I remember years ago at a US Amateur Team where the split first round had an odd number of teams. What was done was to take the bottom 3 teams and pair them against each other. So you had teams b1, b2, & b3 with a very interesting pairing. This was to have Top b1 play top b2, second b1 to play second b2, third b1 to play first b3, fourth b1 to play second b3, third b2 to play third b3, & fourth b2 to play fourth b3. The resulting scores of each game were added up to determine the scores for each team for that round.
I was wondering if anyone had figured a way to avoid byes at a team only scholastic event. Especially one with either weighted board scores and/or an odd number of players on the team? I thought it might be useful for a final round so players did not have to sit out and wait maybe for a second team from the same school to finish.
I have a way that has been used many times. Tell the coaches that the last school to enter multiple teams will be the one that gets all of their teams if that would make it even, and all but one of their teams otherwise.
At an elementary school team tournament that I directed last spring, the advance entry list had an odd number of teams. I told the organizer to find out if any school had requested to have a second team and to take the first team to reply.
At the Pittsburgh Chess League, we have three divisions. The first two divisions have 8 teams, so they play 7 game round robins. The third section is a 5 round Swiss System qualifier event which leads into a final set of quads. If there is an odd number of teams at the start of the year, Division I has one team dropped, so that division plays 6 rounds rather than 7; every team in that division has a bye month. The third division must be even for the purposes of the Swiss System to avoid team byes.
I proposed a similar scheme decades ago in the Illinois Chess Bulletin. I called it a “trye”. Instead of giving 1 team a bye, you give 3 teams a trye, and every player gets a game.
My scheme was, if I do say so myself, a bit more elegant than the above. Top b1 vs top b2, second b1 vs top b3, second b2 vs second b3. And similarly (or perhaps in mirror image) for third and fourth boards. This way, there are no top-vs-third or second-vs-fourth pairings, only top-vs-second and third-vs-fourth.
Even more symmetric (3-way symmetry) would be top b1 vs second b2, top b2 vs second b3, top b3 vs second b1, and similarly (or in mirror image) for third vs fourth.
A little elementary logic will show that this is impossible. With an odd number of teams, each with an odd number of players, the total number of individual players in the event will be odd times odd, which is always odd. So there will always be one player sitting out.