Intransitive dice

A set of dice is intransitive if it contains dice, with the property that rolls higher than more than half the time, rolls higher than more than half the time, and so on, but does roll higher than more than half the time. In other words, a set of dice is intransitive if the binary relation – rolls a higher number than more than half the time – on its elements is not transitive. More simply, normally beats, normally beats, but does normally beat.
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 " does not normally beat " it is now " normally beats ". Using such a set of dice, one can invent games which are biased in ways that people unused to intransitive dice might not expect.

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, 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. The following tables show all possible outcomes for all three pairs of dice.
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 that each die beats the next one in the cycle. The largest possible probability is one over the golden ratio,.

Variations

Efron's dice

Efron's dice are a set of four intransitive dice invented by Bradley Efron.
Image:Efron dice 2.svg|thumb|320px|Representation of Efron's dice. The back side of each die has the same faces as the front except for the 5, 5, 1 die.
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. C beats A with probability, 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 :

Miwin's dice

Miwin's dice were invented in 1975 by the physicist Michael Winkelmann. Miwin's dice are a set of nontransitive dice invented in 1975 by the physicist Michael Winkelmann. They consist of three different dice with faces bearing numbers from one to nine; opposite faces sum to nine, ten or eleven.Miwin's dice facilitate generating numbers at random, within a given range, such that each included number is equally-likely to occur. In order to obtain a range that does not begin with 1 or 0, simply add a constant value to bring it into that range.

1 – 9: 1 die is rolled : P = P =... = P = 1/9
0 – 80: 2 dice are rolled, always subtract 1: P = P =... = P = 1/9² = 1/81

The numbers on each die give the sum of 30 and have an arithmetic mean of five. Miwin's dice have six sides, each of which bear a number, depicted in a pattern of dots. The standard set is made of wood; special designs are made of titanium or other materials.

1/3 of the die-face values can be divided by three without carry over.
1/3 of the die-face values can be divided by three having a carry over of one.
1/3 of the die-face values can be divided by three having a carry over of two.

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
the probability that IV rolls a higher number than V is
the probability that V rolls a higher number than III is

This is because the probability for a given number with all three dice is 11/36, for a given rolled double is 1/36, for any rolled double 1/4. The probability to obtain a rolled double is only 50% compared to normal dice.
The dice in the first and second Miwin sets have similar attributes: each die bears each of its numbers exactly once, the sum of the numbers is 30, and each number from one to nine is spread twice over the three dice. This attribute characterizes the implementation of intransitive dice, enabling the different game variants. All the games need only three dice, in comparison to other theoretical nontransitive dice, designed in view of mathematics, such as Efron's dice. In the first set, each die is named for the sum of its two lowest numbers. The dots on each die are colored blue, red or black. Each die has the following numbers:

Die III	with red dots	1	2			5	6	7		9
Die IV	with blue dots	1		3	4	5			8	9
Die V	with black dots		2	3	4		6	7	8

Numbers 1 and 9, 2 and 7, and 3 and 8 are on opposite sides on all three dice. Additional numbers are 5 and 6 on die III, 4 and 5 on die IV, and 4 and 6 on die V. The dice are designed in such a way that, for every die, another will usually win against it. The probability that a given die in the sequence will roll a higher number than the next in the sequence is 17/36; a lower number, 16/36. Thus, die III tends to win against IV, IV against V, and V against III. Such dice are known as nontransitive.
In the second set, each die is named for the sum of its lowest and highest numbers. The dots on each die are colored yellow, white or green. Each die has the following numbers:

Die IX	with yellow dots	1		3		5	6	7	8
Die X	with white dots	1	2		4		6		8	9
Die XI	with green dots		2	3	4	5		7		9

The probability that a given die in the sequence will roll a higher number than the next in the sequence is 17/36; a lower number, 16/36. Thus, die XI tends to win against X, X against IX, and IX against XI.
In the third set:

Die MW 5	with blue numbers					5	6	7	8							15	16
Die MW 3	with red numbers			3	4							11	12	13	14
Die MW 1	with black numbers	1	2							9	10							17	18

In the fourth set:

Die MW 6	with yellow numbers					5	6			9	10			13	14
Die MW 4	with white numbers			3	4			7	8									17	18
Die MW 2	with green numbers	1	2									11	12			15	16

The probability that a given die in the first sequence or the second sequence will roll a higher number than the next in the sequence is 5/9; a lower number, 4/9.

Other distributions

In the 0 – 90 distribution, the governing probability is P = P =... = P = 8/9³ = 8/729. To obtain an equal distribution with numbers from 0 – 90, all three dice are rolled, one at a time, in a random order. The result is calculated based on the following rules:

1st throw is 9, 3rd throw is not 9: gives 10 times 2nd throw
1st throw is not 9: gives 10 times 1st throw, plus 2nd throw
1st throw is equal to the 3rd throw: gives 2nd throw
All dice equal: gives 0
All dice 9: no score

Sample:

1st throw	2nd throw	3rd throw	Equation	Result
9	9	not 9	10 times 9	90
9	1	not 9	10 times 1	10
8	4	not 8	+ 4	84
1	3	not 1	+ 3	13
7	8	7	7 = 7, gives 8	8
4	4	4	all equal	0
9	9	9	all 9	-

This gives 91 numbers, from 0 – 90 with the probability of 8 / 9³, 8 × 91 = 728 = 9³ − 1. In the 0 – 103 distribution, the governing probability is P = P =... = P = 7/9³ = 7/729. This gives 104 numbers from 0 – 103 with the probability of 7 / 9³, 7 × 104 = 728 = 9³ − 1
In the 0 – 728 distribution, the governing probability is P = P =... = P = 1 / 9³ = 1 / 729. This gives 729 numbers, from 0 – 728, with the probability of 1 / 9³. This system yields this maximum: 8 × 9² + 8 × 9 + 8 × 9° = 648 + 72 + 8 = 728 = 9³ − 1. One die is rolled at a time, taken at random. Create a number system of base 9:

1 must be subtracted from the face value of every roll because there are only 9 digits in this number system
× 81 + × 9 + × 1

Examples:

1st throw	2nd throw	3rd throw	Equation	Result
9	9	9	8 × 9² + 8 × 9 + 8	728
4	7	2	3 × 9² + 6 × 9 + 1	298
2	4	1	1 × 9² + 4 × 9 + 0	117
1	3	4	0 × 9² + 3 × 9 + 3	30
7	7	7	6 × 9² + 6 × 9 + 6	546
1	1	1	0 × 9² + 0 × 9 + 0	0
4	2	6	3 × 9² + 1 × 9 + 5	257