# Intransitive dice

A set of dice is intransitive (or nontransitive) if it contains X>2 dice, X1, X2, and X3... with the property that X1 rolls higher than X2 more than half the time, and X2 rolls higher than X3 etc... more than half the time, but where it is not true that X1 rolls higher than Xn more than half the time. In other words, a set of dice is intransitive if the binary relation – X rolls a higher number than Y more than half the time – on its elements is not transitive. More simply, X1 normally beats X2, X2 normally beats X3, but X1 does not normally beat Xn.
It is possible to find sets of dice with the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time. This is different in that instead of only "A does not normally beat C" it is now "C normally beats A". Using such a set of dice, one can invent games which are biased in ways that people unused to intransitive dice might not expect (see Example).

## Example

Consider the following set of dice.
* Die A has sides 2, 2, 4, 4, 9, 9.
* Die B has sides 1, 1, 6, 6, 8, 8.
* Die C has sides 3, 3, 5, 5, 7, 7.
The probability that A rolls a higher number than B, the probability that B rolls higher than C, and the probability that C rolls higher than A are all ⁠5/9⁠, so this set of dice is intransitive. In fact, it has the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time.
Now, consider the following game, which is played with a set of dice.
* The first player chooses a die from the set.
* The second player chooses one die from the remaining dice.
* Both players roll their die; the player who rolls the higher number wins.
If this game is played with a transitive set of dice, it is either fair or biased in favor of the first player, because the first player can always find a die that will not be beaten by any other dice more than half the time. If it is played with the set of dice described above, however, the game is biased in favor of the second player, because the second player can always find a die that will beat the first player's die with probability ⁠5/9⁠. The following tables show all possible outcomes for all three pairs of dice.
| Player 1 chooses die A Player 2 chooses die C | Player 1 chooses die A Player 2 chooses die C | Player 1 chooses die A Player 2 chooses die C | Player 1 chooses die A Player 2 chooses die C | Player 1 chooses die B Player 2 chooses die A | Player 1 chooses die B Player 2 chooses die A | Player 1 chooses die B Player 2 chooses die A | Player 1 chooses die B Player 2 chooses die A | Player 1 chooses die C Player 2 chooses die B | Player 1 chooses die C Player 2 chooses die B | Player 1 chooses die C Player 2 chooses die B | Player 1 chooses die C Player 2 chooses die B |
| AC | 2 | 4 | 9 | BA | 1 | 6 | 8 | CB | 3 | 5 | 7 |
| 3 | C | A | A | 2 | A | B | B | 1 | C | C | C |
| 5 | C | C | A | 4 | A | B | B | 6 | B | B | C |
| 7 | C | C | A | 9 | A | A | A | 8 | B | B | B |
If one allows weighted dice, i.e., with unequal probability weights for each side, then alternative sets of three dice can achieve even larger probabilities than ${\frac {5}{9}}\approx 0.56$ that each die beats the next one in the cycle. The largest possible probability is one over the golden ratio, ${\frac {1}{\varphi }}\approx 0.62$.

## Variations
### Efron's dice

Efron's dice are a set of four intransitive dice invented by Bradley Efron.
The four dice A, B, C, D have the following numbers on their six faces:
* A: 4, 4, 4, 4, 0, 0
* B: 3, 3, 3, 3, 3, 3
* C: 6, 6, 2, 2, 2, 2
* D: 5, 5, 5, 1, 1, 1
Each die is beaten by the previous die in the list with wraparound, with probability ⁠2/3⁠. C beats A with probability ⁠5/9⁠, and B and D have equal chances of beating the other. If each player has one set of Efron's dice, there is a continuum of optimal strategies for one player, in which they choose their die with the following probabilities, where 0 ≤ x ≤ ⁠3/7⁠:
    P(choose A) = x
    P(choose B) = ⁠1/2⁠ - ⁠5/6⁠x
    P(choose C) = x
    P(choose D) = ⁠1/2⁠ - ⁠7/6⁠x

### Miwin's dice

Miwin's Dice were invented in 1975 by the physicist Michael Winkelmann.
Consider a set of three dice, III, IV and V such that
* die III has sides 1, 2, 5, 6, 7, 9
* die IV has sides 1, 3, 4, 5, 8, 9
* die V has sides 2, 3, 4, 6, 7, 8
Then:
* the probability that III rolls a higher number than IV is ⁠17/36⁠
* the probability that IV rolls a higher number than V is ⁠17/36⁠
* the probability that V rolls a higher number than III is ⁠17/36⁠

## Intransitive dice set for more than two players

A number of people have introduced variations of intransitive dice where one can compete against more than one opponent.

### Three players
#### Oskar dice

Oskar van Deventer introduced a set of seven dice (all faces with probability ⁠1/6⁠) as follows:
* A: 2, 02, 14, 14, 17, 17
* B: 7, 07, 10, 10, 16, 16
* C: 5, 05, 13, 13, 15, 15
* D: 3, 03, 09, 09, 21, 21
* E: 1, 01, 12, 12, 20, 20
* F: 6, 06, 08, 08, 19, 19
* G: 4, 04, 11, 11, 18, 18
One can verify that A beats {B,C,E}; B beats {C,D,F}; C beats {D,E,G}; D beats {A,E,F}; E beats {B,F,G}; F beats {A,C,G}; G beats {A,B,D}. Consequently, for arbitrarily chosen two dice there is a third one that beats both of them. Namely,
* G beats {A,B}; F beats {A,C}; G beats {A,D}; D beats {A,E}; D beats {A,F}; F beats {A,G};
* A beats {B,C}; G beats {B,D}; A beats {B,E}; E beats {B,F}; E beats {B,G};
* B beats {C,D}; A beats {C,E}; B beats {C,F}; F beats {C,G};
* C beats {D,E}; B beats {D,F}; C beats {D,G};
* D beats {E,F}; C beats {E,G};
* E beats {F,G}.
Whatever the two opponents choose, the third player will find one of the remaining dice that beats both opponents' dice.

#### Grime dice

Dr. James Grime discovered a set of five dice as follows:
* A: 2, 2, 2, 7, 7, 7
* B: 1, 1, 6, 6, 6, 6
* C: 0, 5, 5, 5, 5, 5
* D: 4, 4, 4, 4, 4, 9
* E: 3, 3, 3, 3, 8, 8
One can verify that, when the game is played with one set of Grime dice:
* A beats B beats C beats D beats E beats A (first chain);
* A beats C beats E beats B beats D beats A (second chain).
However, when the game is played with two such sets, then the first chain remains the same, except that D beats C, but the second chain is reversed (i.e. A beats D beats B beats E beats C beats A). Consequently, whatever dice the two opponents choose, the third player can always find one of the remaining dice that beats them both (as long as the player is then allowed to choose between the one-die option and the two-die option):
    Sets chosen by opponents Winning set of dice Type Number ABE1 ACE2 ADC2 AED1 BCA1 BDA2 BED2 CDB1 CEB2 DEC1

### Four players

A four-player set has not yet been discovered, but it was proved that such a set would require at least 19 dice.

## Intransitive 4-sided dice

Tetrahedra can be used as dice with four possible results.

### Set 1

* A: 1, 4, 7, 7
* B: 2, 6, 6, 6
* C: 3, 5, 5 ,8
P(A > B) = P(B > C) = P(C > A) = ⁠9/16⁠
The following tables show all possible outcomes:
| BA | 2 | 6 | 6 | 6 |
| 1 | B | B | B | B |
| 4 | A | B | B | B |
| 7 | A | A | A | A |
| 7 | A | A | A | A |
In "A versus B", A wins in 9 out of 16 cases.
| CB | 3 | 5 | 5 | 8 |
| 2 | C | C | C | C |
| 6 | B | B | B | C |
| 6 | B | B | B | C |
| 6 | B | B | B | C |
In "B versus C", B wins in 9 out of 16 cases.
| AC | 1 | 4 | 7 | 7 |
| 3 | C | A | A | A |
| 5 | C | C | A | A |
| 5 | C | C | A | A |
| 8 | C | C | C | C |
In "C versus A", C wins in 9 out of 16 cases.

### Set 2

* A: 3, 3, 3, 6
* B: 2, 2, 5, 5
* C: 1, 4, 4, 4
P(A > B) = P(B > C) = ⁠10/16⁠, P(C > A) = 9/16

## Intransitive 12-sided dice

In analogy to the intransitive six-sided dice, there are also dodecahedra which serve as intransitive twelve-sided dice. The points on each of the dice result in the sum of 114. There are no repetitive numbers on each of the dodecahedra.
Miwin's dodecahedra (set 1) win cyclically against each other in a ratio of 35:34.
The miwin's dodecahedra (set 2) win cyclically against each other in a ratio of 71:67.
Set 1:
| D III | purple | 1 | 2 | | | 5 | 6 | 7 | | 9 | 10 | 11 | | | 14 | 15 | 16 | | 18 |
| D IV | red | 1 | | 3 | 4 | 5 | | | 8 | 9 | 10 | | 12 | 13 | 14 | | | 17 | 18 |
| D V | dark grey | | 2 | 3 | 4 | | 6 | 7 | 8 | | | 11 | 12 | 13 | | 15 | 16 | 17 | |
* D III
* D IV
* D V
Set 2:
| D VI | cyan | 1 | 2 | 3 | 4 | | | | | 9 | 10 | 11 | 12 | 13 | 14 | | | 17 | 18 |
| D VII | pear green | 1 | 2 | | | 5 | 6 | 7 | 8 | 9 | 10 | | | | | 15 | 16 | 17 | 18 |
| D VIII | light grey | | | 3 | 4 | 5 | 6 | 7 | 8 | | | 11 | 12 | 13 | 14 | 15 | 16 | | |
* D VI
* D VII
* D VIII

### Intransitive prime-numbered 12-sided dice

It is also possible to construct sets of intransitive dodecahedra such that there are no repeated numbers and all numbers are primes. Miwin's intransitive prime-numbered dodecahedra win cyclically against each other in a ratio of 35:34.
Set 1: The numbers add up to 564.
| PD 11 | grey to blue | 13 | 17 | 29 | 31 | 37 | 43 | 47 | 53 | 67 | 71 | 73 | 83 |
| PD 12 | grey to red | 13 | 19 | 23 | 29 | 41 | 43 | 47 | 59 | 61 | 67 | 79 | 83 |
| PD 13 | grey to green | 17 | 19 | 23 | 31 | 37 | 41 | 53 | 59 | 61 | 71 | 73 | 79 |
* PD 11
* PD 12
* PD 13
Set 2: The numbers add up to 468.
| PD 1 | olive to blue | 7 | 11 | 19 | 23 | 29 | 37 | 43 | 47 | 53 | 61 | 67 | 71 |
| PD 2 | teal to red | 7 | 13 | 17 | 19 | 31 | 37 | 41 | 43 | 59 | 61 | 67 | 73 |
| PD 3 | purple to green | 11 | 13 | 17 | 23 | 29 | 31 | 41 | 47 | 53 | 59 | 71 | 73 |
* PD 1
* PD 2
* PD 3

## Generalized Muñoz-Perera's intransitive dice

A generalization of sets of intransitive dice with $N$ faces is possible. Given $N\geq 3$, we define the set of dice $\{D_{n}\}_{n=1}^{N}$ as the random variables taking values each in the set $\{v_{n,j}\}_{j=1}^{J}$ with
$\mathbb {P} \left[D_{n}=v_{n,j}\right]={\frac {1}{J}}$,
so we have $N$ fair dice of $J$ faces.
To obtain a set of intransitive dice is enough to set the values $v_{n,j}$ for $n,j=1,2,\ldots ,N$ with the expression
$v_{n,j}=(j-1)N+(n-j){\text{mod}}(N)+1$,
obtaining a set of $N$ fair dice of $N$ faces
Using this expression, it can be verified that
$\mathbb {P} \left[D_{m}<D_{n}\right]={\frac {1}{2}}+{\frac {1}{2N}}-{\frac {(n-m){\text{mod}}(N)}{N^{2}}}$,
So each die beats $\lfloor N/2-1\rfloor$ dice in the set.

### Examples
#### 3 faces

| $D_{1}$ | 1 | 6 | 8 |
| $D_{2}$ | 2 | 4 | 9 |
| $D_{3}$ | 3 | 5 | 7 |
The set of dice obtained in tis case is equivalent to the first example on this page, but removing repeated faces. It can be verified that $D_{3}>D_{2},D_{2}>D_{1}\ {\text{and}}\ D_{1}>D_{3}$.

#### 4 faces

| $D_{1}$ | 1 | 8 | 11 | 14 |
| $D_{2}$ | 2 | 5 | 12 | 15 |
| $D_{3}$ | 3 | 6 | 9 | 16 |
| $D_{4}$ | 4 | 7 | 10 | 13 |
Again it can be verified that $D_{4}>D_{3},D_{3}>D_{2},D_{2}>D_{1}\ {\text{and}}\ D_{1}>D_{4}$.

#### 6 faces

| $D_{1}$ | 1 | 12 | 17 | 22 | 27 | 32 |
| $D_{2}$ | 2 | 7 | 18 | 23 | 28 | 33 |
| $D_{3}$ | 3 | 8 | 13 | 24 | 29 | 34 |
| $D_{4}$ | 4 | 9 | 14 | 19 | 30 | 35 |
| $D_{5}$ | 5 | 10 | 15 | 20 | 25 | 36 |
| $D_{6}$ | 6 | 11 | 16 | 21 | 26 | 31 |
Again $D_{6}>D_{5},D_{5}>D_{4},D_{4}>D_{3},D_{3}>D_{2},D_{2}>D_{1}\ {\text{and}}\ D_{1}>D_{6}$. Moreover $D_{6}>\{D_{5},D_{4}\},D_{5}>\{D_{4},D_{3}\},D_{4}>\{D_{3},D_{2}\},D_{3}>\{D_{2},D_{1}\},D_{2}>\{D_{1},D_{6}\}\ {\text{and}}\ D_{1}>\{D_{6},D_{5}\}$.