That is, all teams should have a fair shot at winning the tournament (and/or qualifying to the next level). No team should have to do something different in the schedule from the other teams because of how they are seeded. Now it may be that some pools are of different size than other pools (there just isn't any number that divides into thirteen!), but in this case an entire pool is playing more games than other pools. No team in the larger pool is doing anything different than any other team in that pool.
There have been some awfully "funky" formats for sectional tournaments before. I can recall one in which the top seeded team completely bypassed the first day of the tourney, and were automatically placed in a "winners" pool of the next day (i.e., different "divisions.") Such tournaments have their place, and may even be desirable in certain circumstances, but not for UPA Series events (unless teams have a choice of the division in which they would like to be placed).
Teams that travel from far away should be able to go home after a tournament feeling like they played their fair share of games, no matter where they finished. This means that teams should be eliminated from the tournament as late as possible, and that if they are eliminated early, there are 'Division II' options available to them. On the other hand,
Four games per day, games to 15, is about the most any team should be asked to play. (Limited exceptions for five games to 13). People who are into triathlons and marathons can go compete in those other competitions. Furthermore, these games should be scheduled in two-hour time blocks (with a cap of "plus two" after 1 hour and 40 minutes). Time caps are needed, not only for scheduling purposes, but so that a team does not have to play more than eight hours of Ultimate in a day. In those games where the wind is strong, both teams have good defenses, or neither team has a good offense, games to 15 can drag out for a long time. A succession of long, killer games often results in unwarranted injuries.
There are exceptions to this rule, but they should be limited to Regional and post-Regional play. For example:
Another set of exceptions come into play when a tournament is faced with an undesirable number of teams. For example, thirteen teams is just a difficult number to work with. Many of these situations are dealt with by eliminating the number of teams to eight teams after the first day, and then a double elimination format the second day. The double elimination format, however, results in five games, and so games can be no longer than to the score of thirteen.
Should this be in here? What is fair? Can anybody help out at all here?
Ideally, it shouldn't matter where or how teams are seeded. Indeed, when the entire tournament is in one round-robin pool, it doesn't matter how the teams are seeded. Attempts have been made, in every format in this manual, to minimize the effect of seeding. A team might get eliminated earlier or later because of where they were seeded, but whether or not the team qualifies for the next level tournament should not depend on seeding.
Anybody want to elaborate on 'general principles' of seeding here?
Six Teams 1 2 3 4 5 - - - - - 1 - 5 4 3 2 6 2 - 4 3 6 1 5 3 - 6 2 1 5 4 4 - 2 1 5 6 3 5 - 1 6 4 3 2 6 - 3 5 2 4 1In the schedule below, each row is a round, and each column can be a field. So, in the first round, 1 plays 5 on field one, etc.
1 v 5 2 v 4 3 v 6 1 v 4 2 v 3 5 v 6 1 v 3 2 v 6 4 v 5 1 v 2 3 v 5 4 v 6 1 v 6 2 v 5 3 v 4
Nevertheless, while the UPA tries to minimize such situations, they all can't be avoided. The following are the UPA Standard set of round-robin tie breaker rules.
Rule 1a. If, after the application of a given rule, all of the teams are still tied, go to the next rule. For an example of this application, see Example 3.3.
Rule 1b. If not all teams, but one or more subgroups of the teams are tied, separate these teams into groups and go back to rule 2 with each of the groups individually. For an example of this application, see Example 3.2.
Example 2.2. A, B, and C, are tied for first place; they are all 3-2 after the six team round-robin. A has beaten both B and C, while B has beaten C. The records among the three teams only are: A is 2-0, B is 1-1, and C is 0-2. A finishes first, B finishes second, and C finishes third.
Example 2.3. A, B, and C are in a three-way tie. A has beaten B, B has beaten C, C has beaten A. The relevant records for all three teams are 1-1. This tie-breaker won't work, and you must go on to tie-breaker #3.
Example 3.2. A, B, C are in a three-way tie for first place. A has beaten B 15-11, B has beaten C 15-12, and C has beaten A, 15-13. A's point differential, then, is +4 and -2, which equals +2. B's is -1 and C's is -1. A takes first place. B and C are still tied. When, after the application of a rule, there are still teams that are tied, we go back to rule 2. Since B beat C, B takes 2nd place, and C takes 3rd. At this point we do not go onto rule 4.
Example 3.3. A, B, C are in a three-way tie for first place. A has beaten B 15-13, B has beaten C 16-14, and C has beaten A, 15-13. A's point differential, then, is +2 and -2, which equals +0. B's is 0 and C's is 0. This tie-breaker can not be applied, go on to tie-breaker rule 4, unless there are only three teams in the pool to begin with, in which case you should have played an extra point. See the discussion on three-team pools in Section 3.1.
Rule 4a. Multiple games against common opponents are averaged.
Example 4.2. Assume all of example 1, but that, for some reason, B beat D twice, 15-7 and 15-12. Take the average of the scores and only count it once, thus, we would calculate the point differential as though B beat D once, by a score of 15-9.5. Then A takes first (point differential of +6), B takes second (we count, as a point differential, the score of 15-9.5, which is 5.5), and C takes third (beating D by three goals).
Strictly speaking, mathematically, the computation regarding losses should be subtracted. Here's an example:
A beats W 15-12 B beats W 15-8 A beats X 15-4 B beats X 15-8 A lost to Y 15-8 B lost to Y 15-12 A lost to Z 15-8 B lost to Z 15-4Suppose, further, that these were the only games involving common opponents of A and B, and that all the tie-breakers are the same through the first eight rules. Then, applying this rule, A's differentials are 3, 11, -7 and -7. The tie-breakers, then, are
A's tie breakers = sqrt(3) + sqrt(11) - sqrt(7) - sqrt(7) B's tie breakers = sqrt(7) + sqrt(7) - sqrt(11) - sqrt(3)A's tie breakers are approximately -0.2 while B's tie breakers are approximately +0.2.
Example 3.2. The same as above except that D's two losses were to B, 19-9 and 17-14, and D never played A. Averaging the scores would give a score of 18-11.5, or a point differential of 6.5. C's point differential is -6, and so C takes third place.
Example 3.3. The same as above except that D's two losses were to B and to E. The loss to B was 19-9, the loss to E was 17-14.
Should it be relevant if C has played E ? I don't know the answer, and so I am not finishing this example.