Hello Roger,
In the ECF official grading statement it is said that grading "points are allocated in respect of each game." and that one's final "grade is calculated by dividing the total number of points scored by the number of games played" which is an average of grading points scored in each game.
Various rules for allocating grading points with respect of each game can be expressed with the following formulae:
Code: Select all
a2 = a + ka*(q - p);
b2 = b + kb*((100 - q) - (100 - p));
where 'a' is your grade, 'b' grade of your opponent, 'p' your expected performance (expected performance of your opponent is then '100 - p'), 'q' your actual performance (actual performance of your opponent is then '100 - q'), 'a2' your new grade (your grading points allocated for the game) and 'b2' your opponent's new grade (your opponent's grading points allocated for the game).
(if the players played only one game in the season 'q' is either 100, 0 or 50, if they played more than one game it can be a number between 0 and 100 inclusively)
What makes the rules different is a choice of factors 'ka' and 'kb' and function 'p = f(d)'.
The difference between GS rule and AGS3 rule is in factors 'ka' and 'kb' only, for GS 'ka = kb = 1' and for AGS3 'ka = kb = 1/2'. Both GS and AGS3 use the same linear 'p = f(d)' shown in the figure 1 below (green line).
Figure 1: Relationship between expected performance 'p' and grade difference 'd' as defined in GS (green line), CGS, AGS and AGS2 (blue line), ÉGS, ÉGS2, ÉGS3 and ÉGS4 (red line), ÉGS5 and ÉGS6 (yellow line), and (normal relationship 'p = 100*(1 + Erf[d/50])/2', where the error function Erf[z] is the integral of the Gaussian distribution) as originally defined by Élo (brown line above yellow). Expected performance 'p' is a function of grade difference 'd', i.e., 'p = f(d)'. Note that both FIDE and USCF switched from normal (brown line) to logistic (yellow line) relationship 'p = f(d)' which they found provides a better fit for the actual results achieved.
Although grading points are allocated with respect of each game, this is done once a season after the players had played all of their games. This fact can be utilized to asses how to distribute penalty and reward when allocating grading points in respect of each game, and instead of using 'ka = kb = 1/2' we could calculate 'ka' and 'kb' using the following formulae:
Code: Select all
ca = Abs[(qa - 50 - da)/2];
cb = Abs[(qb - 50 - db)/2];
c = ca + cb;
ka = If[c > 0, ca/c, 1/2];
kb = If[c > 0, cb/c, 1/2];
where 'da' is grade difference between your grade and average grade of your opposition (positive if your grade is above average grade of your opposition), 'db' is grade difference between your opponent's grade and average grade of his opposition (positive if your opponent's grade is above average grade of his opposition), 'qa' your actual performance against your opposition, 'qb' your opponent's actual performance against his opposition (if your grade differs from average grade of your opposition by more than 40 points, it is taken to be exactly 40 points above (or below) yours, if your opponent's grade differs from average grade of his opposition by more than 40 points, it is taken to be exactly 40 points above (or below) his.
AGS4: We define AGS4 (Amended Grading System four) to be AGS3 with factors 'ka' and 'kb' calculated using above formulae (rather than taking 'ka = kb = 1/2').
Figure 2a: Factor 'ka' (used in AGS4) as a function of 'qa' and 'qb' where 'da = -40' and 'db = 0'.
Figure 2b: Factor 'ka' (used in AGS4) as a function of 'qa' and 'qb' where 'da = -20' and 'db = 0'.
Figure 2c: Factor 'ka' (used in AGS4) as a function of 'qa' and 'qb' where 'da = 0' and 'db = 0'.
Figure 2d: Factor 'ka' (used in AGS4) as a function of 'qa' and 'qb' where 'da = 20' and 'db = 0'.
Figure 2e: Factor 'ka' (used in AGS4) as a function of 'qa' and 'qb' where 'da = 40' and 'db = 0'.
Roger de Coverly wrote:Robert Jurjevic wrote:Can you prove or show why do you think that AGS3 would change grades too little (or equivalently why is GS a system which changes the grades for the exactly right amount)?
Suppose you consider the very top of the food chain. Suppose for many years the top players are first amongst equals at around 250. Suppose a player comes along who can score 75% against all these 250's. How does his grade ever reach 275 in AGS3? He scores 15 draws and 15 wins. First season his grade is ( 250 *30 + 25 * 15) /30 = 262.5. Second season (30*( 250 + 262.5)/2 +25 *15) /30 = 268.75. Third season ( 30*(250+268.75)/2 +25*15) = 271.88. As he gains fewer and fewer points each season, he never reaches 275. I'm assuming that the 250s manage to retain their 250 status by beating someone else.
If the 250 players played approximately as expected (i.e. approximately at their level of 250, say every 250 player performed on average 50% against approximately 250 field) AGS4 would assign to the player who scored 75% a grade of 275, i.e. AGS4 would reward the 75% player for the maximum amount in every game he played against the 250's, but it would not penalize (nor reward) either of the 250 players for his game against the 75% player (games the 250 players played against the 75% player wouldn't count in grading of the 250 players).
GS would assign to the player who scored 75% a grade of 275, i.e. AGS4 would reward the 75% player for the maximum amount on average in every game he played against the 250 players,
but it would penalize each of the 250 players for the maximum amount on average for his game against the 75% player (though this penalty may be relatively small as it may be done for relatively few games in respect to number of games each 250 player did play, it is still a penalty which should not have been done, also there are cases where this error may not be small at all, say the example of the lighthouse keepers).
AGS3 would assign to the player who scored 75% a grade of 262.5, i.e. AGS3 would reward the 75% player for a half of the maximum amount on average in every game he played against the 250 players, and it would penalize each of the 250 players for a half of the maximum amount on average for his game against the 75% player (AGS3 simply does not have a means to asses how to split the penalty and the reward, so it does it 50/50).
Note that it is a coincidence that in GS rule you do not need to know your grade in order to calculate your new grade (it is enough to know your actual performance and grades of your opponents), for all other rules you need to know your grade as well, and in general you need to know your grade. GS rule can also be expressed in a way so that you need to know your grade, say rules 1a and 1c below are equivalent:
Rule 1a: For a win you score your opponent's grade plus 50; for a draw, your opponent's grade; and for a loss, your opponent's grade minus 50. Note that, if your opponent's grade differs from yours by more than 40 points, it is taken to be exactly not 40 points above (or below) yours. At the end of the season an average of points-per-game is taken, and that is your new grade.
Rule 1c: For a win you score your grade plus 50 minus grade difference; for a draw, your grade minus grade difference; and for a loss, your grade minus 50 minus grade difference. Note that, if your opponent's grade differs from yours by more than 40 points, it is taken to be exactly not 40 points above (or below) yours. At the end of the season an average of points-per-game is taken, and that is your new grade.
So, there is a maximum allowed amount (in grading points) for which you can reward o penalize the 75% player (as if he is not un-graded he must have a grade), if you choose to reward the 75% player for the maximum allowed amount then you should not penalize the 250 players for their games they have played against the 75% player (as already all it could have been given or taken was given to the 75% player), but you can choose to splilt the reward and the penalty 50/50, which AGS3 does. AGS4 is more sophisticated and would spit the reward and the penalty depending on the circumstances (using actual performances of the 250 players against their oppositions AGS4 tries to asses how to distribute the penalty and the reward in each game the 75% player has played).
Roger de Coverly wrote:There's no point in modifying the ECF system to slow down grading changes by averaging everyone against their previous grade. Fundamentally if 3 players all play equivalent people over at least 30 games with a performance of 175 (based on the most recent published grading list), then they should all get grades of 175 regardless of whether (a) they had a previous grade of 175 or (b) they had a previous grade of 150 or (c) whether they had a previous grade of 200 or (d) they are a player new to English chess.
Assuming that (as stated above) there is a maximum allowed amount (in grading points) for which you could reward or penalize players, the answer depends on the rule one did use and how the rule splits the reward and the penalty. 'Correct' rules should IMHO obey the axiom which states that
grading points should be allocated in each game so that the total grade correction equals the difference between expected and actual performance (which is to me a perfectly logical and natural requirement), i.e, '(a2 - a) + (b - b2) = q - p' for any 'a', 'b', 'p' and 'q'.
Why in GS you do not need to know your grade in order to allocate grading points in each game? One allocates 'a2' grading points to you in each game according to the formula:
where 'a' is your grade, 'b' grade of your opponent, 'p' your expected performance (expected performance of your opponent is then '100 - p'), 'q' your actual performance (actual performance of your opponent is then '100 - q'), 'a2' your new grade (your grading points allocated for the game) and 'b2' your opponent's new grade (your opponent's grading points allocated for the game).
GS's 'p = f(d)' (for 'd <= 40') is:
so 'a2 = a + ka*(q - p) = a + ka*(q - 50 + d)' and as in GS 'ka = 1' one has 'a2 = a + (q - 50 + d) = a + d + q - 50' and as 'a + d = b' is your opponent's grade you do not need to know your grade 'a' in order to calculate 'a2', as 'a2 = b + q - 50', where 'b' is your opponent's grade. Nevertheless, the correction one applies to your grade is 'ka*(q - p)', and the correction one apples to your opponent's grade is 'kb*((100 - q) - (100 - p)) = kb*(p-q)'. The question is which 'ka' and 'kb' to choose, i.e. how much to correct your grade and the grade of your opponent?
It can be proven that if '(a2 - a) + (b - b2) = q - p' for any 'a', 'b', 'p' and 'q'. then 'ka + kb = 1'. Unfortunately, GS's 'ka = ka = 1', so 'ka + kb = 2'.
Kind regards,