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?
