Okay, so I just kicked off a 100 player round-robin knockout tournament (the "Caro-Kann Open") with group size set to 5. This means for round 1 we have 20 groups of 5 players each. With one player advancing from each group, I was assuming that this would result in 20 group winners (assuming no ties) advancing to the second and final round. This means in round 1 each player has 8 simultaneous games, but then those who make it to round 2 will have to play 36 simultaneous games. Or so I was assuming. But now I'm thinking I have it wrong.

Are there just 2 rounds? OR do the winners then get put into groups of 5, with the winners of the 4 groups of 5 in round 2 advancing to yet another round? That would make more sense. Is that how it is?

I've got players in my tournament who are wondering...

-Mac

 

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Yes, your group sizes will never exceed what you set it to, in this case 5. So round 2 will consist of multiple groups yet again, and round 3, etc etc until only 1 group is left.

jay wrote: Yes, your group sizes will never exceed what you set it to, in this case 5. So round 2 will consist of multiple groups yet again, and round 3, etc etc until only 1 group is left.

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 Excellent! That makes much more sense than what I originally thought. Thanks!

 

Then what if you have 85 players in your tournament with a group size of five? You would have 17 players for round 2... What kind of groups would you get?