(2012-11-26) Standard spots on cubic dice :
How many different ways to mark a die,
if pips on opposite faces must add up to 7?
Answer : 16. (It's
240 if opposite faces needn't add up to 7.)
My local bargain store sells two types of dices from China,
for 99¢ a pack.
Five transparent red cubes with white pips, 19 mm on a side.
Twelve 15 mm cubes (white, green & red) with black or white pips.
Besides the obvious difference in sizes and colors, I noticed that the
dices in the two packs were not alike because of the arrangement
of the spots.
(2012-11-29) Unrestricted Dice
Let's drop the rule that requires opposite faces to add up to 7.
In this section, a "configuration" is understood to be merely a labeling of
the faces of the die, regardless of the orientation of the labels...
To each such configuration correspond
8 different orientations of traditional markings with spots.
Thus, the 30 configurations enumerated below correspond to
240 ways to manufacture a die.
If the three faces
2-3-6 share one vertex (which they always do in a traditional die)
then they can be arranged in two configurations (left-handed and right-handed).
For either of those, there are 6 ways to arrange the "hidden" faces
1,4,5 (in only one of those is the sum of opposing faces always equal to 7).
That accounts for a total of 12 possible configurations (so far).
If the 2-3-6 faces do not meet at a vertex, then two of them must be opposing
each other (with the third sharing an edge with the other two).
There are just three possible such configuration
(think of it as choosing which of the 2-3-6 is the middle one).
With respect to any such base, the other three faces can be arranged in 6
different configurations. That's a total of 18 possible configurations.
Adding that to the 12 configurations described in the previous paragraph,
we obtain a grand total of 30 possible configurations.
Configurations in which opposing faces never add up to 7 :
For either of the two ways the 2-3-6 faces can meet at a vertex,
we have to put either 4 or 5 opposite to 6, then we have no further leeway.
That accounts for 4 configurations (so far).
For any of the three ways two of the 2-3-6 faces are opposing each other,
we have two choices for the face opposite to the "middle" face and two further
choices for setting the remaining two faces (that are opposite to each other).
That's 12 configurations to add to the 4 enumerated
in the previous paragraph, for a grand total of 16.
All told, in the 30 ways to number a dice, there are two ways (right-handed
and left-handed) in which opposing sides always add up to seven and
16 ways in which they never do.
(2013-02-06) A 9-hedron stable on only 3 faces (3-spindle)
An elongated n-gonal dipyramid
is a substitute for a fair n-sided die.
If its pyramidal components aren't too flat,
such a 3n-hedron is only stable on an horizontal
plane if its rests on one of its n lateral faces. If the die
has an order-n axis of symmetry, all those faces are equiprobable because they're
equivalent
(assuming uniform mass density).
(2012-11-28) Nontransitive set of dice.
Rock-paper-scissors
(Rochambo)
with dice.
Here's a nontransitive set of three 3-sided dice
(or 6-sided dice with the same spots on opposite sides) due to
Dr. Nathaniel Hellerstein,
CCSF :
Red = {3,5,7}. Yellow = {2,4,9}. Blue = {1,6,8}.
With probability 5/9 in every case,
red beats yellow, yellow beats blue and blue beats red:
Red beats Yellows
3
5
7
2
4
9
Yellow beats Blue
1
6
8
3
5
7
Blue beats Red
2
4
9
1
6
8
Transitivity, or lack thereof...
Transitivity is a fundamental property of
ordering relations which formally states
that if A ranks B and B ranks C, then A must rank C.
We're so used to ranking things transitively that not being able to do so is disturbing.
None of the above three dice is preferable to both of the others.
There are many other entertaining examples with dice.
A far more serious nontransitive case is democratic majority voting. Indeed,
it's possible that voters who would prefer A to B and B to C would
actually prefer C to A. That's
Condorcet's paradox. An unfortunate fact of life.
Scams based on nontransitivity will fool people who don't know what to look for.
One nice example is the so-called Penny Ante game which
Walter Francis Penney (1913-2000)
introduced in October 1969 (in 10 lines).
Martin Gardner (1914-2010) discussed it in the
Mathematical
Games column of the October 1974 issue of Scientific American :
Each player predicts that a sequence of three heads (1) or tails (0) will occur in a sequence
of flips of a fair coin before what the other has predicted.
That game would be fair if both predictions were picked
at random (as if inscribed on two balls drawn from an urn with 8 balls).
However, the opportunity to pick a
prediction knowing the choice of the first player always gives the second player
an opportunity for a 2:1 advantage or better.
As is often the case, an informed choice is better than a random one.
Probability of a second-player win in Penney's game with optimal strategy
Prediction of Player A
000
001
010
011
100
101
110
111
Best Prediction for Player B
100
100
001
001
110
110
011
011
Winning Probability for B
7/8
3/4
2/3
2/3
2/3
2/3
3/4
7/8
To justify the above optimal strategy,
we have to compute the probability of a win for all possible responses of B to all choices of A.
This is best done using an algorithm due to
John Conway (1937-2020)
presented in the paper of Humble and Nishiyama
quoted in the footnotes below. Conway's algorithm applies to the generalized Penney
game, where the two predictions need not be of length 3
(they don't even have to be of the same length).
With predictions of the same length k≥4
János Csirik (1946-)
found in 1992 that Player A's best choice to limit the advantage of B is a sequence whose bits
are all indentical except 3 (1 at the beginning and 2 at the end).
This leaves a option with winning odds (1+2k-1):(1+2k-2).
"Optimal strategy for the first player in the Penney ante game" by J.A. Csirik,
Combinatorics, Probability and Computing, Volume 1, Issue 4 (1992), pp 311-321.
Steve Humble,
MBE and
Yutaka Nishiyama (1948-)
prefer to play Penney Ante with a
deck of cards instead (26 black cards and 26 red ones).
Be that as it may, the exact probabilities involved in the "Humble-Nishiyama randomnsess game"
differ slightly from those of the original Penney game and they're tougher to work out,
unless a computer is used.
In certain cases, there's even a nonzero probability of a tie
(undecided game) which can't happen in Penney Ante
with an unlimited number of coin flips. For example, when A bets "001"
and B wisely replies with "100"
the game is tied if "00" never occurs, which happens if and only
if red and black always keep alternating, as they do with a probability of:
2 / C(52,26) = 4.03292... 10-15 which is minute but nonzero.
(2012-11-28) Sicherman's Pair of Dice
Any total, from 2 to 12, has the same probability as with standard dice.
One die has pips 1,2,2,3,3,4 and the other is marked 1,3,4,5,6,8.
These dice were invented by Colonel George L. Sicherman
(then of Buffalo, New York) whose discovery was reported by
Martin Gardner, in one of his legendary
Scientific American columns (1978).
The best way to investigate this matter involves generating polynomials.
Besides proving the basic claim, this approach can establish the
uniqueness of Sicherman's dice among 6-sided
dice with nonzero markings:
Proof :
To a face with n spots, we assign the monomial xn.
To the whole die correspond the sum of the polynomials associated with its faces.
For example, the polynomial associated to a standard die is:
S = x + x2 + x3 + x4 + x5 + x6
The number of ways we can obtain a total of n pips when we roll several
dice is the coefficient of xn in the product of
their polynomials (HINT: to obtain that
term, you must sum up all the ways there are to pick one term from each factor
so that the exponents of x add up to n).
Therefore, with two standard dice (a red one and a green one, say)
the number of ways to roll a total of n pips is the coefficient
of xn in the square of the above
polynomial. Namely:
Now, the interesting remark is that S can be factored:
S =
x ( 1 + x ) ( 1 - x + x2 ) ( 1 + x + x2 )
We may regroup the factors of the square S2 in the following way:
S2 =
[x ( 1 + x ) ( 1 + x + x2 )]
[x ( 1 + x ) ( 1 - x + x2 )2 ( 1 + x + x2 )]
The two square brackets expand respectively as follows:
x + 2x2 + 2x3 + x4
=
x + x2 + x2 + x3 + x3 + x4
and
x + x3 + x4 + x5 + x6 + x8
Those correspond to 6-sided dice marked 1,2,2,3,3,4
and 1,3,4,5,6,8.
What's somewhat miraculous is that we end up with
a pair of 6-sided dice.
To match what's done with traditional dice, those dice should be built
with opposite faces adding up to
5 for the lower die and 9 for the upper die.
Other groupings of the above factors of S2
yield proper dice only when every resulting polynomial has nonnegative coefficents.
We obtain:
1-2-4-5 tetrahedron with 1-2-3-3-4-5-5-6-7 enneahedron.
1-4 coin and
1-2-2-3-3-3-4-4-4-5-5-5-6-6-6-7-7-8 octadecahedron.
(2013-02-05) Rolling several dice instead of one.
When does the sum of two dice give equiprobable totals?
The standard set of seven polyhedral dice made popular by
Dungeons & Dragons
consists of the five platonic
solids and a pair of 10-sided pentagonal deltohedra.
One is marked from 0 to 9 and the other from 00 to 90.
Those two are known as percentile dice.
When rolled together, the percentile dice give any total from 0 to 99 with equal probability (1/100).
In traditional role playing games (RPG)
a total of zero (0+00) is interpreted as 100.
This exception wouldn't be needed if the low die was marked 1-10
instead of 0-9 (all other standard dice do start with 1).
Some decimal dice are available which allow just that; they're simply not popular.
More generally, we may consider a set of p n-sided fair dice
where the j+1st face of the i+1st
die is marked j.ni ).
When those dice are rolled, they give
any total from 0 to np-1
with equal probability. Let's generalize:
For a prescribed integer M, what are the sets
fair dice marked integers that will give any total between 0 and M-1 with probability 1/M ?
Well, the polynomial approach introduced in the previous section
reduces this question to the factorization of the polynomial:
(1-xM ) / (1-x) =
1 + x + x2 + x3 + ... + xM-1
The factors of those polynomials are called cyclotomic
("cycle-splitting") and they've been studied and cataloged by generations of mathematicians.
Dismissing as trivial the type of splitting described in the above introduction, the first
non-trivial factorization is for M = 6:
A factorization gives a legitimate set of dice only if all the factors are polynomials whose
coefficients are nonnegative integers. In this case, only three possibilities exist:
A single six-sided die marked (0,1,2,3,4,5).
A 3-sided (curved) die marked (0,1,2) and a coin marked (0,3).
A coin marked (0,1) and a 3-sided (curved) die marked (0,2.4).
More generally, we can devise such a set of marked dice for any ordered
factorization of the integer M.
If M is prime, there's only one solution (a single die with M sides).
(2013-04-14) Polyhedral Dice
Dice people actually use, for divination or recreation.
Rôle playing games (RPG) call for
a variety of dice besides the traditional 6-sided cubic dice (D6).
The most popular sets have 7 dice:
Icosahedra were already used in Antiquity, for divination purposes.
The large (52 mm) glass die
shown at left is one of the most famous extant examples (c. AD 100).
It was auctioned off at
Christie's
for $17925 on December 11, 2003.
Prior to that auction, it had drawn little attention and
was expected to fetch between $4000 and $6000.
It would be worth a lot more now.
For 12-sided dice, the regular dodecahedron
has eclipsed the rhombic dodecahedron,
which was apparently mass-produced only once, around 1978, for the
Ask
Astro-Dice fortune-telling game.
For the shot at left, I got an old set from eBay
($15 on 2013-03-29).
New ones
use regular dodecahedra, unfortunately...
Note that Pluto (top symbol on the center die)
was still a planet back then.
In recent years, two distinct isohedra with 24 faces
have been mass-produced as dice by Louis Zocchi
(hear Lou's pitch).
One is the isohedral tetrakis hexahedron or tetrakis cube
pictured at left.
The other is a large die in the shape of a
deltoidal icositetrahedron
(strombic icositetrahedron or
trapezoidal icositetrahedron ).
It's marketed by GameScience (Zocchi's company)
under the name of D-Total, featuring fancy markings
that are meant to facilitate the use of the die as a substitute
for dice with
2, 3, 4, 5, 6, 7, 8, 10, 12, 20, 24, 30, 40, 50, 60, 70 or 80 sides.
This is jointly credited to
Dr. Alexander F. Simkin, Frank Dutrain (of LD Diffusion) and Louis Zocchi (2009).
(2013-04-26) Commercial Dice Sizes
Some of the most common sizes for cubic dice are:
5 mm micro,
8 mm tiny,
12 mm mini ( less than 1/2 '' )
15 mm regular ( 5/8 '' standard
RPG size, in the US )
16 mm medium ( largest backgammon size )
19 mm large ( 3/4 '' casino size )
25 mm or 28 mm jumbo ( 1'' )
35 mm or 38 mm giant ( 1½ '' )
55 mm monster ( 2'' or more )
Polyhedral dice
are loosely matched with 6-sided dice of similar bulk:
Caliper Sizes (between any face and its opposite element)
Mini 12 mm
Regular 15 mm
Large 19 mm
Jumbo 28 mm
Giant 35 mm
D4
17.4 mm
D6
12 mm
15.0 mm
19 mm
25 mm
D8
14.8 mm
24 mm
D10
15.9 mm
29 mm
D12
18.5 mm
26 mm
D20
19.3 mm
D24
19.5 mm
24 mm
D30
32 mm
Oversized dice could damage dice trays.
They're best tossed on carpets.
(2013-04-14) Convex Isohedra are fair dice.
All the faces of an isohedron are equivalent.
An isohedron is a polyhedron whoses faces
are all equivalent.
That's to say that every face can be transformed into any other through some spatial isometry
(rotation or reflection) that maps the polyhedron onto itself.
The dual of an isohedron is an isogonal polyhedron,
and vice-versa (duality being understood
with respect to the sphere inscribed in the isohedron or circumscribed to its dual).
Sometimes, the dual is more readily understood than the primal.
The dual of a convex solid (polyhedral or not) is convex.
So, every convex isogonal polyhedron is associated to an isohedron and vice-versa.
In particular, every Catalan solid (i.e., the dual of one of the 13
Archimedean polyhedra) is an isohedron.
So are the duals of isogonal prisms and antiprisms (respectively called
amphihedra and deltohedra).
However, there's no requirement that the dual of an isohedron be equilateral.
So, there are isohedra that are not duals of uniform polyhedra
(as uniform means both isogonal and equilateral).
For example, the familiar regular tetrahedron
is not the only isohedral tetrahedron...
Any tetrahedron whose opposing edges have the same length
(as illustrated at right) is an isohedron.
Such a tetrahedron is called a disphenoid.
The dual of a disphenoid is another disphenoid;
disphenoids are both isogonal and isohedral (they're thus
noble).
Therefore, any disphenoid would make a perfectly fair 4-sided die.
A disphenoid is chiraliff the three quantities a, b
and c are distinct.
(2013-02-22) The Secret of Round Dice
A steel ball moves in an isogonal cavity,
dual of the isohedral outer pattern.
Currently, "6-sided" spherical dice are
available (the inner cavity is octahedral ).
Any isohedral die
could be made this way, by carving out a cavity
in the shape of its dual
(the dual of an isohedron is an
isogonal polyhedron ).
However, the fairness of such a die would only be guaranteed if the cavity
was isohedral as well. In other words, it must be
noble (that word simply means
both isogonal and isohedral).
Among convex polyhedra, the only noble ones are the
Platonic solids
and the disphenoids.
The latter type would yield a great way to make 4-sided
dice without any sharp corners (in fact, without any corners at all).
If a scalene disphenoid is used, the artefact would be a sphere where
the outer markings would seem asymmetrically distributed.
Yet, it would be a perfectly fair die and, therefore,
a great conversation piece...
(2013-02-14) Introducing the
scalene isogonal tetradecahedron :
There's only one such solid (besides the 6-antiprisms and 12-prisms).
If d is the degree of every vertex in an
isogonal
polyhedron, it has at most d different edge lengths and
d non-congruent faces. When both maxima are achieved, the
isogonal polyhedron is said to be scalene.
On the other hand, an isogonal polyhedron where at least two adjoining edges
have the same length can be called isosceles.
Any polyhedron where all edges have the same length is said to
be equilateral.
A polyhedron that's both isogonal and equilateral is said to
be uniform
There are 75 or 76
nonprismatic uniform polyhedra
(the convex ones are the
5 Platonic solids and the
13 Archimedean solids).
Consider now the three distinct types of isogonal tetradecahedra
obtained by cutting off the eight corners of a cube with increasing severity:
In the first two cases, the truncation must be the same at all corners
of the cube, or else we wouldn't obtain a polyhedron with isogonal symmetry.
However, if the truncation planes intersect
inside the cube (along a new edge) then the isogonal symmetry persists
when different truncations are used for nonadjacent
vertices of the cube.
The resulting scalene isogonal tetradecahedron
is a zonohedron featuring 3 edge lengths,
6 rectangular faces and two distinct types of hexagonal faces.
At right is my own old school cardboard model
with voided rectangular faces, where edges are in a 1:2:3 proportion.
(This photo may fool the eye if you're not aware that surfaces with pencil patterns
are inner ones.)
If all such truncations are alike, then the resulting isosceles solid has
square faces and is more readily obtained by "classical" truncation
of a regular octahedron, without creation of new edges
(the scalene version can't be so constructed, because
the 8 planes supporting the hexagonal faces
don't form an octahedron).
Therefore, the locution isogonal truncated octahedron
can only denote the tetradecahedron with square faces
(and congruent hexagonal faces) discussed next
as the shape of traditional Korean dice.
By definition, an isohedron is said to be scalene if its
dual is (the dual of an isohedron
is an isogonal polyhedron and vice-versa).
For example, the dual of the polyhedron just described is
a scalene isohedron (a fair die) with 24 faces congruent to the same
triangle (scalene
tetrakis hexahedron).
(Gerard Villarreal, TX. 2013-02-05)
Juryeonggu
What polyhedron is this 14-sided Korean die ?
Uniform Truncated Octahedron
The juryeonggu is precisely an
isogonal
truncated octahedron.
Because its tetragonal faces are square,
it's a special case of the
above solid.
The shape is fully specified by
the length (x) of the edges between hexagons,
assuming the square faces have unit sides.
For x = 1, we would obtain the
uniform truncated octahedron
(pictured above at left)
which features regular hexagonal faces.
(Don't call it a cuboctahedron, which is
another uniform tetradecahedron, as is the
truncated cube.)
This Archimedean solid can tile space without voids
(a non-uniform juryeonggu can't).
It's the basis for the near-optimal foam described
by Lord Kelvin in 1887
(Kelvin's cell has the same vertices as the tetradecahedron but all its edges are curved
and so are its hexagonal faces).
For any x, a juryeonggu has the following dimensions (derived below).
The locution "radius to" denotes the distance from the polyhedron's center.
The radius to any vertex is the radius R of the circumscribed sphere.
Dimensions of an Isogonal Truncated Octahedron (Korean Juryeonggu)
Definition & Symbolic Name
Value
Side of a square face (basic unit)
a4
1
Length of an edge between hexagonal faces
a6
x
Diagonal of a square face
Ö2
Diagonal of an hexagonal face
1 + x
Width of an hexagon (between parallel sides)
Ö3/2 ( 1 + x )
Radius of the circumscribed sphere
R
( 1 + x + ½ x2 )½
Radius of a square face
r4
Ö2/2
Radius of an hexagonal face
r6
Ö3/3 ( 1 + x + x2 )½
Radius to [center of] a square face
h4
Ö2/2 ( 1 + x )
Radius to [center of] an hexagonal face
h6
Ö6/3 ( 1 + ½ x )
Caliper ratio ( h4 / h6 )
Ö3 (1+x) / (2+x)
Radius to [centers of] sides of squares
R4
½ ( 3 + 4x + 2x2 )½
Radius to [centers of] other edges
R6
1 + x/2
Solid angle of a square face
W4
4 Arcsin [ 1 / (3+4x+2x2 )]
Solid angle of an hexagonal face
W6
p/2
- 3 Arcsin (3+4x+2x2 )-1
Surface area of a square face
A4
1
Surface area of an hexagonal face
A6
Ö3/4 (1 + 4x + x2 )
Total surface area ( 6A4+8A6 )
A
6 + 2Ö3 (1+4x+x2 )
Volume of a square pyramid ( A4 h4 / 3)
V4
Ö2/6 ( 1 + x )
Volume of an hexagonal pyramid ( A6 h6 / 3)
V6
Ö2/24
(1+4x+x2)(2+x)
Total Volume
V
6 V4 + 8 V6
Derivations for the above (outline) :
R can be obtained as the radius of the equator circumscribed to an
isogonal octagon whose sides are either
the diagonal of a square face
(length Ö2) or an edge
of length x. In that equatorial plane, we also
find h4 and R6 :
R2 = [ 1 + (1+x)2 ] / 2 = 1 + x + ½ x2
h4 = ( 1 + x ) / Ö2
R6 = 1 + x/2
An hexagonal face is obtained by
subtracting three equilateral triangles of side x from an equilateral triangle of side 1+2x,
so its surface area is:
The diagonal of an isogonal hexagon with sides 1 and x is
(1+x). The width between parallel
sides is equal to that diagonal multiplied by the sine of 60°.
The least obvious quantity is r6 which can be obtained from
planar cartesian coordinates (that's what we use when all else seems to fail).
We then obtain h6 from the Pythagorean theorem:
R2 = r62 + h62
R4 is the height of an isosceles triangle of base 1 with two sides equal to R.
R42 = R2 - ¼ =
( 3 + 4x + 2x2 ) / 4
We may check that R4 = R6 in the
uniform case (x = 1).
The solid angle subtended by a square face is obtained immediately from
R4 using the formula we've established elsewhere
on this site. For hexagonal faces, we just use
the fact that the solid angles subtended by all faces add up to 4p.
Looking for the Perfect Juryeonggu :
The
official blog
of the city of Gyeongju
states that, in a traditional juryeonggu,
all faces have the same surface area.
This entails a quadratic equation in x, whose positive root is:
x = Ö( 3+4/Ö3 )
- 2 = 0.304213765421624907891...
Gerard Villarreal (private communication)
advocates a juryeonggu with
h4 = h6
so that all faces are tangent to the same inner sphere
(which may then be called the
inscribed sphere,
by analogy with the isohedral concept).
This entails a quadratic equation whose positive solution is:
x = (Ö3 - 1) / 2
= 0.36602540378443864676372317...
In either case, it was guessed that
endowing an isogonal truncated octahedron with a particular
property that isohedra possess would endow them with the same fairness as
isohedral dice. It ain't quite so...
Oversimplifying the conclusions of the discussion below,
the latter guess turns out to correspond to the situation
where the die bounces a very large number of times.
At the other extreme is a die that doesn't bounce at all
(think of a randomly oriented die immersed in glycerol an dropped just above a
sticky surface).
Such a die would simply land on any face with a probability proportional
to the solid angle subtended by that face
from the center of the polyhedron. All faces subtend the same solid angle
(2p/7) when:
Since x can't be both
0.3660 and 0.3217 (obviously)
no isogonal truncated octahedron
can be unconditonally fair (like an isohedron would be).
However, for any given set of physical conditions and casting style,
there's one isogonal truncated octahedron
that looks fair...
Empirically Fair Dice :
As dice are actually rolled in specific conditions that are somewhere between
the two (contrived)
extremes described above, a juryeonggu that looks fair in practice
would have to correspond to a parameter x determined empirically under
those given conditions.
One way to do so is to build two isogonal truncated octahedra
with different parameters x1
and x2 (not too far from a guess of 0.35).
Cast both 700 times and count how many times they land on an
hexagonal face (N1 and N2 respectively).
By linear interpolation, we'd approach a perfect score of 400 for a die having the
following value of x:
x =
(N2 - 400) x1 + (400 - N1 ) x2
N2 - N1
The two most interesting dice to build are the ones mentioned above:
If you build a third die using the above interpolation, you may roll your three dice
many times and plot with good precision the curve giving the probability of an hexagon as a function of x
(with just 3 known points, you may as well assume the curve is a parabola or
a circle).
Use that information to build a fourth die, if you must.
That last die may not be quite fair (with your own particulars)
but it's unlikely that anybody will ever detect that!
The precision of the above method is limited by the standard deviation
on N, which is about 13 for nearly-perfect dice with 700 trials
(it's proportional to the square root of the number of trials).
I've carved a 23.9 mm orthohedral
(x = 0.366)
juryeonggu to a precision of 0.05 mm
and obtained the results tabulated below.
They show that an hexagonal face is about twice
as likely as a square one,
although an hexagon subtends only 9.4%
more solid angle than a square.
Orthohedral Juryeonggu on Felt-Covered Wooden Dice Tray
(2013-03-27) Quasi-fair and blatantly unfair dice
Discussing non-isohedral commercially available dice.
"Colonel" Lou Zocchi first earned a solid reputation
in the manufacture and sale of isohedral
dice which he pioneered in 1974, riding the wave of the increasing
popularity of role playing games (RPG).
For lack of symmetry among their faces, the fairness of non-isohedral
dice may depend critically on the way they are cast and/or on the resilience
of the landing surface.
Such dice may roll true on plexiglass but not on felt (say)
or the imparted spin (long roll or not) may bias them.
Nevertheless, those things can substitute for fair dice with odd numbers of sides
(all isohedra have an even number of faces).
A good example is the 7-sided die pictured above
(16.3 mm thick, as devised by Lou Zocchi) which
I found to roll true under typical conditions
(two dice tossed together 392 times from a ribbed cup onto a circular felt-covered wooden tray).
For non-isohedral dice to give at least the illusion of rolling true,
they must be designed by trial and error and tested extensively.
Sometimes, the necessary rigor isn't exercised in commercial endeavors...
The worst mass-produced offender is probably the 5-sided die (D5)
pictured at left, a flawed design for which
Lou Zocchi was granted
US Patent 6926275 in 2005.
In his patent application, Zocchi stated that the device
had been tested by rolling it 10163 times on plexiglass
and was found to be practically fair under such conditions.
I find this pretty hard to believe after the following field-test:
Using my trusted ribbed cup and felt-covered dice tray,
I cast 5 such dice together 100 times
(for a total of 500 individual outcomes) and saw the dice land 282 times on
one of the two triangular faces and only 218
times on one of the three rectangular sides.
So, the 28.2 % share of each triangular face was nearly
twice the 14.5 % share of each rectangular face. Bad.
If needed in actual gaming,
a fair D5 die can be nicely simulated by rolling a good D6 cube
until the outcome isn't 6
("casino" dice are machined from
extruded cellulose acetate
to a precision better than
0.0005"). The average number N
of actual rolls required for a valid outcome is only 1.2, since :
N = 1 + N/6
yields N = 6/5
Other good alternatives exist which forego extra rolls entirely,
including the use of a 5-sided spindle
or a 10-sided isohedron with duplicate labels.
The celebrated Zocchihedron pictured at left is an interesting
randomizing device but it's not a solid die at all.
It can't be rigorously tested as proper dice can because it has memory...
It contains a spherical cavity partially filled with sand which helps it come
to a full stop but also makes it impossible to grasp its true dynamical state
from its outer appearance.
US Patent D303553 (1989) |
US Patent 6926276 (2005)
(2013-01-27) The Fair Dice Problem
Isohedra are intrinsically fair dice. Are there any others ?
Fair dice do not create randomness or uniform probability distributions;
they merely conserve it.
When thrown from a truly random orientation, a fair die is equally likely to land on
any face.
A sufficient condition for an homogeneous convex polyhedron to
be a fair is to be isohedral, for the above is then satisfied by reason of symmetry.
An isohedron is a
face-transitive polyhedron.
This is to say that every face is the image of any other in at least one isometric
transformation of the entire polyhedron (i.e., a rotation or a mirror reflection
mapping the polyhedron onto itself).
An isohedron is thus a polyhedron where all faces are congruent to each other.
The converse need not be true.
For example, the convex deltahedra with
12,
14 or
16 faces aren't isohedral.
Isohedral symmetry is precisely what guarantees that all faces
are strictly equivalent. In any type of damped motion
(including, but not limited to, inelastic shocks with fixed objects of any shape)
if all initial orientations of an homogeneous rigid
isohedral die are equiprobable,
then it will certainly come to rest on any on its faces equiprobably.
This conclusion certainly holds for the Newtonian mechanics
of rigid bodies,
which will be our only
concern in the rest of the discussion. It would also hold for
other mechanical laws, including special and general
relativity, that can deal with the fiction of homogeneous and isotropic matter
and are insensitive to chirality
(this last restriction comes from the fact that chirality-changing transformations
are allowed as isohedral transformations).
The symmetry argument is otherwise general enough to deal with fragile isohedra made
from (amorphous) gelatin or rubber.
Whenever the final integrity of such an isohedral die is sufficient to
identify it as resting on some face, it will be any face with equal probability!
Revisit now the argument we used for
isogonal truncated octahedra
to prove that one value of the single parameter describing those shapes
must correspond to a fair die which isn't isohedral
(no truncated octahedron can possibly be isohedral, since squares and hexagons can't be congruent).
That theoretical argument
(and/or the practical recipe we gave to determine something close to the
correct shape using linear interpolation) was based on the implicit assumption
that dice are always cast the same way
(on an horizontal table covered with a particular shock-absorbing material, say).
(2013-02-07) Necessary conditions for polyhedral dice to be fair
What two extreme ways of casting dice imply for fair dice.
The thesis [master's thesis |
pdf]
filed in 1997 by Ed Pegg, Jr. for his
Master's degree at
UCCS
was entitled A Complete List of Fair Dice.
In it, the famous recreational mathematician actually classified
isohedra. He was fully aware of the remote possibility
that some fair dice
might exist besides isohedra (which are fair by symmetry)
but he clearly estimated (rightly so) that the less technical
term was more suitable for a title.
Elsewhere, he answered the
question "Can a non-isohedral fair die exist?"
by using the example of a square pyramid with isosceles lateral faces, thrown as a die under some set of standard
conditions. The four triangular faces have the probability by symmetry.
The probability of the square face is above that if the pyramid is tall and below that if the pyramid is short.
Therefore, an intermediate height must
exist for which all faces have the same probability.
He adds:
However, once the conditions changed, the die would no longer be fair.
(I have a strong argument for this, but no proof.)
What follows can serve as the proof that Ed Pegg, Jr. is calling for.
The main difficulty was that the lack of fairness of something like the
aforementioned square pyramid can only be established if we analyze
theoretically at least two sets of physical conditions.
The trick is to consider two limiting cases rather than any
realistic motion. A fair die could theoretically
be thrown in any possible and couldn't show a difference
in probabilities between those two limits, which turn out to
be simple enough to analyze.
(2013-02-07) Quasistatic Dice Casting
Slow dead-cat bounce on a sticky surface.
In 1981, David Singmaster (b. 1939) discussed the proposal
that an homogeneous die would land on a face with a probability
proportional to the solid angle subtended
by that face (as seen from the center of gravity).
At first sight, the idea looks silly...
For one thing, this would assign nonzero probabilities to unstable faces (whenever
the orthogonal projection of the center of gravity on the plane of the face isn't inside the face,
the die cannot rest on that face at all).
Also, it would seem to overestimate the probabilities of lateral faces in flat prisms
(our physical intuition is that a coin can stand on its edge but will never land on it).
Nevertheless, we can describe a contrived quasistatic regime
that leads to that conclusion.
Admittedly, dice are never cast this way but it's an idealized limit
of a physical situation and fair dice ought to be fair
under any conditions, including those (like isohedra are).
Therefore, fair dice must be equispherical
(all faces must subtend the same solid angle).
First, we assume that the horizontal plane on which dice land is infinitely sticky,
so that a die can pivot about a vertex or an edge but will never roll.
Thus, it can come to rest on a surface that couldn't serve as a stable resting place
without such stickiness (a spindle could land on its pyramidal
extremities).
Second, we assume dynamical (inertial) effects are negligible.
This would happen if the die was moving at low speed in a very viscous fluid.
We may also assume that this fluid is only slightly less dense than the homogeneous
stuff the die is made from.
The die is fully immersed in the fluid, at rest in a random orientation.
Upon release, it will essentially fall at constant speed
(its terminal velocity) without rotating.
discussed below is an idealized way dice
could actually be cast. What's crucial is the fact that this describes
an actual physical situation (or, at least, the idealized limit of such situations).
The conclusions derived from this contrived model would thus be valid for any die
which is intrinsically fair (i.e., unconditionally fair) in the same
way convex isohedra are: A fair die lands with equal probability on any of its
faces regardless on the surrounding conditions, provided it starts at rest in a random
orientation.
Under such contrived conditions, the die will simply land on whichever face is directly
below its center of gravity.
The assumption that the die is initially randomly oriented
means that the downward vertical through the center of gravity crosses a
face with a probability proportional to the solid angle
it subtends, as seen from the center of gravity.
That same probability is also the probability that the die will land on the
prescribed face.
For example, an isogonal truncated octahedron
makes a fair die under such quasistatic consitions
(dead-cat bounce) when the ratio of an edge between hexagons to the side of a square face is:
x =
0.32173356003298450750124...
(2013-02-18) Thermal Tossing of Dice
The bottom face of a die tends to be closest to its center of gravity.
When placed on an agitated horizontal plate mimicking
thermal motion, dice would naturally
tend to orient themselves in the most energetically favorable way.
That's achieved by lowering the center of gravity as much as possible.
By reducing the amplitude of the agitation gradually until motion freezes,
we effectively cast the dice in a way that definitely favors,
for the bottom position, the faces which are closest to the center of gravity.
(2013-04-07) Mesohedron (Michon, 2013)
A mesohedron is an equispherical orthohedron.
An equispherical die that's fair under the previously described
quasistatic conditions
cannot be fair with the thermal tossing method as well,
unless all its faces are also
equally distant from the center of gravity.
Thus, all the faces of a fair die should be equally distant from the
center of gravity and also subtend the same solid angle.
Isohedra clearly meet both conditions
by symmetry.
Below are examples of
non-isohedral dice that satisfy this restriction
(the solids that don't cannot possibly be fair dice).
Let's establish the vocabulary:
An orthohedron is defined
as a polyhedron with an inscribed sphere; the center of that sphere is
equally distant from all the faces.
If all the faces of an orthohedron subtend the same solid angle
(i.e., if it's equispherical ) then we call it a
mesohedron.
If the mass distribution of a solid orthohedron is such that the
aforementioned center is
at the center of gravity, we say it's balanced.
The above shows that a fair die must be a balanced mesohedron.
A balanced isohedron with spherical inertia
(i.e., its three principal moments of inertia are equal)
is a fair die under all tossing conditions, by reasons of symmetry.
So is a balanced isohedron with an axis of symmetry and mere cylindrical
intertia about that axis (the moment of inertia about the axis of symmetry can
differ from the moments about perpendicular axes).
This later case applies to
bipolar dice
(amphihedra or
deltohedra).
Whichever relevant condition is automatically satisfied for
isohedral solids of uniform mass density (ordinary unloaded dice).
By contrast, the fairness of a balanced mesohedron is
only guaranteed for the two extreme casting methods described above
(quasistatic or thermal
regimes) which we may loosely think of as dead cat bounce
and high resilience, respectively.
For intermediate conditions, no such guarantee exists.
The so-called fitness dice depicted at left
are a pair of 10-sided latex rubber dice meant to be rolled together
to suggest a type of physical exercise and
a number of repetitions to perform. The dice are large enough
(7" height) to be tossed on a gym floor. A pair retails for about $30.
Those seem to be shaped roughly like 10-sided mesohedra.
Let's describe what a perfect 10-sided mesohedron would be:
A die with that general appearance will be equispherical
if and only if both square faces subtends a solid angle of
2p/5 each.
Indeed, by symmetry, the rest of the spat (4p)
is shared equally among the eight other faces, so each of those also
subtends a solid angle of 2p/5.
On the other hand, a polyhedron is orthohedraliff one point (the center)
belongs to all bisectors of its
face angles
(i.e., the dihedral angles formed when two faces meet at an edge).
Technically, a bisector is a set of two orthogonal planes
consisting of all points equally distant from two intersecting planes.
In the case of a convex polyhedron, we may focus our attention to only a quarter
of that figure (a half-plane bordered by an edge of the polyhedron and
featuring a nonempty intersection with its interior).
A mesohedron, is an equispherical orthohedron.
Putting both of the above conditions together yields the following cross-section
(in a plane perpendicular to half of the horizontal edges).
The angle q is determined by the aforementioned
equisphericity condition:
q =
Arctg 1/5¼ = 33.7722424...°
Dimensions of a Mesohedral Decahedron ( of height 5¼ )
The above area of a trapezoidal face is equal to its height
f/2 multiplied into the half-sum of its bases
(1+f)/2. It could also be obtained with
Brahmagupta's formula
(since an isosceles trapezoid is indeed a cyclic quadrilateral ).
The total volume is twice the
volume of a conical frustum.
Note one similarity with the geometry of a sphere:
The surface area of this polyhedron happens to be four times the area of its
equatorial cross-section.
(2013-04-23) Balanced Mesopentahedron ( M5 )
Making an orthohedral rhombic pyramid equispherical and balanced.
Consider a tetragonal pyramid
with two vertical planes of symmetry.
Its horizontal base is a rhombus
(i.e., an equilateral quadrilateral) and it has an inscribed sphere
by reasons of symmetry, since the bisectors at the four horizontal edges
do intersect at a point on the vertical axis.
Such a shape depends on three parameters; the vertical height
h and the half-diagonals of the horizontal rhombus,
x and y.
Alternately, we could consider as parameters the three different
side lengths, a, b and c :
a 2 = x 2 + h 2b 2 = y 2 + h 2c 2 = x 2 + y 2
One parameter determines the size. We may adjust the other two to make
a uniform-density solid both equispherical and balanced.
Here it goes:
The tangent of the dihedral angle based on an horizontal edge is hc/xy.
(HINT: xy/c is the altitude of the right triangle of sides x and y.)
The inclination of the bissecting plane (with respect to the horizontal) is half
the angle corresponding to that...
Now, in a straight pyramid of uniform mass density, the center of gravity is at a height
z = h/4. Thus, the tangent of the bissector's inclination must be ¼
which makes the original dihedral angle's tangent 2t/(1-t2) = 8/15.
All told, the solid is balanced when:
hc / xy = 8 / 15
or
z (x2+y2 )½ / xy = 2 / 15
The solid is equispherical with respect to the center at altitude z
if and only if the base subtends a solid angle of
4p/5. Indeed, since the other 4 faces share equally
the rest of the spat (4p) by symmetry,
this make them subtend the same solid angle of 4p/5.
Using our expression for the
solid angle subtended by a rhombus,
this condition translates into:
(2013-04-07) Balanced Mesoheptahedron ( M7 )
Seven-sided mesohedron with a ternary axis of symmetry.
To obtain an heptahedron with a vertical ternary axis of symmetry,
we may truncate off horizontally one pole of a
bipolar polyhedron with
a vertical ternary axis.
That's to say, either a trigonal dipyramid
or a trigonal deltohedron.
In either case, the final solid can't be an orthohedron
unless the untruncated side is flatter than the side of the truncated pole.
"Predicting a Die Throw
(Science Daily, 2012-09-12) from an AIP press release for:
"The three-dimensional dynamics of the die throw" in
Chaos, 22, 8 pages (December 2012)
by Marcin Kapitaniak, Jaroslaw Strzalko, Juliusz Grabski &
Tomasz Kapitaniak
(Lodz, Poland)
(2013-04-19) Rolling Two-Dimensional Dice
A simplified analysis...
Let's consider a polygonal wheel bouncing on an horizontal track.
(2012-11-28) Nontransitive set of dice.
(strombic icositetrahedron or
trapezoidal icositetrahedron ).
It's marketed by GameScience (Zocchi's company)
under the name of D-Total, featuring fancy markings
that are meant to facilitate the use of the die as a substitute
for dice with
2, 3, 4, 5, 6, 7, 8, 10, 12, 20, 24, 30, 40, 50, 60, 70 or 80 sides.
This is jointly credited to
Dr. Alexander F. Simkin, Frank Dutrain (of LD Diffusion) and Louis Zocchi (2009).
(
(2013-03-27) Quasi-fair and blatantly unfair dice
q =
Arctg 1/5¼ = 33.7722424...°
