2010-2011 Grades?

[Nicky Chorley]

I've also been playing in the Hammersmith internal rapid tournament and have a record of 0-8. I don't imagine it getting any better if I play more games, as all the players are a lot stronger than me. There just aren't players at a similar strength to me. It's not really much fun to lose every game :/.

If I manage to secure employment, I think I'll invest in coaching. Maybe then I'd get to play through a game and discuss it at the same time and that may help with learning.

[Roger de Coverly]

Maybe then I'd get to play through a game and discuss it at the same time and that may help with learning.

You could put the odd game up on this site in "Chess Questions". It might indicate where things are going wrong. What strength opposition do you face in the Hammersmith club rapid-play?

[E Michael White]

Do tell how a player can get a grade above 296 by losing all nine games.

The way I was looking at this was for an ungraded player to lose 9 games to the top player on the previous list who must be at least 60 points higher than the second player. The former first player must also lose games to very low rated players. So there are two parts to this a) getting the top rated player in one list say 2010 at least 60 points ahead of the 2nd player and b) ensuring the former first player loses at least 60 points and other players grades remain roughly the same.

Part b) In 2010 Player A is graded 356 (artificially high for illustration), player B ungraded. Player A wins 9 games against player B. Player B only plays those 9 games. Player A however loses 23 games to players more than 40 points below him, ie less than 316.

The new starter process, which does not require the reiterative procedure, will allocate player B a final 2011 grade of (356 - 60) = 296 ie A’s grade less 60. For those games, Player A will be allocated an extra 10 points per game as player Bs estimated start grade will be more than 40 points lower than As.

Player As eventual 2011 grade is (23x(356-90) +9x366)/32 = 294
Player Bs 2011 grade is 296 as above putting him in first place if other players grades apart from A and B remain roughly the same and don’t exceed 296.

Part a) Getting the top player 60 points ahead of the second can be achieved in several ways eg a) over a few years scoring 100% and using the 40 point rule b) as a new starter the previous year with results which build a pyramid of results against other ungraded players, discussed many times on this forum c) several other ways.

If the membership proposal comes about, leaving players off the list who don’t pay subs, then another way will be to build a pyramid of new starter results where A is graded 346 and B 296, then A refuses to become a member of the ECF leaving B number 1.

Second, is this a Schroedinger-type cat?

I dont think this works due to the uncertainty surrounding some of his work. Barbara Woodhouses K9 factor will also fail. However a Sherzinger group of Cats could succeed but she will need to ask papa first.

I then set up a 9 game match between the winning GM and my neighbours cat.

This comes within a whisker of succeeding but when I tested it out the neighbours cat withdrew to the apple tree as he had misuderstood that a category 9 event did not necessarily involve fun with 8 other cats.

[timlawson]

The above madness over generating grades does reinforce my personal thoughts towards our excellent system!

[Simon Dixon]

The above madness over generating grades does reinforce my personal thoughts towards our excellent system!

The funny thing is, using the + - 40 point rule, if you enter a couple of opens, it is possible to get a high grade without winning a game.

[Roger de Coverly]

The funny thing is, using the + - 40 point rule, if you enter a couple of opens, it is possible to get a high grade without winning a game.

I wouldn't be so sure about that. Suppose in each of two Opens you face four players graded 175 and one graded 150. You score just half a point in each open. Provided you have a published grade of at least 135, your performance is (8*175 +2* 150 - 400)/10 = 130. If you don't have a grade, I don't think you get convergence in the looped iteration until your grade calculates as 110. In other words (8*150 + 2*150 - 400) /10 = 110

A published grade of 125 would get a performance of (8*165+2*150-400)/10 = 122

If you manage an extra half point, your performance converges immediately to (8*175 + 2*150 - 350) /10 = 135.

So if you are a new player, a single dropped half point over 10 games can cost you 25 on your published grade.

[Ben Purton]

Ive been told I have a grade of 197 (sigh) and 172 rapidplay(YAY).

At least the later means I am extremely "Well in" for the 4NCL rapidplay handicap prizes.

So upset that the loss on final day of season has made me go from 201.4 to 197.3 . Bit gutted.

[Brian Valentine]

I wouldn't be so sure about that. Suppose in each of two Opens you face four players graded 175 and one graded 150. You score just half a point in each open. Provided you have a published grade of at least 135, your performance is (8*175 +2* 150 - 400)/10 = 130. If you don't have a grade, I don't think you get convergence in the looped iteration until your grade calculates as 110. In other words (8*150 + 2*150 - 400) /10 = 110A published grade of 125 would get a performance of (8*165+2*150-400)/10 = 122If you manage an extra half point, your performance converges immediately to (8*175 + 2*150 - 350) /10 = 135.So if you are a new player, a single dropped half point over 10 games can cost you 25 on your published grade

Roger,
I may have misunderstood the brief description of the process, but I get different numbers for the new player in your example, though I agree the general point. My understanding is as follows.

A new player loses all games then the iteration process uses 175*8+2*150-50*10= 1200/10=120 and then plugs this grade into the final stage including the 40 point rule to give 160*8+150*2-500=1080/10=108

If the player manages 1 draw then the iteration process gives 175*8+150*2-450=1250/10=125 and the final stage gives 165*8+150*2-450=1170/10=117

[Nicky Chorley]

Maybe then I'd get to play through a game and discuss it at the same time and that may help with learning.

You could put the odd game up on this site in "Chess Questions". It might indicate where things are going wrong. What strength opposition do you face in the Hammersmith club rapid-play?

I suppose I could do that, yes. The regulars at the club are graded between 110 and 191 (slow play). If I get some time at the weekend, I'll put one of my games up from this week.

Thanks!

[Roger de Coverly]

If the player manages 1 draw then the iteration process gives 175*8+150*2-450=1250/10=125 and the final stage gives 165*8+150*2-450=1170/10=117

I was assuming two draws, so first process gave 175*8 + 150*2 -400 = 1300 /10. Second process gives 170*8 + 150*2 -400 = 1260/10.

It was my impression that iteration continued until convergence.

From http://grading.bcfservices.org.uk/help.php#notes

Estimating a starting Grade for an ungraded player
A Rapid Grade, where available, will be used in default of a Standard Grade; and vice versa. If the player has no Grade at all, a starting grade is calculated as follows.

Stage 1 is to calculate a 'grade' for each ungraded player on his games against graded opponents. The 40-point rule is not used. If all his opponents are graded, it stops there and the result will be used as his starting grade.

Stage 2 brings in games between the ungraded players. Once again the 40-point rule is not used. The players are 'graded' on all their games, using as starting grades the figures obtained from Stage 1.

The resulting 'grades' will not be very accurate. So they are fed in again as new starting grades, and Stage 2 is repeated. This continues, with increasing accuracy each time, until the figures (more or less) stop changing. The starting grades can then be considered accurate.

These starting grades are then used in the grading proper.

The 10% example converges very slowly

so 3rd (8*166 + 150*2 -400) -> 123
4th (8*163 -100) -> 120
5th (8*160 -100) -> 118
6th (8*158 -100) -> 116
7th (8*156 -100) -> 115
8th (8*155 -100) -> 114
9th (8*154 -100) -> 113
10th (8*153 -100) -> 112
11th (8*152 -100) -> 112

So if you ignore decimals, it converges at 112

Not ignoring decimals, it continues to slowly spiral downwards to 110.

There's presumably a proof that where you get 10% or below or 90% and above , that the iteration can give dubious results. At a guess, you might get more reliable results if you estimated the new players ignoring the outliers and then brought them in at no more than a handful of iterations. Perhaps they now do this. There was someone two years ago playing on about board 3 in a lower London league division who came out at around 220 in the first draft.

[Simon Dixon]

175*8+2*150-50*10= 1200/10=120

175*8+2*150-50*10=1650/10=165

[E Michael White]

Roger
I read it that:- If all the opponents of an ungraded player are graded the results don’t go through step 2 but the "starting grade" is fed into the "grading proper".

A player scoring 0% against graded opposition gets a "start grade" of 50 less than the average grades of his opps. This is fed in as a "starting grade" to the "grading proper" and will normally be reduced by a further 10 as a result of the 40 point rule. If the grades of opps are spread, not all results are modified by the 40 rule.

The result for an ungraded player scoring 0% against the same player or players with closely packed grades is then their average grade less 60. This was the basis for the comical scenario of a 0% loser being top of the ECF rating list.

If an ungraded player is guaranteed graded opps then entering 2 opens and scoring 0% is likely to give a higher final grade than a minimal score in a lower tournament. I suppose the player could then deliberately default if drawn against an ungraded player which would otherwise mean taking his chances with the loopy procedure

[Alex Holowczak]

175*8+2*150-50*10= 1200/10=120

175*8+2*150-50*10=1650/10=165

No, 175*8+2*150-50*10=1200

[IM Jack Rudd]

Please put some brackets in those expressions. Mathematics written like that makes my head hurt.

[E Michael White]

([As] (( )) you ([(wish)]) Jack)