Intransitive dice#Efron's dice

{{Short description|Game variations and descriptions of intransitive die and their behaviour}}

{{Multiple issues|

{{More citations needed|date=May 2024}}

{{Original research|date=May 2024}}

}}

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 – {{math|X}} rolls a higher number than {{math|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).{{cite web|last1=Weisstein|first1=Eric W.|author-link=Eric W. Weisstein|title=Efron's Dice|url=https://mathworld.wolfram.com/EfronsDice.html|access-date=12 January 2021|website=Wolfram MathWorld|language=en}}{{cite web|last=Bogomolny|first=Alexander|author-link=Alexander Bogomolny|title=Non-transitive Dice|url=https://www.cut-the-knot.org/Probability/NonTransitiveDice.shtml|url-status=live|archive-url=https://web.archive.org/web/20160112225243/https://www.cut-the-knot.org/Probability/NonTransitiveDice.shtml|archive-date=2016-01-12|website=Cut the Knot}}{{cite journal |last1=Savage |first1=Richard P. |title=The Paradox of Nontransitive Dice |journal=The American Mathematical Monthly |date=May 1994 |volume=101 |issue=5 |pages=429–436 |doi=10.2307/2974903 |jstor=2974903 |url=https://www.jstor.org/stable/2974903}}{{cite journal |last1=Rump |first1=Christopher M. |title=Strategies for Rolling the Efron Dice |journal=Mathematics Magazine |date=June 2001 |volume=74 |issue=3 |pages=212–216 |doi=10.2307/2690722 |jstor=2690722 |url=https://www.jstor.org/stable/2690722 |access-date=12 January 2021}}

Example

Image:Intransitive dice 2.svg

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 {{sfrac|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.

  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.

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 {{sfrac|5|9}}. The following tables show all possible outcomes for all three pairs of dice.

class="wikitable" style="border:none; text-align:center;margin:0 auto;"

| colspan="4" style="border:none;"|Player 1 chooses die A
Player 2 chooses die C

rowspan="5" style="border:none;width:2em;"|

| colspan="4" style="border:none;"|Player 1 chooses die B
Player 2 chooses die A

rowspan="5" style="border:none;width:2em;"|

| colspan="4" style="border:none;"|Player 1 chooses die C
Player 2 chooses die B

{{diagonal split header|C|A}}249

! {{diagonal split header|A|B}}

168

! {{diagonal split header|B|C}}

357
3

| style="background:#ffc;"|C || A || A

! 2

| style="background:#ffc;"|A || B || B

! 1

| C || C || C

5

| style="background:#ffc;"|C || style="background:#ffc;"|C || A

! 4

| style="background:#ffc;"|A || B || B

! 6

| style="background:#ffc;"|B || style="background:#ffc;"|B || C

7

| style="background:#ffc;"|C || style="background:#ffc;"|C || A

! 9

| style="background:#ffc;"|A || style="background:#ffc;"|A || style="background:#ffc;"|A

! 8

| style="background:#ffc;"|B || style="background:#ffc;"|B || style="background:#ffc;"|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.{{cite journal |last1=Trybuła |first1=Stanisław |title=On the paradox of three random variables |journal=Applicationes Mathematicae |date=1961 |volume=4 |issue=5 |pages=321-332 |url=https://eudml.org/doc/264121}}

Variations

=Efron's dice=

Efron's dice are a set of four intransitive dice invented by Bradley Efron.

Image:Efron dice 2.svg

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 {{sfrac|2|3}}. C beats A with probability {{sfrac|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 {{nowrap|0 ≤ x ≤ {{sfrac|3|7}}}}:

:P(choose A) = x

:P(choose B) = {{sfrac|1|2}} - {{sfrac|5|6}}x

:P(choose C) = x

:P(choose D) = {{sfrac|1|2}} - {{sfrac|7|6}}x

=Miwin's dice=

Image:Miwin Wuerfel Titan.gif

{{main|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 {{sfrac|17|36}}
  • the probability that IV rolls a higher number than V is {{sfrac|17|36}}
  • the probability that V rolls a higher number than III is {{sfrac|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 {{sfrac|1|6}}) as follows:{{cite web|last=Pegg|first=Ed Jr.|author-link=Ed Pegg Jr.|date=2005-07-11|title=Tournament Dice|url=http://www.mathpuzzle.com/MAA/39-Tournament%20Dice/mathgames_07_11_05.html|url-status=live|archive-url=https://web.archive.org/web/20050804234631/http://www.maa.org/editorial/mathgames/mathgames_07_11_05.html|archive-date=2005-08-04|access-date=2012-07-06|website=Math Games|publisher=Mathematical Association of America}}

  • A: 2, {{0}}2, 14, 14, 17, 17
  • B: 7, {{0}}7, 10, 10, 16, 16
  • C: 5, {{0}}5, 13, 13, 15, 15
  • D: 3, {{0}}3, {{0}}9, {{0}}9, 21, 21
  • E: 1, {{0}}1, 12, 12, 20, 20
  • F: 6, {{0}}6, {{0}}8, {{0}}8, 19, 19
  • G: 4, {{0}}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:{{cite web |last1=Grime |first1=James |title=Non-transitive Dice |url=http://grime.s3-website-eu-west-1.amazonaws.com/ |archive-url=https://web.archive.org/web/20160514125245/http://grime.s3-website-eu-west-1.amazonaws.com/ |archive-date=2016-05-14 |url-status=dead}}{{Cite journal |last=Pasciuto |first=Nicholas |date=2016 |title=The Mystery of the Non-Transitive Grime Dice |url=http://vc.bridgew.edu/undergrad_rev/vol12/iss1/18 |journal=Undergraduate Review |volume=12 |issue=1 |pages=107–115 |via=Bridgewater State University}}

  • 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"
align="center"

! colspan="2" rowspan="2" | Sets chosen
by opponents

! colspan="2" | Winning set of dice

align="center"

! Type

! Number

align="center"

| A

BE1
align="center"

| A

CE2
align="center"

| A

DC2
align="center"

| A

ED1
align="center"

| B

CA1
align="center"

| B

DA2
align="center"

| B

ED2
align="center"

| C

DB1
align="center"

| C

EB2
align="center"

| D

EC1

= Four players =

It has been proved that a four player set would require at least 19 dice.{{Cite journal|last1=Reid|first1=Kenneth|last2=McRae|first2=A.A.|last3=Hedetniemi|first3=S.M.|last4=Hedetniemi|first4=Stephen|date=2004-01-01|title=Domination and irredundance in tournaments|url=https://www.researchgate.net/publication/266258217|journal=The Australasian Journal of Combinatorics [electronic only]|volume=29}} In July 2024 GitHub user NGeorgescu published a set of 23 eleven sided dice which satisfy the constraints of the four player intransitive dice problem.{{Cite web |last=Georgescu |first=Nicholas |title=math_problems/intransitive.ipynb at main · NGeorgescu/math_problems |url=https://github.com/NGeorgescu/math_problems/blob/main/intransitive.ipynb |archive-url=https://web.archive.org/web/20250327230902/https://github.com/NGeorgescu/math_problems/blob/main/intransitive.ipynb |archive-date=27 March 2025 |access-date=2025-03-27 |website=GitHub |language=en}} The set has not been published in an academic journal or been peer-reviewed.

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) = {{sfrac|9|16}}

The following tables show all possible outcomes:

class="wikitable"

! {{diagonal split header|A|B}} !! 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.

class="wikitable"

! {{diagonal split header|B|C}} !! 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.

class="wikitable"

! {{diagonal split header|C|A}} !! 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) = {{sfrac|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:

border="1" class="wikitable"

|D III

purple125679101114151618
D IVred134589101213141718
D Vdark grey234678111213151617

Standard-Dodekaeder-D III.gif|D III

Standard-Dodekaeder-D IV.gif|D IV

Standard-Dodekaeder-D V.gif|D V

Set 2:

border="1" class="wikitable"

|D VI

cyan1234910111213141718
D VIIpear green12567891015161718
D VIIIlight grey345678111213141516

Standard-Dodekaeder-D VI.gif|D VI

Standard-Dodekaeder-D VII.gif|D VII

Standard-Dodekaeder-D VIII.gif|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.

border="1" class="wikitable"

|PD 11

grey to blue131729313743475367717383
PD 12grey to red131923294143475961677983
PD 13grey to green171923313741535961717379

Primzahlen-Dodekaeder-PD 11bf.gif|PD 11

Primzahlen-Dodekaeder-PD 12bf.gif|PD 12

Primzahlen-Dodekaeder-PD 13bf.gif|PD 13

Set 2: The numbers add up to 468.

border="1" class="wikitable"

|PD 1

olive to blue71119232937434753616771
PD 2teal to red71317193137414359616773
PD 3purple to green111317232931414753597173

Primzahlen-Dodekaeder-PD 1bf.gif|PD 1

Primzahlen-Dodekaeder-PD 2bf.gif|PD 2

Primzahlen-Dodekaeder-PD 3bf.gif|PD 3

Generalized Muñoz-Perera's intransitive dice

A generalization of sets of intransitive dice with N faces is possible.{{cite web|last1=Muñoz Perera|first1=Adrián|author-link=Adrián Muñoz Perera|title=A generalization of intransitive dice|url=https://pereradrian.github.io/doc/adrian_munnoz_perera_generalized_intransitive_dice_2024.pdf|access-date=15 December 2024|language=en}} 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 ==

border="1" class="wikitable"

|D_1

168
D_2249
D_3357

The set of dice obtained in this 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 ==

border="1" class="wikitable"

|D_1

181114
D_2251215
D_336916
D_4471013

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 ==

border="1" class="wikitable"

|D_1

11217222732
D_22718232833
D_33813242934
D_44914193035
D_551015202536
D_661116212631

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\}.

See also

References

{{reflist}}

Sources

{{refbegin}}

  • {{cite book|authorlink=Martin Gardner|last=Gardner|first=Martin|title=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|url=https://archive.org/details/colossalbookmath00gard|url-access=limited|edition=1st|location=New York|publisher=W. W. Norton & Company|date=2001|page=[https://archive.org/details/colossalbookmath00gard/page/n302 286]–311}}{{ISBN needed}}
  • {{cite book|title=Spielerische Mathematik mit Miwin'schen Würfeln|publisher=Bildungsverlag Lemberger|isbn=978-3-85221-531-0|language=German}}

{{refend}}