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BGonline.org Forums
Swiss tournament playoffs
Posted By: bob koca
Date: Saturday, 7 June 2008, at 2:26 a.m.
At the recent Chicago ABT tournament I was discussing this with Sam Pottle and Chris Yep. The main problem we wanted to solve was a fair transparent way to setup a playoff bracket in a Swiss tournament. I like the following system:
1) Count an undefeated player as having 4 entries, a once defeated player as hvaing 2, and a twice defeated player as having 1. Start the playoff rounds the first time the total number of entries is 8 or fewer. Another way of looking at this is that an undefeated player is automatically in the finals and a once defeated player is automatically to the semis. The two loss players may get lucky and get a bye to the semis if the total entries is not 8.
2) Pay 4 places in the main 4, 2, 1, 1. Note that this may need to be adjusted to 4, 2, 1 if there happened to be an no loss player entering the playoffs.
A 2 loss player who then loses in the round of 8 and thus is not guaranteed money enters the consolation round one round advanced. This makes sense becasue he would then be at an equal place with other 3 loss players and would have the same number of wins.
Depending on number of players either pay 2, 1 or 2, 1, .5, .5 in the consolation round.
Concerns: 1) If the field is quite large then this pays too much to the first place, and would be imbalanced as far as ABT points go compared to other tournaments. 2) I don't like how the %age to the winner can change. Perhaps if it turns out to be 4, 2, 1 Should put that 1 down to the consolation. Changing 2, 1, into 2, 1, .5, .5 or changing 2, 1, .5, .5, to 2, 1, .5 , .5 .25, .25, .25, .25 3) If there are exactly three 2 loss players then there is too much importance to whom gets the bye.
Bob Koca
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