What is the most non-transitive set of 5 dice I've seen some of the other questions on non-transitive dice, but none of them answered the question: For what set of 5 6-sided dice is there the most non-transitivity? I define "most non-transitive" to mean maximizing the minimal probability in the chain A>B>C>D>E>A, and after that, maximising the second lowest probability, then the third, etc. Also, I want to know how many dice are needed to extend non-transitivity to three people, so for any two dice they pick, you can pick a die that beats both of them at the same time with probability greater than 50%?
To summarize:
Question 1: For what set of 5 6-sided dice is there the most non-transitivity?
Question 2: How many dice are needed such for any two dice they pick, you can pick a die that beats both of them at the same time with probability greater than 50%?
 A: Answer to Question 2:

How many dice are needed such for any two dice they pick, you can pick
a die that beats both of them at the same time with probability
greater than 50%? probability:

Source for info:
See
https://en.wikipedia.org/wiki/Intransitive_dice#Three_players for more info. It's not entirely up to date - there are new developments it doesn't mention - but the info it has should otherwise be correct.
I don't know if it's been proved what the lowest number is. With some extra rules, it can be done with 5 dice (though the 3rd player also gets to choose whether all the players roll their die once, or twice (or equivalently, get 2 of the same die, that they chose). Without extra rules, it can be done with a set of 7 dice. (Where each player chooses 1 die.)

Oskar dice[edit] Oskar van Deventer introduced a set of seven dice
(all faces with probability  1 / 6 ) as follows:[7]A: 2, 2, 14, 14, 17, 17
B: 7, 7, 10, 10, 16, 16
C: 5, 5, 13, 13, 15, 15
D: 3, 3, 9, 9, 21, 21
E: 1, 1, 12, 12, 20, 20
F: 6, 6, 8, 8, 19, 19
G: 4, 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.

