# Nontransitive dice - what happens if all three are played?

**URL:** <https://boards.straightdope.com/t/nontransitive-dice-what-happens-if-all-three-are-played/760289>\
**Category:** Factual Questions\
**Created:** [July 19, 2016, 6:47am UTC](https://boards.straightdope.com/t/nontransitive-dice-what-happens-if-all-three-are-played/760289 "2016-07-19T06:47:51Z")\
**Posts on this page:** 3\
**Page:** 1

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**Author:** ![Mangetout](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/mangetout/32/19_2.png) [@Mangetout](https://boards.straightdope.com/u/Mangetout)\
**Post date:** [July 19, 2016, 6:47am UTC](https://boards.straightdope.com/t/nontransitive-dice-what-happens-if-all-three-are-played/760289/1 "2016-07-19T06:47:51Z")

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Didn’t want to hijack [this](http://boards.straightdope.com/sdmb/showthread.php?t=798713)thread with what might be a simple and possibly stupid question:

Nontransitive dice are (in the common example) dice that are so configured that, in a game of multiple rolls, A will tend to beat B; B will tend to beat C; but C will tend to beat A. ([linky](https://en.wikipedia.org/wiki/Nontransitive_dice))

Now, I know it’s not magic - it’s statistics, so I am not expecting answers like ‘the universe implodes’, but all of the discussion seems to be about two player games using only two of the three dice.

What happens (game-wise), if three players play? One die each. They obviously can’t all beat each other.

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**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [July 19, 2016, 7:58am UTC](https://boards.straightdope.com/t/nontransitive-dice-what-happens-if-all-three-are-played/760289/2 "2016-07-19T07:58:08Z")

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You can work it out yourself easily (albeit tediously) enough. One well-known set of nontransitive dice has player A’s die equally likely to be 1, 6, or 8; player B: 2, 4, or 9; and player C: 3, 5, or 7.

This gives 3 \* 3 \* 3 equally likely possibilities:

[ul]  
[li]123 (A \< B \< C)[/li][li]125 (A \< B \< C)[/li][li]127 (A \< B \< C)[/li][li]143 (A \< C \< B)[/li][li]145 (A \< B \< C)[/li][li]147 (A \< B \< C)[/li][li]193 (A \< C \< B)[/li][li]195 (A \< C \< B)[/li][li]197 (A \< C \< B)[/li][li]623 (B \< C \< A)[/li][li]625 (B \< C \< A)[/li][li]627 (B \< A \< C)[/li][li]643 (C \< B \< A)[/li][li]645 (B \< C \< A)[/li][li]647 (B \< A \< C)[/li][li]693 (C \< A \< B)[/li][li]695 (C \< A \< B)[/li][li]697 (A \< C \< B)[/li][li]823 (B \< C \< A)[/li][li]825 (B \< C \< A)[/li][li]827 (B \< C \< A)[/li][li]843 (C \< B \< A)[/li][li]845 (B \< C \< A)[/li][li]847 (B \< C \< A)[/li][li]893 (C \< A \< B)[/li][li]895 (C \< A \< B)[/li][li]897 (C \< A \< B)[/li][/ul]

So, out of 27 equally likely possibilities, we have the following frequencies:  
[ul]  
[li]B \< C \< A: 8[/li][li]A \< B \< C: 5[/li][li]A \< C \< B: 5[/li][li]C \< A \< B: 5[/li][li]C \< B \< A: 2[/li][li]B \< A \< C: 2[/li][/ul]

Of course, with other specific sets of nontransitive dice, you can get other specific distributions.

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**Author:** ![hammos1](https://avatars.discourse-cdn.com/v4/letter/h/ba8739/32.png) [@hammos1](https://boards.straightdope.com/u/hammos1)\
**Post date:** [July 19, 2016, 9:11am UTC](https://boards.straightdope.com/t/nontransitive-dice-what-happens-if-all-three-are-played/760289/3 "2016-07-19T09:11:40Z")

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Efron’s dice are a set of four dice with non-transitive properties. The sides on the dice are as follows:

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

So A beats B 2/3 of the time, B similarly beats C, C beats D, and D beats A.

You can work out what happens when all four dice are rolled together quite easily. For example, for D to win, it has to come up with a 5 (which happens 1/2 the time), and dice C has to come up with a 2 (which it does 2/3 of the time). So D will win overall 1/2 \* 2/3 = 1/3 of the time. Similarly:

C wins 1/3 of time  
B wins 1/2 \* 2/3 \* 1/3 = 1/9 of time  
A wins 2/3 \* 2/3 \* 1/2 = 2/9 of time