Soccer Tournament

Let $\ell$ denote the total score of losing team and $t_1, t_2, t_3$ the total scores for other three teams.

Since $\ell$ team lost all matches, then $0 \leq \ell \leq (4 - 1)(4 - 1) = 9$ where $9$ is the maximum total score the losing team can get.

---

For $t$ teams, since we want their score sums to be less than $\ell$ , the interval of all possible $t$ values is $0 \leq \text{score sum of }t\text{ teams} \leq 8 < 9$ Since each of $t$ teams earns $4$ points from beating $\ell$ team, $0 \leq \text{score sum of other two matches} \leq 8 - 4 = 4$ Carefully observe the graph with black segments. Since there must be a winner for each match, then one of the teams automatically earns $4$ points, which forces the point value for another match against different team to be $0$ . However, since the graph with nodes $t_1$ , $t_2$ and $t_3$ is a cycle graph, then each of these teams earns $4$ points in the same manner. Thus, they earn $8$ points in all, which is less than the maximum total score of the losing team.

So the losing team's total score is $9$ points, whereas the three team's total scores are each $8$ points. So the losing team can $\text{win the tournament}$ .

simply possible.

as,there are 4 teams and each team will face every other team exactly one match , so, there will be totally 6 matches.

suppose, four teams are-A(won their 3 matches out of 3 ), B(won their 2 matches out of 3) ,C(won their 1 match out of 3) and D(didn't win any match or the looser)

now, if A win 3 matches they will get at least $4+4+4=12$ points.

and if B win 2 matches they will get at least $4+4=8$ points

and if C win 1 match they will get at least $4$ points.

if they all loose their other matches by nil ,their point will not increase or remain the same.

now, if D(the looser) can score at least 3 points in a match and lose,after 6 matches ,their score will be= $3 \times 6=18$ , the most score in the tournament.

so, by loosing all the matches, it is still possible for the loosing team to win the tournament.

Yes, by example:

• A:3 B:4
• A:3 C:4
• A:3 D:4
• B:4 C:0
• B:0 D:4
• C:4 D:0

So A wins even though they lost every game.

It would be helpful to explain what a 4-point game is, though. It can be interpreted as there are a total of 4 points, so the possible outcomes are 4-0, 3-1, or 2-2. In that case, there would be no possibility of losing all three games and still winning the tournament.