Does anyone know a 12 man double elimation tournament setup?

Question

I'm looking for a way for 12 people to compete in a tournament without any overlaps or vacancies. Every player should be able to play an equal amount of rounds and the double elimination bracket should not lead to the final unfairly; the champion should always be able to lose one game.

Also, I'm looking for how to rank the players when 8 of the players have known records and 4 of the players are invited.

Answer 1

Your condition that everyone play an equal number of rounds may break down if, when players begin to be eliminated, there is an odd number of survivors left at the end of any given round. Someone must receive a bye for the next round. I suggest that every time this occurs, all the surviving players with the fewest losses who have not yet been given a bye (it would be unfair for a player to get a second bye while others have been playing in every round so far) draw lots for the bye in the next round. Here is a tournament that requires at most two byes. It is difficult to describe a tournament without a diagram, but I will try.

After the first round we have six players who are 1-0 and six who are 0-1. Pair all the winners with one another and also all the losers with one another for the second round.

After the second round we have three players who are 2-0, three who are 0-2 and thus eliminated, and six who are 1-1. Since there are nine players left, choose one of the undefeated players by lot to receive a bye and pair the other two for round 3. Pair the six 1-1 players so that each of the second-round losers plays a first-round loser. (All pairings should be done so that two players do not face one another a second time until it is unavoidable.)

After round 3, three more players are gone with 1-2 records. Remaining are one player with a 2-0 record (the player who got the bye), one player who is 3-0, and four players who are 2-1(the three winners from the losers' bracket plus the loser to the player who is now 3-0). For round 4, the two undefeated players are paired together, and the four 2-1 players play in two pairs.

After round 4 two more players are out, with 2-2 records, and there are two possibilities for the remaining players:

a) Three players are 3-1 and one is 3-0 ; then the 3-0 player faces a 3-1 player other than the one he just played and the other two 3-1 players face one another. If the undefeated player wins again, then only he and one other player will be left, and they play each other until one is eliminated. But if the 3-0 player loses his game, the player who beat him gets a bye while he (the player who was 3-0 but who just lost) plays the only other remaining player. After that game there will be only two players left, both with one loss, and they will play for the championship.

b) Two players are 3-1, one is 2-1, and one is 4-0; then the 4-0 player goes against one of the 3-1 players and the 2-1 player goes against the other. This latter game will eliminate one player for sure. If the undefeated player wins again, the player he beat is out and he plays the only other survivor until one of them is eliminated. If the 4-0 player loses, the player who beat him sits out while he (the player who just lost his first game) plays the only other player left. The winner of that game eliminates the loser and then plays the one who just sat out for the championship.

As far as ranking the players,in the absence of any other information, it seems to me that the eight players of known records should be ranked in the first eight spots in the order of their records. The four invitees should draw lots for the last four places.

Answer 2

wah

Answer 3

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Answer 4

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