Sumo Wrestler partial round robin tournament A coworker of mine is a big fan of Sumo. He recently came to me with a problem he's been wondering about for years:
In a professional sumo tournament there are 42 wrestlers. A tournament lasts 15 days. Each wrestler fights once per day. Each bout is with a different opponent. There are no repeat matches. Each bout results in either a win or a loss, there are no ties. Each wrestler will finish with a win/loss record that totals 15 (e.g., 15-0, 0-15, 8-7, 7-8).
What is the maximum number of wrestlers out of the 42 that can finish with a winning record (8-7 or better)? What is the maximum number that can finish with a perfect 15-0 record?
For the purpose of this problem we'll ignore differences in skill and assume that all wrestlers are equal to each other. In reality the wrestler ranked #1 is almost certainly going to beat #42 every time. To keep it simple we'll treat matches as if they were random coin tosses. Additionally the exact pairings can be random. In a real tournament wrestlers only fight other wrestlers around their own rank; #1 would never have to suffer the indignity of fighting #42. Since we're ignoring skill that fact can also be overlooked. We must also assume no wrestler drops out of the tournament before the 15 days are finished.
I'm fairly certain that the answer to the second part of his questions is n/2 or 21. If after the first day all of the winners keep fighting losers and keep winning then I believe the tournament will be over before all matches are exhausted. I'm not sure how to prove this though.
As for the first part of the problem I'm not even sure where to begin. I was able to find some information about efficiently scheduling partial round robin tournaments but nothing about characterizing their outcomes.
 A: Part 1:
We split the team into 3 groups of 13 and one group of 3, respectively A, B, C and D.
In each of the first 3 groups each player competes against all the other players, so each player has 12 competitions. We make sure that each player has 6 wins and 6 losses.
Then Ai plays against Bi and loses, Ai plays against Ci and wins and Bi Plays against Ci and loses. Now each competitor has 7 wins and 7 loses.
All of Group A wins against 1 of Group D, All of Group B wins against 2 of group D and all of group C wins against 3 of Group D. We now have 39 people with 8 wins and 7 loses.
Each player in Group D plays against the others and get to 15 competes (they had 13 earlier).
It cannot be more than 39, because if we have 40 then for those 40 there will be 40 more wins than losses and the remaining 2 players cannot have enough losses.
Part 2:
It is clear that the number of players with 15 or more wins is limited to 21, because if there are 22 players who won 15 matches then we need 22 players who have lost at least one match.
The way to prove that it is 21 is by splitting the group into 21 As and 21 B, i.e A1...A21 and B1...B21.
Ai competes against Bi in the first round and against B(i+j-1)mod21 in the j round. This would cause a repeat only after more than 21 rounds.
All the 21 As have 15 wins
A: Your answer to the second question looks correct, so long as $\frac{42}{2} \ge 15$, which is indeed true.
For the first question, you might consider trying to have $w$ fighters with results of $8-7$, and $42-w$ fighters with results $0-15$.  That would require  $w \le 15(42-w)$, leading to $w \le 39.375$.  So it looks as if the maximum number of $8-7$ winners is $39$, leaving the other three fighters $21\times 15  - 8 \times 39 =3$ wins and so $42$ losses between them.
This looks eminently possible. Have the last three fight each other (it does not matter what happens in these three fights, but it will involve $3$ wins and $3$ losses) and have them lose all their fights across with the set of $39$ (divide that group into three subsets of $13$, each subset figting an individual loser), while having each of the $39$ have a $7-7$ record within that set: for example arrange them in a circle so they win against the $7$ in front of them and lose to the $7$ behind them. 
