After the previous post, at least one person have seen the light. I find Najib is quite a reasonable person if you present the facts accurately , unlike some other people around.
Najib has announced changes to the Renegade Rapid Weekender due to be held on 30th January. The changes are:-
1) An Unrated prize has been added for players who are unrated.
2) Tie Break for the event is also advertised to avoid controversies.
Go here to read the full details. The announced tie-breaks are
1. Bucholz with variable [37]- For Best and Weakest game dropped are counted as "0",
- For not played games (forfeit, bye, etc) is "compute with real points"
- For adding own points is "No"
- For dropped players is counted as "No points (for all rounds)"
2. SB with variable [52]- variables are the same as above
3. Bucholz with variable [37]
- variables are the same except - For not played games (forfeit, bye, etc) is "compute as draw against the player himself"
Here is an explanation of what the tie-breaks actually mean. The first tie-breaker is the normal Buchholz that we all know. Under this tiebreak the 17th Rapid placing would be Fadzil, Ian, Abdullah.
The second tie-break is normal Sonneborn-Berger. Under this tiebreak the 17th Rapid placing would be Fadzil and Ian tie followed by Abdullah.
The third tie-break is the one actually used in 17th Rapid and the result will be Abdullah, Fadzil then Ian according to me but chess-results give Abdullah, Ian , Fadzil.
Subscribe to:
Post Comments (Atom)
2 comments:
The first tie-break looks like Median-Buchholz to me, unless the "0" is actually a boolean value, but given the context, I think it just means the opponents with the highest and lowest points are not counted.
Actually, this tie-break has a weakness too: If say, there are 2 guys tied for first, it is not uncommon that only one of them have played the guy who ended up third (whom we will assume is the 3rd strongest player in the tournament), and then beaten him. His extra effort would be discounted in the tiebreak.
But it is better than the draw-yourself tiebreak, and hey, nothing's perfect.
I would also suggest that tiebreak #3 be the "normal" Buchholz, since at the end of the day, if you played one opponent less than your rival, you've had it easier. Keep in mind that this would only be considered if the first two fail.
Yes agreed with ur comments except for the first. The way I understand it, it is not Median-Buchholz. Hard to explain, the programmer is Italian :)
Post a Comment