Combinatorics of a Tournament 
$8$ people participate in a tournament, so that each person plays with all the other people once. If a person wins against another, then the winner gets $2$ points, while the losing team will get none. If they tie, they each get $1$ point, respectively. When finished, the people are ranked depending on how many points they have in total. How many points does a person need to have, to secure their spot in the best four players.

I know that the maximum possible amount of points one can achieve is $14.$ I also know a score of $12$ gurantees you to be in the top $5.$ I'm not really sure how to continue with this information, how would I do this problem?
 A: Note that there are a total of $56 = 2\binom{8}{2}$ total points given out in the tournament. If anyone gets at least $11$ points, then they are guaranteed to be in the top 4. Suppose for contradtiction otherwise, that there are at least $5$ people who earn $11$ points. Then that would mean there's only  $56 - 5\times 11 = 1$ point left over that could have been won by one of the 3 remaining players. This is impossible though because these players also play against each other, and so within these $3$ players there must have been $6 = 2\binom{3}{2}$ points allocated from these games.
Now we show that getting $10$ points isn't enough to be in the top 4 (assuming arbitrary tie-breaking). We show that there is a scenario where there are five players who get $10$ points each. Label the players 1 through 8.

*

*Player 1 beats 2, 3, 6, 7, 8

*Player 2 beats 3, 4, 6, 7, 8

*Player 3 beats 4, 5, 6, 7, 8

*Player 4 beats 5, 1, 6, 7, 8

*Player 5 beats 1, 2, 6, 7, 8

*Player 6 beats 7

*Player 7 beats 8

*Player 8 beats 6

This scenario works.
