Football:What is the minimum number of points to guarantee qualification in a group of 4 teams?

Question

In a group of 4 teams, each team plays 3 matches ( against the 3 other teams). A win gives a team 3 points, a draw 1 point, and a loss 0 points. In the end the top two teams of the group qualify.

It is clear that the minimum number of points that guarantees a team the qualification is 6. The issue is showing this mathematically and without playing around with words and intuition.

Answer 1

It is not true that 6 points guarantees qualification. Consider the scenario where three teams achieve 2 wins and 1 loss and the fourth team loses all their games:

Team A beats teams B and D, loses to C

Team B beats teams C and D, loses to A

Team C beats teams A and D, loses to B

Team D loses to teams A,B,C.

Then teams A, B, and C all have 6 points but there are a maximum of 2 spots.

It is clear to see that if you get 7 points, you will go through- simply table the possible forced results and list all ways to fill in the remaining results, or note that the largest possible number of points to be had from 6 games is 18, and if one is to be in third place on points, the teams in first and second both have at least as many points. So this would lead to 21 points being had, which is a contradiction.

Ahh, and it appears that quid beat me to the punch. In the spirit of the tournaments going on now, fair play to him...

Answer 2

It is clear that the minimum number of points that guarantees a team the qualification is 6.

This is not true. It is possible that three teams each have $6$ points, leaving a team with $6$ points not qualified: A,B,C all beat D, and A beats B, and B beats C, C beats A. Then A,B,C all have two victories and thus $6$ points.

The minimal number of points that guarantees qualification is $7$.

To see this suffices, note that the total number of points is at most $3$ times the number of games, that is $3 \times 6= 18$. Suppose a team has $7$ points, than it is not possible that it is at position three or worse, as the two teams ahead of it have at least as many points each. So we have a total of at least $3 \times 7 = 21$ points, which is impossible.

Tangentially, the minimal number of points that can suffice to qualify is $2$. When $A$ beats $B,C,D$ and $B,C,D$ among them have all draws.