If each of 7 teams played 3 times, that would be a total of 21 games (7.3) if they didn't play each other. I believe your question intends for the teams to play each other. In this case, we would divide by 2 (since team 1 playing team 2 would be a single game, not two games). Therefore we have 21/2=10.5 games, which is clearly impossible. Most tourneys are a minimum of 3 game guarantees. I don't think teams are going to pay very much for a two game guarantee tourney. Three game guarantees would be one of the following: 1. Pool play with three teams in a pool, play 2, followed by a single elimination tourney where everyone makes the single elimination, seeded on pool play records. 6 Team 3 Game Guarantee Bracket. Fill out, securely sign, print or email your 8 team pool play 3 game guarantee form instantly with SignNow. The most secure digital platform to get legally binding, electronically signed documents in just a few seconds. Available for PC, iOS and Android. Start a free trial now to save yourself time and money!
Nine teams is a lousy number to have. For initial play you must either make two unequal pools of 4 and 5, or play a nine-team round-robin over two days. Actually, this latter alternative, commonly overlooked, is a pretty good format. It's a great way to get lots of games for everyone, if each team is willing to make the committment, ahead of time, to finish the schedule. If you have time for five rounds each day, you can schedule each team for four games and one bye. See the schedule, Table 9.1.3.Note, the round-robin is not an acceptable format if less than three teams are qualifying, because of the three-way tie for first place possibility. See Section 9.6 for more details about this option.
Three pools of three is never acceptable. Download wifi hacker for android 100 working. Simply too much depends upon seeding in such a format.
On the second day, the top four teams in each pool play an eight-team single elimination. (Note, the brackets and the initial pairings for the single elimination is the same as the upper half of the 8-team double elimination format, see Table 8.1.1).
Note, it is not acceptable to take only the top two teams from each pool and have semi's and finals, because of the possibility of a three-way tie for first in one of the pools.
On the second day, the top team in each pool play each other. The winner takes first place. The loser still has another game to play. The second and third teams in each pool play a four-team loser's bracket; i.e., a semis and a final. This takes two rounds. The winner of this bracket, then, plays the loser of the game for first place. The winner of this game takes second place. The brackets for this second day are shown in Table 6.5.
Start with two unequal pools of four and five (see Section 9.5).
On the second day, the top four teams in each pool play an eight-team double elimination. (See Table 8.1.1).
This involves an awful lot of games for some of the teams. It is possible that one or two teams will play five games the second day. Note, too, that one of the pools will have played four games on the first day. (In women's play, the first round of the double elimination could be an abbreviated game to 11, (since no one gets eliminated after only one game). Otherwise, all of the games should be to 13 (except for the winner's bracket final, which is to 15).
On the second day, the top three teams in each pools play a six-team modified double elimination (see Table 6.4).
On the second day, the top four teams in each pool play an eight-team double elimination. (See Table 8.1.1).
This involves an awful lot of games for some of the teams. It is possible that one or two teams will play five games the second day. Note, too, that one of the pools will have played four games on the first day. (In women's play, the first round of the double elimination could be an abbreviated game to 11, (since no one gets eliminated after only one game). Otherwise, all of the games should be to 13 (except for the winner's bracket final, which is to 15).
To ease things, you might consider eliminating the very last round of the double elimination. The last round is for second and third place, which may not be very relevant if four or more teams are qualifying anyway. If this game is not played, use the tie-breakers for double elimination (Section 2.6). In the Mid-Atlantic, a format similar to this was used, and this game was optional. I.e., the team that wanted to play was given second place; if both wanted to, they played; if neither wanted too, or it was too late, we used the tie-breakers.
Certainly it is unfair that all the teams in Pool A have to play one more game than those in Pool B, but there is no alternative, other than to have a round-robin of nine teams, or to have three pools of three. A mitigating factor, however, is that the two weakest teams are in Pool A. Furthermore it is acceptable to play to Pool A games to two fewer points than Pool B games (i.e., play Pool A games to 13 and Pool B to 15, or, for women, 11 and 13).
Another potential problem is that Pool B's games require only three rounds, and Pool A's games require five. One can, however, stretch Pool B's games over four rounds, using a schedule like the following:
When a team does drop out, the only fair course of action is to not count any game in which the offending team has played. This will surely elate those teams which now get to remove a loss from their record (and frustrate those teams which must remove a win from their record), but the alternative, giving a win-by-forfeit to any team that has not played the team that has dropped out, is worse. This is unfair to any team which has lost to the offending team. And for teams getting a win-by-forfeit, what should the score be? Suppose a team getting a win is involved in a tie-breaker situation and the score of the win-by-forfeit comes into consideration?
Lastly, it should be restated that it is official UPA Policy that if a team does not finish the tournament (i.e., drops out and loses by forfieting), they can not qualify for the next event. So, for example, if a team at sectionals is 5-1, then drops out, thinking that they will finish at 5-3 and qualify for regionals, forget it. None of their games count. They aren't even 0-8, they are 0-0, and it's almost as if they didn't even show up to the tournament. (Their appearance will count, however, for the purposes of dues, memberships, and 'years of qualification,' etc.).
A round-robin of six teams must be spread out over two days. Play four rounds the first day, and the fifth game as the first round on the second day. Because no team has a bye during the round-robin, no team has an unfair advantage for the games that follow the round-robin. For some of the formats, where there are lots of games, you might consider making the round-robin games to 13.
Commonly, tournament schedulers like to pit the top two seeds against each other during the last round of a round-robin. This is not a good idea, however, when the round-robin is spread over two days because then the top two seeds might be playing each other twice on the second day. Make sure that these two teams play each other on the first day.
Be sensitive to the fact that some teams might be sticking around overnight only to play in a single round-robin game when they have already been eliminated on the first day. Such a team may not show up, and this creates all sorts of problems. One way to avoid this is to have a leisurly start on the first day and only play three round. Then, the second day involves two games of round-robin, and two rounds to finish the four-team single elimination. Another possibility, is to use the option below.
There are two advantages to this format. First, a team can only get eliminated by losing a game (i.e., there is no possiblity of a team going home an account of a point-differential). Second, a team can lose all four games on the first day and still have an incentive to show up the second day, because, theoretically, they can still win the tournament.