Some years ago we were setting up rules for a team tournament. Winning team would get a promotion to a higher league. The following case was considered:

Teams consist three players. Before last round there is a three way tie in the top. Team A has beaten team B, Team B beat Team C and Team C beat team A. We ended in doing nothing special and letting them play against lower teams and use the usual team tie breakers.

My idea was to play a following tie break:
Players in teams are numbered 1,2,3. Teams A,B,C are in order by other tie breakers.
Play games:
A1 vs. B2
B1 vs. C2
C1 vs. A2
A3 vs. B3
C3 free
Winning two games gives team a win in the tournament.

The idea was rejected as not following the spirit of a team tournament. What would you think?

What if two teams both win two games?

That means that team C must win both its games but teams A and B only need to win two out of three. So this inherently unfair to team C.

_________________
Still officially AGA 5d but I play so irregularly these days that I am probably only 3d or 4d over the board (but hopefully still 5d in terms of knowledge, theory and the ability to contribute).

The treatment of the 3's seems to give an asymmetrical result. Don't you have to give all three a bye? Alternatively play three Micky Mouse (sorry, blitz) games among them while the bigs are playing with slower time controls? The trouble is that any truly fair system would seem to run the risk of also ending in a tie.

_________________
Dave Sigaty
"Short-lived are both the praiser and the praised, and rememberer and the remembered..."
- Marcus Aurelius; Meditations, VIII 21

C was already the last with other tie breakers. This way it might get a better chance.

Quicker games are an option. Anyway, my question was whether you consider my suggestion as improvement over the regular system, just continuing the tournament and pairing all top three teams down against a new team.

In that case, no. The proposal anticipates a problem that might not exist after the final round. Also it already employs the normal tiebreakers to disadvantage one of the teams. This seems like a case where the cure is worse than the disease.

_________________
Dave Sigaty
"Short-lived are both the praiser and the praised, and rememberer and the remembered..."
- Marcus Aurelius; Meditations, VIII 21

I should have been a bit more specific, but it was many years ago. We specified separate systems for various number of teams. We had 5 rounds. With 5 or 6 teams the system was round robin. With 8 teams swiss. With 7 teams we had a problem. We might get into above mentioned scenario after 4 rounds. We would have three teams winning each other in a circle and winning against all other teams. The other four teams might all have had a bye. If we continue the fifth round with usual swiss, then one of the top three teams would get a bye. So the question is, whether it is better to assign weaker random opponents a bye or try to get maximal number of games against new playing opponents within the tied teams.

First, note that under the proposed solution if B loses all its games and C1 beats A2, both A and C end with two wins. We have eliminated B but failed to select a winner.

On the other hand, the goal of playing more games to select a winner seems nice. So I changed my mind and think it would be better to have some special solution rather than give one of the teams in the lead a bye in the last round.

However, what I would be tempted to do given seven 3-person teams in a five round tournament would be different again.

(Disclaimer: I've never arranged a serious tournament of any kind. On the other hand for a number of years I was the person responsible for arranging the 'tournament' at the year-end party of one of my study groups here in Tokyo so I have had a bit of practice at fitting odd numbers of people into some sort of structure that ensures everyone plays and has a good time.)

So I would be upset with a traditional team tournament of team A playing B, C playing D, E playing F, and G having the bye. The reason is that in the end five out of seven teams would only play four matches. The most reasonable idea that I could come up with off the top of my head was a combination of the following.

First, the regular games will be organized as three mini-Swiss tournaments between the seven top boards, the seven middle boards, and the seven bottom boards. The complete teams never sit down against each other. Obviously there is one person left over in each group each round.

So we form seven groups of three: A1B2C3 (top board from team A, the middle board from team B, and the bottom board from team C), B1C2D3, C1D2E3, D1E2F3, E1F2G3, F1G2A3, and G1A2B3. Each round a group of three that is unpaired for regular play has a mini knockout match at half the regular time control. First the middle and bottom board players play each other, e.g. B2 plays C3. Then the winner plays the remaining top board, e.g. A1. In this way everyone plays in every round.

"But wait!", you say, "There are seven triplets and only five rounds so it doesn't add up." You are right but if we only do it five times, it won't be fair. So in round one, three triplets (nine players) play fast games while only twelve players play regular games. In the succeeding four rounds one triplet (three players) plays fast games while eighteen players play regular games. That is a total of seven triplets.

In this way everyone plays four rounds of regular games and every one has a bye in the regular games. Everyone plays either 1 or 2 fast games (the seven winners of the fast game between the middle and bottom boards will play 2 games) when they have the bye from the regular games. The amount of playing is completely symmetrical for all the teams and each individual member of each team. The team tournament results are based on the sum of the members of each team. The fast games can be worked into the basic team result, or they could be used as the first tiebreaker. What do you think?

_________________
Dave Sigaty
"Short-lived are both the praiser and the praised, and rememberer and the remembered..."
- Marcus Aurelius; Meditations, VIII 21

In this case C wins, having lost no games while A has lost one game. I failed to write this in my opening post, but in a reply to Herman.
The original system was in case of seven teams, to have a round robin with slightly shorter thinking times. However I missed one board meeting, when I was sick, and the others decided to change the system. Anyway you system seems workable, but I prefer the original one.