If 9 people play against each other twice for one point each, what is the minimum number of points you would need to be guaranteed in the top 6? If 9 people were going to play a game against each other, two times each (each game would be 1 point), how many points would an individual person need to get to be guaranteed to be in the top 6?
I am trying to figure out the magic number for a tournament at work and we haven't been able to come to a conclusion so I'm turning to you.
 A: A score of $10\;$or more guarantees placement in the top $6$ (allowing ties for $6$-th place).

To see this, suppose player $A$ has a score of at least $10.\;$If player $A$ was not in the top $6$, there would have to be $6$ other players, each having a score of $11$ or more.$\;$But then the total of the scores would be at least $(6{\,\times\,}11) + 10 = 76$, which is impossible, since the number of games played is $72$.

However, if player $A$ has a score of $9$, it's possible for $6$ other players to have scores of $10$, in which case, $A$ will not place in the top $6$. 

To see this, label the players as $1,2,3,4,5,6,A,B,C$. 

We will show a tournament for which the scores are: 


*

*$10\;$for players $1,..,6$.$\\[5pt]$

*$9\;$for player $A$.$\\[5pt]$

*$3\;$for player $B$.$\\[5pt]$

*$0\;$for player $C$.$\\[5pt]$


The above scores can be realized as follows . . .


*

*Player $C$ loses all games played, so player $C$ gets a score of$\;0$, and each of the other players gets $2$ points from player $C$.$\\[8pt]$

*Let players $1,...,6$ tie against each other, tie with player $A$, and beat players $B$ and $C$ on both tries, so they each get scores of $5 + 1 + 2 + 2 = 10$.$\\[8pt]$

*Let player $A$ tie against all other players except player $C$.$\;$Then the score for player $A$ is $7 + 2 = 9$.$\\[8pt]$

*It follows that player $B\;$has a score of $0 + 1 + 2 = 3$.$\\[8pt]$

