Random question. After two rounds, there will be four players in a group with score 1-1. How is it determined who plays with whom in the 3rd and final round?
Looks like my argument zoomed right over your head. I am not saying to match-up players using a random seeding, but to just select 8 players out of 14 at random.
Do you have some sort of proof or argument to show that this is an extremely rare outcome? What odds are we talking about here?
That's not how a well-designed tournament should work either.
You can just select the top 8 players by putting the 14 names in hat, shuffling them and then drawing 8 names out one by one. Yes, that would also be completely random and, in that sense, technically fair. But it any particular...
So, twice in this tournament one person was randomly selected to be matched someone with a higher score, and both times that person randomly happened to be DauT.
Seeing from DauT's perspective, that does not seem very fair or random.
There were 2 matchups in the whole tournament where players with different record were paired and both times it involved Daut playing someone with a better record.
Please tell me more about how the tournament rules make sense.
Here is something interesting: If Tatoh and Slam both deliberately throw their games (slam anyways has basically no chance of qualifying), Tatoh and Daut will both guaranteed qualify.
(Verified by enumerating all possible outcomes on a computer.)
Please tell me more about how the tournament...
Why do we care that tournaments are organized sensibly and not let arbitrary decisions give unfair advantage to one or the other player?
What rule exactly was followed to determine which of the 5 (1-2) players was chosen to play a (2-1) player?
No, the next best thing is the usual...
Neither did Villese, and he can get through if he wins against Vinchester and Daut loses against Tatoh.
Slight favorite means precious little in a best of three.
He probably wouldn't complain at all because his team-mate will qualify by winning against him. But why should that stop us from...
The more I think about it, the dumber tie-breaking based on Buchholz appears to be (at least in the current settings). There is a likely scenario where we might have to decide between two (2-2) players based on which player lost to a 4-0 player and which lost to a 3-1 player. Which of course is...
The thing is having a minimum of 2 wins is much more important than those tie-break scores.
If Daut loses to Tatoh, one of Vinchester, Villese, Dogao or Slam are going to qualify, having faced a much weaker opponent for their fourth match, based on the arbitrary way one (1-2) player was chosen...
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.