Intransitive dice

A set of dice is intransitive (or nontransitive) if it contains three dice, A, B, and C, with the property that A rolls higher than B more than half the time, and B rolls higher than C more than half the time, but where it is not true that A rolls higher than C 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, A normally beats B, B normally beats C, but A does not normally beat C.

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. 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.
 * 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.

Now, consider the following game, which is played with a set of dice. 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.
 * 1) The first player chooses a die from the set.
 * 2) The second player chooses one die from the remaining dice.
 * 3) Both players roll their die; the player who rolls the higher number wins.

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 dice is beaten by the previous dice 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 dice with the following probabilities, where 0 ≤ x ≤ $3⁄7$:
 * P(choose A) = x
 * P(choose B) = $1⁄2$ - $5⁄6$x
 * P(choose C) = x
 * P(choose D) = $1⁄2$ - $7⁄6$x

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 Then:
 * dice III has sides 1, 2, 5, 6, 7, 9
 * dice IV has sides 1, 3, 4, 5, 8, 9
 * dice V has sides 2, 3, 4, 6, 7, 8
 * 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 $17⁄36$

Warren Buffett
Warren Buffett is known to be a fan of intransitive dice. In the book Fortune's Formula: The Untold Story of the Scientific Betting System that Beat the Casinos and Wall Street, a discussion between him and Edward Thorp is described. Buffett and Thorp discussed their shared interest in intransitive dice. "These are a mathematical curiosity, a type of 'trick' dice that confound most people's ideas about probability."

Buffett once attempted to win a game of dice with Bill Gates using intransitive dice. "Buffett suggested that each of them choose one of the dice, then discard the other two. They would bet on who would roll the highest number most often. Buffett offered to let Gates pick his die first. This suggestion instantly aroused Gates's curiosity. He asked to examine the dice, after which he demanded that Buffett choose first."

In 2010, Wall Street Journal magazine quoted Sharon Osberg, Buffett's bridge partner, saying that when she first visited his office 20 years earlier, he tricked her into playing a game with intransitive dice that could not be won and "thought it was hilarious".

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.

Oskar dice
Oskar van Deventer introduced a set of seven dice (all faces with probability $1⁄6$) as follows:
 * A: 2, 2, 14, 14, 17, 17
 * B: 7, 7, 10, 10, 16, 16
 * C: 5, 5, 13, 13, 15, 15
 * D: 3, 3, 9, 9, 21, 21
 * E: 1, 1, 12, 12, 20, 20
 * F: 6, 6, 8, 8, 19, 19
 * G: 4, 4, 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):


 * {| class="wikitable"

! colspan="2" rowspan="2" | Sets chosen by opponents ! colspan="2" | Winning set of dice ! Type ! Number
 * - align="center"
 * - align="center"
 * - align="center"
 * A || B || E || 1
 * - align="center"
 * A || C || E || 2
 * -align="center"
 * A || D || C || 2
 * - align="center"
 * A || E || D || 1
 * - align="center"
 * B || C || A || 1
 * - align="center"
 * B || D || A || 2
 * - align="center"
 * B || E || D || 2
 * - align="center"
 * C || D || B || 1
 * - align="center"
 * C || E || B || 2
 * - align="center"
 * D || E || C || 1
 * }

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:

In "A versus B", A wins in 9 out of 16 cases.

In "B versus C", B wins in 9 out of 16 cases.

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:

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.