Intransitive dice

A set of nontransitive dice is a set of dice for which the relation "is more likely to roll a higher number" is not transitive. See also intransitivity.

This situation is similar to that in the game Rock, Paper, Scissors, in which each element has an advantage over one choice and a disadvantage to the other.

Example

Consider a set of three dice, A, B and C such that

die A has sides {2,2,4,4,9,9},
die B has sides {1,1,6,6,8,8}, and
die C has sides {3,3,5,5,7,7}.

Then:

the probability that A rolls a higher number than B is 5/9 (55.55 %)
the probability that B rolls a higher number than C is 5/9
the probability that C rolls a higher number than A is 5/9

Thus A is more likely to roll a higher number than B, B is more likely to roll a higher number than C, and C is more likely to roll a higher number than A. This shows that the relation "is more likely to roll a higher number" is not transitive with these dice, and so we say this is a set of nontransitive dice.

Efron's dice

Efron's dice are a set of four nontransitive 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

Probabilities

Each die can be beaten by another with a probability of 2/3:

P(A>B)=P(B>C)=P(C>D)=P(D>A)={2 \over 3}

B's value is constant; A beats it on 2/3 rolls because four of its six faces are higher.

Similarly, B beats C with a 2/3 probability because only two of C's faces are higher.

P(C>D) can be calculated by summing conditional probabilities for two events:

C rolls 6 (probability 1/3); wins regardless of D (probability 1)
C rolls 2 (probability 2/3); wins only if D rolls 1 (probability 1/2)

The total probability of win for C is therefore

\left({1 \over 3}\times 1\right)+\left({2 \over 3}\times {1 \over 2}\right)={2 \over 3}

With a similar calculation, the probability of D winning over A is

\left({1 \over 2}\times 1\right)+\left({1 \over 2}\times {1 \over 3}\right)={2 \over 3}

Best overall die

The probability of a randomly selected die beating another randomly selected die from the remaining 3 dice is not equal for all dice.

As proven above, die A beats B two thirds of the time but beats D only one third of the time.
The probability of die A beating C is 4/9 (A must roll 4 and C must roll 2)
So the likelihood of A beating any other randomly selected die is:

{1 \over 3}\times \left({2 \over 3}+{1 \over 3}+{4 \over 9}\right)={13 \over 27}

Similarly, die B beats C two thirds of the time but beats A only one third of the time.
The probability of die B beating D is 1/2 (only when D rolls 1)
So the likelihood of B beating any other randomly selected die is:

{1 \over 3}\times \left({2 \over 3}+{1 \over 3}+{1 \over 2}\right)={1 \over 2}

Die C beats D two thirds of the time but beats B only one third of the time.
The probability of die C beating A is 5/9
So the likelihood of C beating any other randomly selected die is:

{1 \over 3}\times \left({2 \over 3}+{1 \over 3}+{5 \over 9}\right)={14 \over 27}

Finally, die D beats A two thirds of the time but beats C only one third of the time.
The probability of die D beating B is 1/2 (only when D rolls 5)
So the likelihood of D beating any other randomly selected die is:

{1 \over 3}\times \left({2 \over 3}+{1 \over 3}+{1 \over 2}\right)={1 \over 2}

Therefore the best overall die is C with a probability of winning any random game of 0.5185. In this case, this increased chance is reflected by comparing the sums the numbers on every face of each die, but if you change the number on die B to 100, the 4 on A to 101, the 5 on D to 102 and the 6 on C to 103, the relative strength of the dice are unchanged with C as the most likely winner, but the highest average result will be the B die.

Numbered 1 through 24 dice

A set of four dice using all of the numbers 1 through 24 can be made to be non transitive. With adjacent pairs, one die will win approximately 2 out of 3 times.

For rolling high number, B beats A, C beats B, D beats C, A beats D.

A: 1, 2, 16, 17, 18, 19
B: 3, 4, 5, 20, 21, 22
C: 6, 7, 8, 9, 23, 24
D: 10, 11, 12, 13, 14, 15

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 17:16, equal: 3/36
the probability that IV rolls a higher number than V is 17:16, equal: 3/36
the probability that V rolls a higher number than III is 17:16, equal: 3/36

More about them at Miwins dice and www.miwin.com (German).

References

Gardner, Martin. The Colossal Book of Mathematics: Classic Puzzles, Paradoxes, and Problems: Number Theory, Algebra, Geometry, Probability, Topology, Game Theory, Infinity, and Other Topics of Recreational Mathematics. 1st ed. New York: W. W. Norton & Company, 2001. 286-311.

External links