from the set $\{a, b, c, d\}$?
Of the one's I have tried, it at best is two of the three, but never all.
from the set $\{a, b, c, d\}$?
Of the one's I have tried, it at best is two of the three, but never all.
On
If non-transitive means "never transitive for any triple", that is impossible for a complete tournament (irreflexive, never-symmetric relation) with $4$ or more players. The question sounds like an exercise of rediscovering that fact.
If partial tournaments are allowed then it can be done with any number of players, just partition them into 3 categories A,B,C, and have every A player always beat the B players who always beat C who always beat A, and no matches within any category.
Consider the ordered pair, such that $\{(x,y)\,|\,x,y\in\{a,b,c,d\}\}$. The following relation satisfies those conditions:
$$\{(a,b) ,(b,c), (c,d), (d,a)\}$$
Clearly, this relation is not reflexive since there is no ordered pair with same members i.e. $(x,x)$. This relation is anti-symmetric since for instance, there is no ordered pair $(b,a)$. This relation is also not transitive (which is left for you to work out).