If there are $m$ teams playing a transitive game where the better side wins every time, then we can find the best team in $m-1$ games (either play a binary-tree knockout style tournament, or just a "winner stays on" style tournament). However, there is an issue in that the results of such a tournament are not reliable for finding 2nd place - the second best team could, by chance, have been paired against the best team in the first round and lost.
We could find the second best tournament with a "double-elimination" style tournament (essentially play the same tournament again with the $m-1$ losers), giving $2m-3$ total matches, but I'm wondering if there are any more efficient tournament structures. It seems like this tournament wastes some information as we also find the best team by doing this, so there could be room to save a game or two worth of information gathering.
I suppose the general question would be:
Given a transitive game played by 2 teams, what is the fewest possible games necessary to find the $nth$ best team out of $m$ entrants?
For second place, it must be one of the teams that lost to first place, so have an elimination tournament among them. The winning team played $\log_2 m$ matches so you need $\log_2 m -1$ games after the first $m-1$ for a total of $m+\log_2 m -2$. Third place must then be one of the teams that lost to either first or second. As we don't know how many games the second place team played we don't know how many games are required. The maximum would be if the second place team is the one that lost the final. Then each of the first and second place teams won $\log_2 m -1$ games against other teams in determining first place. The teams that lost to the first place team in the original tournament have competed in the tournament for second place, so we only need the $\log_2 m-1$ teams that lost to the second place team in the first tournament and the $\log_2(\log_2 (m-1))$ that lost to the second place team in the second tournament to enter the tournament for third.