Nontransitive dice - what happens if all three are played?

Factual Questions
1

Didn’t want to hijack thisthread 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)

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.

2

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.

3

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