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INTRANSITIVE DICE
arXiv:1311.6511v1 [math.CO] 25 Nov 2013
BRIAN CONREY, JAMES GABBARD, KATIE GRANT, ANDREW LIU, KENT MORRISON
Abstract. We conjecture that the probability that a random triple of dice is intransitive
is 1/4.
1. Introduction
We define an n-sided die to be an n-tuple (a1 , . . . , an ) of non-decreasing integers, a1 ≤
a2 ≤ · · · ≤ an . We denote the standard n-sided die as
Pn = (1, 2, 3, . . . , n).
Notice that the sum of the faces on Pn is
1 + 2 + ··· + n =
n(n + 1)
.
2
We call an n-sided die proper if the sides are whole numbers between 1 and n inclusive and
the sum of the faces is n(n+1)
.
2
Thus, every proper die which is not a standard one has faces with repeated numbers, i.e.
ai = ai+1 for at least one i with 1 ≤ i ≤ n − 1.
Here is a list of proper dice for n ≤ 5:
n=1
n=2
n=3
n=4
n=5
(1)
(1, 2)
(1, 2, 3), (2, 2, 2)
(1, 1, 4, 4), (1, 2, 3, 4), (1, 3, 3, 3), (2, 2, 2, 4), (2, 2, 3, 3)
(1, 1, 3, 5, 5), (1, 1, 4, 4, 5), (1, 2, 2, 5, 5), (1, 2, 3, 4, 5), (1, 2, 4, 4, 4), (1, 3, 3, 3, 5),
(1, 3, 3, 4, 4), (2, 2, 2, 4, 5), (2, 2, 3, 3, 5), (2, 2, 3, 4, 4), (2, 3, 3, 3, 4), (3, 3, 3, 3, 3)
Research supported in part by a grant from the National Science Foundation.
1
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BRIAN CONREY, JAMES GABBARD, KATIE GRANT, ANDREW LIU, KENT MORRISON
Here is a table of the number P r(n) of proper n-sided dice for n ≤ 28 (sequence A076822
in the Online Encyclopedia of Integer Sequences):
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
25
26
27
28
P r(n)
1
1
2
5
12
32
94
289
910
2934
9686
32540
110780
381676
1328980
4669367
16535154
58965214
211591218
763535450
2769176514
10089240974
36912710568
135565151486
499619269774
1847267563742
6850369296298
If we have two proper n-sided dice A = (a1 , . . . , an ) and B = (b1 , . . . , bn ) we say that
A > B if the probability that A “rolls” a higher value than B is greater than the probability
that B “rolls” a higher value than A. Put another way, if the probability that a randomly
selected number from A is larger than a randomly selected number from B, then we say that
A beats B and write A > B. Formally, A > B if and only if
n
X
i,j=1
ai >bj
1>
n
X
1.
i,j=1
bj >ai
If the two sums are equal then we say that A and B tie. A table can help to make the
comparisons. For example, if A = (1, 2, 4, 4, 4, 6) and B = (2, 2, 3, 3, 5, 6) then the table of
INTRANSITIVE DICE
3
comparisons below shows that out of the 36 possible rolls A wins 17 times, B wins 16 times
and there are 3 ties. Therefore, we say A > B.
A\B
1
2
4
4
4
6
2
B
A
A
A
A
2
B
A
A
A
A
3
B
B
A
A
A
A
3
B
B
A
A
A
A
5
B
B
B
B
B
A
6
B
B
B
B
B
-
One might think that given proper n-sided dice A, B and C that f A > B and B > C
then it necessarily follows that A > C, i.e. that “beats” is a transitive relation. But it’s
not! For example, if A = (1, 1, 4, 4), B = (1, 3, 3, 3), and C = (2, 2, 2, 4) then it is easy to
check that A beats B, B beats C, and C beats A. Such a triple of dice is called intransitive.
Specifically, a set of three dice is called “transitive” if there is a labeling A, B, C of the dice
such that A > B, A > C, and B > C; it is called intransitive if there is a labeling such that
A > B, B > C, and C > A.
It is counter-intuitive that intransitive dice exist. But given that they do, one’s intuition
might lead one to believe that they are rare. The purpose of this paper is to give evidence
that suggests that not only do they exist but they are quite common! Suppose one is given
a randomly selected set {A, B, C, . . . , Y, Z} of dice. Suppose you tabulate all but one of the
pairwise comparisons A versus B, A versus C, B versus C and so on of these dice but leaving
the last one Y versus Z unchecked. Surely knowing all of the previous comparisons must
have some effect on the last one? We conjecture that it does not! Rather, independently of
all of the checks made so far it is equally likely that Y beats Z as that Z beats Y , at least
when we are sampling from proper n-sided dice and n → ∞.
2. Triples of dice
Suppose we choose 3 n-sided proper dice at random from the set of all n-sided proper dice.
Assuming there are no ties the triple of dice we get is either transitive, i.e., one die beats
the other two or else it is intransitive, i.e., each die beats one other. What is the probability
that our random triple is intransitive?
As an example, consider the case where n = 4. There are 5 proper dice and so 53 = 10
different triples of dice. Exactly one of those triples is intransitive (namely the one mentioned
above) and so the probability that a triple of proper 4-sided dice is intransitive is 1/10.
We investigated this numerically. Our data, based on 10,000 experiments for each of
10-sided, 20-sided, 30-sided, 40-sided, and 50-sided dice, is below. We used two different
methods (labeled I and II, both of which we believe to be approximately uniform over our
collection of proper dice) to generate proper dice and found
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BRIAN CONREY, JAMES GABBARD, KATIE GRANT, ANDREW LIU, KENT MORRISON
n
10
20
30
40
50
%intrans, I % intrans, II
15.7
15.8
21.3
20.6
22.0
21.9
22.9
23.3
23.7
22.9
We began to suspect that the correct answer is 25% and we formalize that here with
Conjecture 1. Suppose three dice are chosen at random from the collection of n-sided proper
dice. Then, the probability that this triple is intransitive approaches 1/4 as n → ∞.
We believe that this conjecture will be difficult to prove. But we do have some theoretical
evidence, namely if we dilute the sample space (almost to nothing!) then we can prove this
conjecture.
The simplest possible case is when each die in the population from which we sample
deviates from the standard die Pn by as little as possible. This occurs when one face is
increased by one and another face is decreased by one.
For example, (1, 2, 3, 4, 4, 6, 7, 9, 9, 10) is a 10-sided die that differs from the standard die
P10 in that the 8 increases to 9 and the 5 decreases to 4. We denote this die by s(8, 5).
We studied triples of n-sided dice that all have this shape s(a, b) which means that face a
increases by one and face b decreases by one.
We call these “1-step” dice. How many 1-step dice are there? If n = 3 there is only 1:
s(1, 3) = (2, 2, 2)
If n = 4, we have
s(1, 3)
s(1, 4)
s(2, 4)
s(3, 2)
=
=
=
=
(2, 2, 2, 4)
(2, 2, 3, 3)
(1, 3, 3, 3)
(1, 1, 4, 4)
INTRANSITIVE DICE
5
If n = 5, we have
s(1, 3)
s(1, 4)
s(1, 5)
s(2, 4)
s(2, 5)
s(3, 2)
s(3, 5)
s(4, 2)
s(4, 3)
=
=
=
=
=
=
=
=
=
(2, 2, 2, 4, 5)
(2, 2, 3, 3, 5)
(2, 2, 3, 4, 4)
(1, 3, 3, 3, 5)
(1, 3, 3, 4, 4)
(1, 1, 4, 4, 5)
(1, 2, 4, 4, 4)
(1, 1, 3, 5, 5)
(1, 2, 2, 5, 5)
How many n-sided one-step dice are there? We can pick a as the face to increase in (n − 1)
different ways—we can’t allow an increase in face n. And we can decrease (n − 3) different
faces— we don’t allow a decrease in face 1 or in the face that went up or in the face after
the one that went up. Of course, if we increase face number 1 then we can decrease in faces
3, 4, . . . , n.
In total we find that there are n2 − 4n + 4 = (n − 2)2 1-step dice.
So there are
(n − 2)2
3
triples of dice we could select for our experiment. However, in practice we find that almost
all such triples involve ties.
For example, the standard die ties with every other proper die!
The following lemma is easily checked.
Lemma 1. If s(a, b) and s(c, d) are proper n-sided dice and s(a, b) > s(c, d), then s(a, b) >
s(c, d) has one of the following four forms:
s(x, y)
s(x, y)
s(x + 1, y)
s(x, y + 1)
>
>
>
>
s(y, z)
s(z, x + 2)
s(x, z)
s(z, y)
where a, b, c, d are numbers between 1 and n (with some obvious restrictions such as a 6= 1,
a 6= b, etc.)
We give some examples of each of these four types of inequalities. We have circled the
places where “something out of the ordinary” happens, i.e. the pattern of A’s below the
main diagional, − on the main diagonal, and B’s above the main diagonal that one would
see if the standard die were matched against itself.
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BRIAN CONREY, JAMES GABBARD, KATIE GRANT, ANDREW LIU, KENT MORRISON
s(X,Y) > s(Y,Z)
B = s(5,8)
A = s(2,5)
s(X,Y) > s(Z,X+2)
A = s(4,9)
B = s(1,6)
INTRANSITIVE DICE
7
s(X+1,Y) > s(X,Z)
B = s(5,2)
A=s(6,9)
s(X,Y+1) > s(Z,Y)
B = s(8,4)
A=
s(1,5)
Using this lemma we see for example that the triple A = s(X, X + 2), B = s(X, Z),
C = s(X + 1, Y ) produces a triple where A < B by the second of the inequalities above,
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BRIAN CONREY, JAMES GABBARD, KATIE GRANT, ANDREW LIU, KENT MORRISON
B < C by the third inequality and A < C by the third inequality. This triple will be proper
as long as 1 ≤ X ≤ n − 2, 2 ≤ Y ≤ n and 2 ≤ Z ≤ n. The value X = n is not allowed
because then dice A would be s(n, n+2) which has a face labeled n+1 which is not permitted
in n-sided proper dice. Similarly neither Y = 1 nor Z = 1 is permitted since they would
lead to a face labeled 0.
There are actually 64 different cases of triples of proper 1-step dice that are mutually
comparable. The example s(X, X + 2), s(X, Z), s(X + 1, Y ) is one of those 64. In our next
lemma, we list all 64 of these cases, the first one being this example we have just discussed.
It is extremely tedious to find and list all 64 of these cases; we have used a computer to help
make sure that we have found all of them and that they are all different from each other.
Each case leads to a parametric solution of triples, i.e. we can substitute numbers from 1 to
n in for the parameters X, Y, Z and so have numerical triples of proper n-sided 1-step dice.
There are some issues of coincidences and boundaries but the order of magnitude of these
exceptions is n2 amongst the n3 triples produced by one of our parametrizations.
Lemma 2. Suppose we have three proper n-sided dice s(a, b), s(c, d) and s(e, f ) that are
mutually comparable, i.e. between each pair one is better. Then the three dice
{s(a, b), s(c, d), s(e, f )}
can be represented by one of the following 64 triples:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
s(X, 2 + X)
s(X, 2 + X)
s(X, 4 + X)
s(X, Y )
s(X, Y )
s(X, Y )
s(X, Y )
s(X, Y )
s(X, Y )
s(X, Y )
s(X, Y )
s(X, Y )
s(X, Y )
s(X, Y )
s(X, Y )
s(X, Y )
s(X, Z)
s(1 + X, Z)
s(1 + X, Z)
s(X, 1 + Y )
s(1 + X, 3 + X)
s(1 + X, Z)
s(1 + X, Z)
s(1 + X, Z)
s(Y, 2 + Y )
s(Y, Z)
s(Y, Z)
s(Y, Z)
s(Y, Z)
s(Z, X)
s(Z, X)
s(Z, X)
s(1 + X, Y )
s(2 + X, Y )
s(2 + X, Y )
s(1 + X, Z)
s(1 + X, Z)
s(1 + X, 1 + Z)
s(2 + X, X)
s(2 + X, 1 + Y )
s(Y, Z)
s(Y, 1 + Z)
s(1 + Y, X)
s(1 + Y, 2 + X)
s(Z, X)
s(1 + Z, X)
s(1 + Z, 2 + X)
s(1 + Z, 1 + Y )
INTRANSITIVE DICE
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 1 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 2 + Y )
s(X, 3 + Y )
s(X, 3 + Y )
s(X, 3 + Y )
s(X, 3 + Y )
s(X, 3 + Y )
s(X, 3 + Y )
s(X, 3 + Y )
s(X, 3 + Y )
s(1 + X, 1 + Y )
s(1 + X, Z)
s(1 + X, Z)
s(1 + X, 2 + Z)
s(Y, X)
s(Y, X)
s(Y, 2 + Y )
s(Y, 2 + Y )
s(1 + Y, Z)
s(1 + Y, Z)
s(1 + Y, 2 + Z)
s(Z, X)
s(Z, Y )
s(Z, Y )
s(Z, Y )
s(Z, Y )
s(1 + X, Y )
s(1 + X, Y )
s(1 + X, 2 + Y )
s(1 + X, Z)
s(1 + X, Z)
s(1 + X, 1 + Z)
s(Y, 1 + X)
s(Y, 2 + Y )
s(Y, 2 + Y )
s(Y, 2 + Y )
s(Y, 4 + Y )
s(Y, Z)
s(Y, Z)
s(Y, Z)
s(Y, Z)
s(Y, Z)
s(Y, Z)
s(Y, Z)
s(Y, Z)
s(Y, 1 + Z)
s(Y, 2 + Z)
s(Y, 2 + Z)
s(Y, 2 + Z)
s(2 + Y, Y )
s(Y, X)
s(Y, X)
s(Y, 2 + X)
s(Y, 2 + X)
s(Y, 2 + Y )
s(Y, 2 + Y )
s(Y, 4 + Y )
s(Y, 4 + Y )
s(Z, Y )
s(2 + X, Y )
s(Z, Y )
s(Z, Y )
s(1 + Y, Z)
s(Z, Y )
s(1 + Y, Z)
s(Z, Y )
s(2 + Y, Y )
s(Z, Y )
s(Z, Y )
s(1 + Z, Y )
s(1 + Z, X)
s(1 + Z, 2 + X)
s(1 + Z, Y )
s(1 + Z, 2 + Y )
s(Y, Z)
s(Z, 1 + Y )
s(Y, Z)
s(Y, 1 + X)
s(Y, 1 + Z)
s(Y, Z)
s(Z, X)
s(Y, Z)
s(2 + Y, Z)
s(Z, 1 + Y )
s(2 + Y, Z)
s(Y, 1 + Z)
s(1 + Y, X)
s(1 + Y, 2 + X)
s(1 + Y, 3 + Y )
s(2 + Y, Y )
s(2 + Y, 1 + Z)
s(Z, X)
s(Z, 1 + Y )
s(2 + Y, Z)
s(Z, X)
s(Z, 2 + X)
s(Z, 1 + Y )
s(Z, 1 + Y )
s(1 + Y, Z)
s(Z, 2 + Y )
s(1 + Y, Z)
s(Z, 2 + Y )
s(1 + Y, Z)
s(Z, 2 + Y )
s(1 + Y, Z)
s(Z, 2 + Y )
9
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BRIAN CONREY, JAMES GABBARD, KATIE GRANT, ANDREW LIU, KENT MORRISON
In these triples the letters X, Y and Z are allowed to be anything from 1 to n as long as
the dice produced are proper n-sided dice.
We analyze each case as follows. The four reasons for one die beating another may be
labeled as
(1) s(x, y)
> s(y, z)
(2) s(x, y)
> s(z, x + 2)
(3) s(x + 1, y) > s(x, z)
(4) s(x, y + 1) > s(z, y)
Then the reasoning for case 1 may be described as:
A
B
C
B
C
C
=
=
=
>
>
>
s(X, 2 + X)
s(X, Z)
s(1 + X, Y )
A
(2)
B
(3)
A
(3)
Since C > B and C > A this triple is transitive.
We analyzed all of the triples in this way and found (briefly expressed):
1.T : (2)(3)(3) CBA
5.T : (3)(2)(3) CBA
9.T : (1)(2)(1) ACB
13.I : (1)(1)(1) ABC
17.T : (3)(4)(4) BAC
21.I : (1)(3)(1) ACB
25.T : (1)(3)(4) ACB
29.T : (4)(3)(1) CAB
33.T : (3)(1)(2) BCA
37.T : (3)(4)(2) CBA
41.T : (2)(1)(1) BAC
45.T : (2)(3)(1) CBA
49.I : (2)(4)(1) ACB
53.T : (2)(2)(1) CBA
57.T : (1)(3)(2) CBA
61.T : (4)(3)(2) CAB
2.I : (3)(3)(1) ACB
6.T : (3)(4)(3) CBA
10.T : (1)(4)(1) ACB
14.T : (1)(3)(1) CBA
18.I : (3)(3)(4) ACB
22.I : (1)(1)(4) ACB
26.T : (1)(1)(4) ABC
30.T : (4)(3)(2) ACB
34.I : (3)(4)(4) ACB
38.T : (3)(4)(2) BCA
42.T : (2)(4)(4) BAC
46.I : (2)(3)(2) ACB
50.T : (2)(1)(1) BCA
54.I : (2)(2)(2) ACB
58.T : (1)(2)(4) BAC
62.T : (4)(2)(4) ABC
3.T : (3)(3)(2) CBA
7.T : (3)(3)(1) CBA
11.T : (1)(3)(1) CAB
15.I : (1)(3)(2) ACB
19.T : (3)(1)(4) BAC
23.I : (4)(3)(1) ACB
27.T : (1)(2)(4) ACB
31.T : (4)(3)(4) ACB
35.T : (3)(2)(2) CBA
39.T : (2)(4)(1) BCA
43.I : (2)(2)(1) ACB
47.T : (2)(3)(4) CBA
51.T : (2)(1)(4) BAC
55.I : (2)(2)(4) ACB
59.T : (2)(3)(2) CAB
63.T : (4)(3)(2) CBA
4.T : (4)(3)(3) CBA
8.T : (3)(3)(4) CBA
12.T : (1)(3)(2) ACB
16.T : (1)(3)(4) CBA
20.I : (3)(2)(4) ACB
24.I : (4)(1)(4) ACB
28.I : (1)(3)(4) ACB
32.T : (4)(3)(4) CAB
36.T : (3)(1)(2) CBA
40.T : (2)(2)(2) CBA
44.T : (2)(4)(2) CBA
48.I : (2)(1)(1) ACB
52.T : (2)(4)(1) BAC
56.T : (1)(4)(4) ACB
60.T : (2)(2)(4) ABC
64.T : (4)(2)(4) BAC
We give an example of how to read this table. Recall that entry 38 from our lemma corresponds to A = s(X, 2 + Y ), B = s(1 + X, 1 + Z) and C = s(Y, Z). Now entry 38 above says
T : (3)(4)(2) BCA. The reasons (3)(4)(2) are for the comparisons A versus B, B versus C
and C versus A in that order. The BCA together with the T (for transitive) means that
INTRANSITIVE DICE
11
B > C, C > A, and B > A. In entry 46 we have an I for intransitive so that the letters
ACB mean that A > C > B > A.
Now notice that the 17 cases 2, 13, 15, 18, 20, 21, 22, 23, 24, 28, 34, 43, 46, 48, 49, 54 and
55 all give intransitive triples and the other 47 cases give transitive triples.
In general given a triple, we let the variables X, Y , and Z run from 1 to n and in doing
so produce n3 triples of mutually comparable n-sided proper dice. There are some boundary
issues for values of X, Y and Z near to 1 or to n. But the number of such boundary issues
is O(n2 ) and so does not effect the main term which is of order of magnitude n3 .
However, two of our 64 triples are special, namely numbers 13 and 54
s(X, Y )
s(Y, Z)
s(Z, X)
and
s(X, Y )
s(Y, Z)
s(Z, X)
in that they have automorphisms. Each of these produces 3 repeats when X, Y, and Z are
allowed to run from 1 to n. So really each should be counted with a weight 1/3.
This means that our 64 cases are really 62 23 cases. And 15 23 of these cases are intransitive.
Notice that 15 23 is exactly 1/4 of 62 23 .
Thus, we have found
188 3
n + O(n2 )
3
triples of mutually comparable n-sided 1-step dice. Of these,
47 3
n + O(n2 )
3
are intransitive. We take take the ratio as n → ∞ and get 1/4. Thus, we have proven
Theorem 1. Consider the space of all triples of proper n-sided one step dice which have the
property that no two of the dice on the triple tie each other. Then, in the limit as n → ∞
we have 1/4 of the triples are intransitive.
3. 4 or more dice
We have a more general conjecture about what happens with larger sets of proper dice. To
explain this we look at the case of triples in a different light. There are 8 possible orderings
among 3 dice A, B, and C (we assume there are no ties). Namely
A>B
B>C
C>A
A>B
A>B
A>B
A<B
A<B
A<B
B
B
B
B
B
B
C
C
C
C
C
C
>C
<C
<C
>C
>C
<C
<A
>A
<A
>A
<A
>A
12
BRIAN CONREY, JAMES GABBARD, KATIE GRANT, ANDREW LIU, KENT MORRISON
A<B
B<C
C<A
Two of these are intransitive (the first and last) and all the others are transitive. We
conjecture that all 8 of these cases are equally likely. Then the 1/4 of our earlier conjecture
is just a reflection of this more general conjecture together with the fact that 2 of the 8
scenarios above are intransitive.
If we have four n-sided proper dice A, B, C, and D, then we expect every possible relation
between any two dice to be equally likely. Now there are
4
2( 2 ) = 26 = 64
different possibilities. We expect each choice of relation to occur 1/64th of the time.
What are the possible scenarios now? There are
I. Completely Transitive. For example, A > B, C, D; B > C, D; C > D;
II. One dominant plus intransitive loop. A > B, C, D; B > C > D > B;
III. One inferior plus an intransitive loop. A < B, C, D; B > C > D > B;
IV No dominant, no inferior. A > B, C; B > C, D; D > A; C > D.
How do the 64 assignments split up into these categories?
(3,2,1,0) Completely Transitive. For example A > B, C, D; B > C, D; C > D . A has 3
wins; B has 2 wins; C has 1 win; and D has 0 wins. With relabelings there are are 4! = 24
of these. So the probability of getting this configuration is 24/64 = 3/8.
(3,1,1,1) One dominant plus intransitive loop: A > B, C, D; B > C > D > B. There
are 4 ways to pick the big winner, then two ways to cycle around the intransitive loop.
This accounts for 4 × 2 = 8 with relabelings; the probability that 4 random dice are in this
configuration is 1/8.
(2,2,2,0) One inferior plus an intransitive loop: A < B, C, D; B > C > D > B. Just as
the previous case there are 4 × 2 = 8 of these.
(2,2,1,1) No dominant, no inferior A > B, C; B > C, D; D > A; C > D. Choose the two
winners in 42 = 6 ways. Then choose which of the “winners” beats the other and which of
the “losers” beats the other. There are 24 cases. This configuration has 2 intransitive loops
and occurs in 3/8 random choices of 4 dice.
It’s possible to do similar analysis with any number of dice. The big conjecture is that
when choosing k dice from a large collection of proper n-sided dice, the probability of each
k
of the 2( 2 ) possible assignments of < or > to each pair of dice is equally likely in the limit
as n → ∞.
INTRANSITIVE DICE
13
In practice to calculate the probability that k dice will roll to give a specified configuration,
i.e., a tournament graph on k-vertices, all one needs to do is to calculate the probability of
that tournament graph (or an isomorphic copy) arising from a random assignment of edge
directions on the complete graph on k-vertices.
Here is an example with 5 dice:
5 dice example
(2,2,2,2,2)
Intransitive cycles
(ABE)
(BCE)
(CDE)
(DEA)
(EAB)
(ABEC)
(ADBE)
(ADEC)
(ADBC)
(BECD)
(ABCDE)
(ADBEC)
A=(1,1,3,5,5)
B=(1,3,3,4,4)
A
E
C=(2,2,2,4,5)
D=(1,2,4,4,4)
B
E=(2,2,3,3,5)
D
C
The probability of getting this configuration
from 5 randomly chosen dice should be
5!/2^10=15/128=0.117
We conclude with a possible way to generalize the conjecture to functions. Suppose we
have the space of monotonic functions f on [0, 1] with f (0) = 0 and f (1) = 1. If f and g are
two such functions then we say that f > g if
Z 1
Z 1
sign(f (x) − g(x)) dx >
sign(g(x) − f (x)) dx.
0
0
Then, with an appopriate measure on this space we conjecture that if three such functions
are chosen at random, the probability that they form an intransitive triple is 1/4. And
similarly for random choices of k such functions.
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