Hello Roger,
Your and your opponent's ECF grade with respect of each game is updated using the following formulae:
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a2 = a + ka*(q - p)/na;
b2 = b + kb*((100 - q) - (100 - p))/nb = b - kb*(q - p)/nb;
where 'a' is your grade (from previous season), 'b' grade (from previous season) of your opponent, 'p' your expected performance in the game (expected performance of your opponent in the game is then '100 - p'), 'q' your actual performance in the game (actual performance in the game of your opponent is then '100 - q'), 'a2' your new corrected grade (due to the game), 'b2' your opponent's new corrected grade (due to the game), 'na' the game count taken into calculation of your grade (if you played 30 or more games in the season then 'na' is the number of games you have played in the season), 'nb' the game count taken into calculation of grade of your opponent (if your opponent played 30 or more games in the season then 'nb' is the number of games your opponent has played in the season).
What makes ECF grading systems different is a choice of factors 'ka' and 'kb' and function 'p = f(d)'.
Total grade change with respect of each game is:
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(a2 - a) + (b - b2) = ((kb*na + ka*nb)*(q - p))/(na*nb);
We will examine two types of grading systems, grading systems for which it holds that 'ka + kb = 1' and grading systems for which it holds that 'ka + kb = 2'.
Total grade change with respect of each game for 'ka + kb = 1' grading systems is
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(a2 - a) + (b - b2) = (na + nb)/(2*na*nb) * (q - p) = c1 = c2/2;
Total grade change with respect of each game for 'ka + kb = 2' grading systems is
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(a2 - a) + (b - b2) = (na + nb)/(na*nb) * (q - p) = c2 = 2*c1;
!
As
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(na2 + nb2)/(na2*nb2) = (na1 + nb1)/(2*na1*nb1);
if 'na1 = na2/2' and 'nb1 = nb2/2', though 'c1(na1,na2) = c2(na1,nb1)/2' one could argue that 'c1(na1,na2) = c2(na2,nb2) = c2(na1/2,nb1/2)', i.e., that the total grade change with respect of each game for 'ka + kb = 1' grading systems for 'na' and 'nb' games equals the total grade change with respect of each game for 'ka + kb = 2' grading systems for 'na/2' and 'nb/2' games, consequently making the grade correction independent of the 'ka + kb' and thus denying any intrinsic difference between 'ka + kb = 1' and 'ka + kb = 2' grading systems (i.e., you can convert 'ka + kb = 2' into 'ka + kb = 1' grading system by halving 'na' and 'nb').
Reductio ad absurdum is a logical argument often used in mathematics. You basically start from negative statement of the statement you want to prove to be true and if by assuming that the negative statements is true you arrive at a contradiction (fallacy) you can claim that your original (positive) statement is true.
Say you want to prove that kingfisher can fly. You start assuming that kingfisher cannot fly and say that (assuming that kingfisher cannot fly) you arrived at a conclusion that the earth is flat (which is a contradiction or fallacy) you can claim that kingfisher can fly.
You can prove that square root of two is not a rational number (is irrational number) by assuming that it is a rational number and (taking that assumption as true) getting that say '1 = 2' (a contradiction).
I will use reductio ad absurdum logical argument to prove that statement 'c1(na1,na2) = c2(na2,nb2) = c2(na1/2,nb1/2)' is false, i.e., that statement 'c1(na1,na2) /= c2(na1/2,nb1/2)' is true.
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If 'c1(na1,na2) = c2(na2,nb2) = c2(na1/2,nb1/2)' then 'na1/2' and 'nb1/2' (which must be whole numbers) may not be whole numbers (which is a contradiction), so, 'c1(na1,na2) = c2(na2,nb2) /= c2(na1/2,nb1/2)'.
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So, there is an intrinsic difference between 'ka + kb = 1' and 'ka + kb = 2' grading systems and their corrections 'c1' and 'c2' are genuinely different in quality, as they cannot be made equal by choosing suitable 'na' and 'nb'.
For the cases where you have played against your opponent a match of 'na = nb = n' games total grade change with respect of each game for 'ka + kb = 1' grading systems is
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(a2 - a) + (b - b2) = 1/n * (q - p) = c1 = c2/2;
For the cases where you have played against your opponent a match of 'na = nb = n' games total grade change with respect of each game for 'ka + kb = 2' grading systems is
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(a2 - a) + (b - b2) = 2/n * (q - p) = c2 = 2*c1;
If you have played against your opponent a match of 'na = nb = n' games in each game scoring 'q' (either 0%, 50% or 100%) total grade change with respect of all 'n' games for 'ka + kb = 1' grading systems is
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n * ((a2 - a) + (b - b2)) = n * (1/n * (q - p)) = q - p;
If you have played against your opponent a match of 'na = nb = n' games in each game scoring 'q' (either 0%, 50% or 100%) total grade change with respect of all 'n' games for 'ka + kb = 2' grading systems is
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n * ((a2 - a) + (b - b2)) = n * (2/n * (q - p)) = 2 * (q - p);
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Answering the question which are 'ka + kb = 1' or 'ka + kb = 2' grading systems better should be the same as answering the question if you have played against your opponent a match of 'n' games in each game scoring 'q' (either 0%, 50% or 100%) should total grade change with respect of all 'n' games be 'q - p' or '2 * (q - p)'.
Kind regards,