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BGonline.org Forums
Madison - I'm close to a method for a 100% Swiss
Posted By: neilkaz In Response To: Madison - I'm close to a method for a 100% Swiss (Coolrey)
Date: Friday, 5 September 2008, at 7:14 p.m.
Ray, you've played almost as many of these Swiss's as I have. So I don't need to remind you how long they usually go. But I will anyhow.
The point I think I see you trying to make in a couple of your posts is that you'd like to see the T end after rd 11. That's a consideration unless, as will be common with 50ish players there's two 9-2's left (OK they play one more match as the finals). If there's a 10-1 and a 9-2 and we end after 11 rds, then the 10-1 has won and the 9-2 is 2nd, but that isn't true Swiss Triple Elim ! So what if the event then requires the 9-2 to beat the 10-1 twices. That could go 13 rounds, but this event always has gone 13 rounds. The conso last weekend meant a 13 rd event. Adam and I were both 9-3 playing the finals ! The main went 12 rounds. When Indy was larger sometimes the finals, set up after 8 rounds, actually was a 5 rd bracket (lots of byes however) and certainly the conso went 13 rounds.
So lets see how many players this past weekend were still playing in round 12 !
Tim and Steve were playing the finals match of the main in rd 12. But in the conso, Neil was playing Carter in rd 12, and Adam was playing his victim. That's 6 players still alive in rd 12 and two played rd 13.
My proposed method will result in fewer players playing rd 12 than the current method ! Thus time is less of a consideration and will be even less so by keeping all matches 9 pts. My proposed method would end the event with a clear 1st place player, the only one with less than 3 losses, in 11 or more usually 12 rounds, quite more often than it will require 13 rounds, if we continue to have about 50 players. Thus my proposed 100% Swiss, no conso, 8-3 gets a prize, last players w/out 3 losses wins, is actually a bit shorter than past Labor Day events on average !
With 48ish players I don't think my proposed method will result in 3 9-2's, and it certainly won't result in a 10-1 with two 9-2's thus there will be no need to keep the 8-3's playing to avoid super-byes (if there's an odd number of players better than 8-3 and thus able to win 1st place) and so that the 8-3's can try to become 9-3 and get a higher prize.
If somehow there were three 9-2's, all the 8-3's could stop playing and we could very from the Swiss ever so slightly and use the fair method of resolving a 3 player RR I proposed in my reply to Gregg C, which he, having given these matters and ABT stuff much thought, IMHO, agrees with.
So, my proposed method of 100% Swiss(no conso) with the 8-3's getting a prize and stopping after rd 11 actually results in a having a couple FEWER players tied up after rd 11 than what is currently being done. So this is not more work for the directors, (they do have to do rd 10 and 11 pairings but that's not going to exhaust them when there's only a few players left) and also is not slower than what was done in Madison last weekend or what's been done in past Labor Day Swisses.
.. neilkaz ..
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