fair dice?

#2  Postby Pulsar » Dec 20, 2012 12:28 pm

Very nice :thumbup:
I never knew you could make a dice from a non-platonic solid.
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Re: What kinds of fair dice?

#3  Postby CarlPierce » Dec 20, 2012 12:38 pm

i've been trying to derive the equations for the probability of a cuboid dice landing on its six faces if the X,Y,Z lengths are not equal.
I don't think it is so simple as being proportional to the area of the face.
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Re: What kinds of fair dice?

#4  Postby Pulsar » Dec 20, 2012 12:48 pm

CarlPierce wrote:i've been trying to derive the equations for the probability of a cuboid dice landing on its six faces if the X,Y,Z lengths are not equal.
I don't think it is so simple as being proportional to the area of the face.

Interesting question. The probabilities will depend on the position of the center of mass. The dice is more likely to land on the side for which the center of mass is closest to the ground. I don't know how to calculate the odds, though.

EDIT: I found a discussion about this: http://physics.stackexchange.com/questions/41297/how-to-determine-the-probabilities-for-a-cuboid-die
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Re: What kinds of fair dice?

#5  Postby Microfarad » Dec 20, 2012 1:11 pm

Using magnetic properties, would be possible to make a unequal-faces fair dice?
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Re: What kinds of fair dice?

#6  Postby Cito di Pense » Dec 20, 2012 1:30 pm

Pulsar wrote:The probabilities will depend on the position of the center of mass.


Time for a seminar in group theory.
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Re: What kinds of fair dice?

#7  Postby tuco » Dec 20, 2012 1:51 pm

To be isohedral, each face must have the same relationship with all other faces, and each face must have the same relationship with the center of gravity. - http://www.mathpuzzle.com/Fairdice.htm

This makes sense, and can, in limited fashion, be observed when playing with toys as a kid. So how many possible there are?
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Re: What kinds of fair dice?

#8  Postby CarlPierce » Dec 20, 2012 3:53 pm

Pulsar wrote:
CarlPierce wrote:i've been trying to derive the equations for the probability of a cuboid dice landing on its six faces if the X,Y,Z lengths are not equal.
I don't think it is so simple as being proportional to the area of the face.

Interesting question. The probabilities will depend on the position of the center of mass. The dice is more likely to land on the side for which the center of mass is closest to the ground. I don't know how to calculate the odds, though.

EDIT: I found a discussion about this: http://physics.stackexchange.com/questions/41297/how-to-determine-the-probabilities-for-a-cuboid-die


Very interesting.
The answer isn't the solid angle I'm sure of that. Even for the simpler problem with a perfectly rigid square cuboid onto a perfectly flat surface.
I'm sure the real world problem includes the elasticity of the material which will impart harmonic 'wobbles' as the objects rotation momentum tends to zero meaning that beyond a certain threshold ratio a long thin shaped cuboid will never land on its end if tossed.

I.e if you toss a domino giving it a fair wack of rotation momentum I think it will never land on its end. To get it to stand vertically you must carefully place it on its end. Easy experiment to try.
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Re: What kinds of fair dice?

#9  Postby lpetrich » Dec 20, 2012 9:26 pm

Microfarad wrote:Using magnetic properties, would be possible to make a unequal-faces fair dice?

One need not use magnetic properties. One may have to adjust the sizes of the faces to get equal probabilities.

But there are some different-face semiregular polyhedra that one can use. The quasi-spherical ones are the Archimedean solids.

Originals:
T: 4(3) ... C: 6(4) ... O: 8(3) ... D: 12(5) .... I: 20(3)

Truncate the vertices:
T: 4(3)+4(6)=8 ... C: 8(3)+6(8) = 14 ... O: 6(4)+8(6)=14 ... D: 20(3)+12(10)=32 ... I: 12(3)+20(6)=32

New vertices of vertex truncations merged:
T: 4(3)+4(3)=8 (regular octahedron) ... C/O: 8(3)+6(4)=14 ... D/I: 20(3)+12(5)=32

Truncate the vertices and edges:
T: 4(6)+4(6)+6(4)=14 (truncated octahedron) ... C/O: 8(6)+6(8)+12(4)=26 ... D/I: 20(6)+12(10)+30(4)=62

New vertices of vertex and edge truncations merged:
T: 4(3)+4(3)+6(4)=14 (vertex-merged truncated cube/octahedron) ... C/O: 8(3)+6(4)+12(4)=26 ... D/I: 20(3)+12(5)+30(4)=62

Snub polyhedra:
T: 4(3)+4(3)+12(3)=20 (regular icosahedron) ... C/O: 8(3)+6(4)+24(3)=38 .... D/I: 20(3)+12(5)+60(3)=92

So one can get sizes 14, 26, 32, 38, 62, 92

Notation: # of (face with # of vertices) + ... = total #
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Re: What kinds of fair dice?

#10  Postby Zwaarddijk » Dec 20, 2012 9:39 pm

Unfair dice can be used to obtain fair results, though:

Cast twice. count the results, such that the results of the second cast are counted as a reversal of the first. So, say heads come up first, tails after: heads win, tails first, heads after: tails win. Any other sequence is discarded.
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Re: What kinds of fair dice?

#11  Postby lpetrich » Dec 20, 2012 11:03 pm

Using permutations? H-T vs. T-H? That would work. One may have to keep throwing if one gets H-H's or T-T's.

One can extend that to fancier dice, like use a combination of one kind of face and another kind of face. Consider a cuboctahedron. It's a vertex-truncated cube or octahedron with merged vertices, also a merged-vertex vertex-and-edge-truncated tetrahedron. A regular cuboctahedron has 6 squares and 8 triangles.

So if one uses a square and a triangle, one gets 6*8 = 48 possibilities, or including permutations (sqr-tri, tri-sqr), 6*8*2 = 96 possibilities.
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Re: What kinds of fair dice?

#12  Postby lpetrich » Dec 20, 2012 11:06 pm

tuco wrote:To be isohedral, each face must have the same relationship with all other faces, and each face must have the same relationship with the center of gravity. - http://www.mathpuzzle.com/Fairdice.htm

This makes sense, and can, in limited fashion, be observed when playing with toys as a kid. So how many possible there are?

I had listed them, and there are 2 infinite families and 18 quasi-spherical ones (Platonic and Catalan solids). If one uses only the equatorial belts of prisms and antiprisms, one gets 2 more infinite families.
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Re: What kinds of fair dice?

#13  Postby Zwaarddijk » Dec 20, 2012 11:19 pm

Also, I realize I mixed up coins and dice there; coming up with a way of doing that for dice is more complex and might not always be possible, I guess, alternatively, you'll have to obtain permutations that are of equal length to the number of sides you have.

Maybe something like "any unbroken sequence of all N numbers where none of them is repeated, the last number in the sequence is the result"
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Re: What kinds of fair dice?

#14  Postby lpetrich » Dec 21, 2012 2:08 am

As to coins and dice, coins can be interpreted as d2 dice.

A sequence of all distinct for d(n)? The probability that any one sequence will have all distinct will be 1*(1-1/n)*(1-2/n)*...*(1/n) = n!/(nn) ~ e-n*sqrt(2*pi*n).

Number of faces, probability of an all-faces sequence:
2 0.5
3 0.222222
4 0.09375
5 0.0384
6 0.0154321
7 0.0061199
8 0.00240326
9 0.000936657
10 0.00036288
11 0.000139906
12 0.0000537232

So this method quickly becomes impractical.
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Re: What kinds of fair dice?

#15  Postby Zwaarddijk » Dec 21, 2012 8:02 am

lpetrich wrote:As to coins and dice, coins can be interpreted as d2 dice.

A sequence of all distinct for d(n)? The probability that any one sequence will have all distinct will be 1*(1-1/n)*(1-2/n)*...*(1/n) = n!/(nn) ~ e-n*sqrt(2*pi*n).

Number of faces, probability of an all-faces sequence:
2 0.5
3 0.222222
4 0.09375
5 0.0384
6 0.0154321
7 0.0061199
8 0.00240326
9 0.000936657
10 0.00036288
11 0.000139906
12 0.0000537232

So this method quickly becomes impractical.

It quickly becomes impractical even with heavily unfair coins as well, as the likelihood of HT or TH popping up becomes very low.
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Re: What kinds of fair dice?

#16  Postby lpetrich » Dec 21, 2012 11:21 am

Related to these polyhedra is List of convex uniform tilings, covering regular and some semiregular ones.

The regular ones are triangular, square, and hexagonal tilings; they have all three transitivities. That article also lists semiregular ones that are face-transitive or vertex-transitive; those two categories also have some edge-transitive ones.

As with their polyhedral counterparts, one can make the semiregular tilings from the regular ones by making pyramids on faces, edge rhombi, kites, and irregular pentagons.

The face-transitive ones I think can be called fair dartboards.


The mathematics behind them is interesting.

One starts with a theorem that mathematician Leonhard Euler had stated around 1750's: Euler's Formula For number of vertices V, edges E, and faces V, a polyhedron has this constraint on them:
V - E + F = 2

More generally, V - E + F = X

where X is a "topological invariant" of the surface, the "Euler characteristic". It is equal to 2(1-g), where g is the "genus" or number of "holes in the surface, or the number of cuts of the surface needed to make it sphere-like. A torus or doughnut shape needs one cut, thus, g = 1 and X = 0 for it. Two spheres need to be joined to make a single sphere-like shape, and that's a negative cut. g = -1 for them and X = 4, the sum of X for the individual spheres.

For a surface where V, E, and F may be infinite, that formula may not be very useful, but if the surface was formed by the tiling of some rectangle, then one can use that rectangle's properties. It has periodic boundary conditions, top edge mapping onto bottom edge and left edge mapping onto right edge. That gives this rectangle a torus-like topology, or g = 1 and X = 0. Thus, for it,
V - E + F = 0


The completely regular case is fairly easy. Let each face have n vertices/edges, and let each vertex have "rank" r faces/edges. Notice that duality interchanges r and n. One can easily show:

{V, E, F} = {1/r, 1/2, 1/n} * X / (1/r + 1/n - 1/2)

Solutions, in form (k0,k1) = (smaller, larger) of (r,n):

Polyhedron: X = 2 > 0
(2,k1) for any k1
(3,3), (3,4), (3,5)

Plane: X = 0
(3,6), (4,4)

Hyperbolic surface (infinite): X < 0
(3,k1), k1 >= 7
(4,k1), k1 >= 5
(k0,k1), k0 >= 5, k1 >= k0
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