Seeing as we have been having fun with these...
At the Start of the Winter Chess Festival. What is the maximum number of norms possible for the event?
6
But that is incredibly unlikely because it would require .........
(a) One the 7 norm chasers to lose all games against the other norm chasers.
(b) each IM and GM to lose all games against those 6 remaining norm chasers.
Which puts 6 norm chasers on 4/4
© Those 6 remaining norm chasers would then need to score 2.5/5 against the other 5. Not necessarily with 15 drawn games though.
'Fraid to say it, You are wrong!
The answer is 7 Norms. Phils maths was spot on but Fiona can get 2 norms!
Andy Howie Wrote:Phils maths was spot on!
Andy Howie Wrote:but Fiona can get 2 norms!
Are you sure Andy?
Fiona is already a WIM - so surely there would not be much point in her getting an extra WIM norm.
So was 6 the correct answer ?
;P
Can she not get a WGM and a GM norm though? =|
Andrew,
just looked up paragraph 0.5 of the Fide Handbook.
IM norm requirement is 2450 performance against opponents average rating of at least 2230
WGM norm requirement is 2400 performance against opponents average rating of at least 2180
So if Fiona acheives IM norm she also achieves WGM norm.
Surprisingly that Pat didn't point that out.
Must be the Christmas spirit taking effect.
The field is a bit short of the 3 GMs needed for a GM norm.