This sounds really cool. How does it work?
If I would create a (correspondence) Rengo tournament on OGS, I would draw up the pairings and everything here on the forums and create the actual games on DGS. 
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I have a really old DGS account I could use. My rank might legitimately still be tpk, or at least close to it since it was so long ago. 
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Well, if we could define parings and handicaps by ourselves, so we could take the OGS ranks, even if the games would take part on DGS.
(I also have a DGS account - with one game still active, and I planned to stop playing there when the game is over… because I like OGS much better. But the support for Rengo and other multiplayer games is cool.)
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I’d like to avoid handicaps if possible; if having a lot of tight games is important to many people, I would encourage just pairing the strongest player with the weakest, the second strongest with the second weakest, &c…
I don’t get why you guys think pairing strong players with weak players is gonna be any fun. Imagine 3k paired with 22k. It’s gonna be boring as hell, probably for everyone involved.
If you want some freak games, just get AquaBot (if they’re still around) and make two TPKs play rengo paired with superhuman AI.
Not to mention you just might not like your pair.
That’s why I offer randomized pairs. You get to play with different people.
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Pairs should be rotated, and perhaps the difference in rank shouldn’t be too large. But with randomised pairs, you’ll also get stuff like pair of TPK’s playing a pair of SDK’s, which is even less fun than pairing each SDK with one of the TPK’s.
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As I’ve said before, I’m happy with just about any system for deciding partners, strongest with weakest is just the one I recommend if balancing is a major concern for people. It won’t be perfect by a long shot, I expect the players in the middle of the rank distribution will have the best shots, but I think it will be good enough for government work.
Agreed. I’ve performed some quick napkin calculations using current ranks of 10 people who wrote to this topic. If, say, 1d prefers playing with/against 7k and stronger, only ~12% of random pairings are ok. If 1d accepts 10k and stronger, ~25% of random pairings are fine. Including up to 15k gives ~66%.
With completely random pairings, most of us couldn’t afford to be too picky.
Even better, of course, if a list of participants expands manifold.
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Sometimes I enjoy playing rengo with a much weaker player. Gives me the same rush as “Russian Roulette” (aka one bullet in six-shooter revolver, spin the barrel and hope).
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I like the idea of randomized pairs for each round, but in that case how would we manage points? It seems difficult to run a pairs tourney if the pairs don’t stay together.
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Each partner scores whichever result their team that round scored.
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Seems obvious, not sure why that didn’t occur to me 
Double round robin: each possible pair plays every other possible pair. Given N as number of participants, how many matches would it take?
(2 over n) * (2 over n-2) / 2 if I’m nor mistaken.
Edit: for example 8640 matches with 6 players if I haven’t mis-calculated.
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It’s not 6x5x4x3/2x3x4 ? (=3x5=15)
Combination of 4 between 6
Old memories, should be adjusted if order is important or not and I guess it is because 2 pairs, I hope someone can fix that.
First you want to find the number of possible groups of 4 out of N, which is
(N choose 4) = N! / ( 4! * (N-4)! )
For each group of 4 people, ABCD, there are three possible pairings: you only have to look at the person A is paired with (since the remaining two form the other pair), thus AB vs CD, AC vs BD and AD vs BC.
So in total there are 3 * (N choose 4) possible matches.
For example, with 8 people, we’d have 3 * 8! / (4! * 4!) = 3 * 8 * 7 * 6 * 5 / 4 * 3 * 2 * 1 = 210 matches.
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My ‘solution’ counted many possible pairings multiple times, so it was nonsense. This looks correct to me.
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In fact, your solution is also correct. It’s the other way of looking at it: first choose the first pair, then choose the second pair, and then divide by 2, since each pair is chosen twice.
Look:
Although it does seem you made a mistake in your calculation
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Wow, I have not been following this thread, but I’m quite pleased to see, upon entering it just now, that it has devolved into pairing randomization pedantry
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