2v2 Random OPEN -- let's try again.

[aaronINdayton]

Format for this is going to be tough. Instead of pairing people against individual opponents (which may be the best way to do it,) I would try to lump people into groups of 4 or 8 and have everyone play a game with every other person as a partner once and against every other person exactly twice. When a team wins each player gets 1 point and the losing team get 0, and ties .5 as originally planned.

This is easy to do with groups of 4, but what do you do with the result? What if 3 people go 2-1 and one person goes 0-3? I can't think of a good way to progress groups of 4.

That leaves groups of 8, here's a link to someone who setup a single round tennis tournament of this style for 8 people with the pairings explained: http://www.livewild.org/tennis_combinatorics.html

Now the question is, how many groups of 8 do you have in the first round? For a 24 person tourney you have 3, so what do you do for round 2? take the top... 2 from one group and 3 from the other two? It'd be easier to have either 2 groups of 8 and take the top 4 from each, or 4 groups of 8 and take the top 2, to go to the next round which is a new 8 person pool. Then maybe do a 4 person group for the top 4 there if there isn't a clear winner after the second round. This would require exactly 16 or exactly 32 people.

This may not be the best idea, I haven't thought through the possible records this would generate and how many similar records could exist after each round.