One for all and all for one 5

Lucky, as a mathematician, I can further confirm that mathematically, more balanced teams will be formed with an even number of participants. Captains are chosen by voting, and the number of votes determines their ranking; for example, Gomunguls might be ranked number 1, and LuckyF might be ranked number 20 among the captains. Next, the captains choose their teams, with LuckyF choosing first, and Gomunguls choosing last. Essentially, the first two teams chosen are roughly equal in strength; the weakest captain chooses the best second team, and Gomunguls chooses last. With an even number of people on each team, the teams will be approximately equal in strength (assuming each captain takes the team selection seriously). With an odd number of people on each team, Gomunguls will have an advantage. This advantage is more significant with 3 participants than with 5. Lucky, your idea about reducing the number of people in the pool is also correct. However, it would then need to be limited to perhaps 5-8 people. This would result in 5-8 teams. I don't think this is a good idea for a team tournament.