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How to prevent rating lag?

# How to prevent rating lag?

### 18.08.2013 /   Andrejic, Vladica (2210)

If the FIDE wants to run blitz rating list then rating lag must be prevented. Here is my suggestion...

## Rating lag

Four year ago, the debate began on whether to increase development coefficient, the famous K - the number which FIDE used to calculate rating. The biggest advocate of these changes was Polish grandmaster Bartlomiej Macieja, who felt that K need to be increased because the FIDE began to publish rating lists more frequently. The problem is phenomenon known as the rating lag, and it reflected in the fact that a player earns new rating points but has an older, much lower Elo rating, because there is a lag between the tournaments played and publishing of rating lists. If the tournaments were immediately rated then a player who earned higher Elo rating whould also have higher expectations, hence his future earnings whould be lower. I liked the opinion of John Nunn, who said that the rating lag is a defect which should be eliminated as soon as possible and it was not the inherent part of the Elo system. To put is simply, the rating system should not be adjusted so as to fit the anomalies, i.e. this cannot be the only reason to increase coefficient K.

As we well know, the FIDE introduced the rating lists for both rapid and blitz tournaments. We see that some players managed to play over 100 blitz games in the rating period, and that is perfectly normal. However, if we want to rate the blitz tournaments in the spirit of the FIDE rating system, we must consider the issue of the rating lag question, becuase it is much more prominent here. Let us recall our example from the previous article. Imagine a player whose actual playing strength is about 200 Elo points ahead of his rating at the time of our little experiment; alternatively, he can simply be much better in blitz than in classical chess (let's set it at 200 Elo-points difference in favor of his blitz strength), because the current official FIDE blitz rating system currently “borrows” one's classical Elo as the initial value for the blitz rating calculation. Simply put, for such (i.e. blitz vs. classical rating) 200-point strength difference, our player can expect to score (on the average) a win and a draw in two games against equally rated opposition; a completely regular outcome of our hypothesis is that, under given circumstances, the player will earn about 500 Elo points in 100 games, which would then be 300 points ahead of his real strength in blitz.

How can rating lag be prevented? At the beginning I whould ask the FIDE not to seek a solution from Jeff Sonas. There is no rating lag on the Internet chess servers, because the players get new ratings immediately after a played game, however, such interactive calculation is not in the spirit of the FIDE calculations. Our goal is to find a way to eliminate the rating lag, but in such way that we can rate a lot of games from the rating period, without the specific order of played games must not affect the calculations. This can be done by imitating the interactive system through the following mathematics. People who are not familiar with mathematics may skip the next section and read the conclusion at the end.

## Rating lag mathematics

Let us suppose that a player with an Elo $R$ has played $n$ games, so that in the $i$th game he played against opponent with the Elo $R_i$ and scored $P_i\in\{0,\frac{1}{2},1\}$. His earnings of Elo in $i$th game can be expressed as $$Z_i=K(P_i-f(R-R_i)),$$ where $K$ is development coefficient, and $f$ is the rating expectation function.

I believe that the nature of chess rating system preserves corresponding ratios, and therefore $f(x)=\frac{1}{1+C^x}$, with usual $C=10^{-\frac{1}{400}}$. In its Handbook, the FIDE states that such $f$ is a close approximation to real values, however the FIDE tables are to be exclusively used for calculating. Anyway, for small values $x$ holds \begin{equation*}\label{aprox} f(x)=\frac{1}{2}+ \frac{\log 10}{1600}x+ o(x^2)\approx \frac{1}{2}+ \frac{x}{695}, \end{equation*} so we can aproximate $f$ by linear function. We shall often use the corollary $$f(x+d)\approx f(x)+\frac{d}{695}.$$

The classical way of the FIDE calculation simply sum all individual changes, so we get the total change $$Z=\sum_i Z_i=Z_1+Z_2+...+Z_n.$$

Let us now analyze what happens if we rate game by game. After the first game the change is the same and it is $$Z'_1=Z_1.$$ Before the second game we have new rating so the the change for the second game is $$Z'_2=K(P_i-f(R-R_2+Z'_1))\approx K\left(P_i-f(R-R_2)-\frac{Z'_1}{695}\right)= Z_2-\frac{K}{695}Z_1.$$ The total change after two games will be $$Z'_1+Z'_2\approx \left(1-\frac{K}{695}\right)Z_1+Z_2.$$ For the third game we have $$Z'_3=K(P_i-f(R-R_3+Z'_1+Z'_2))\approx Z_3-\frac{K}{695}Z'_1-\frac{K}{695}Z'_2= Z_3- \frac{K}{695}\left(\left(1-\frac{K}{695}\right)Z_1+Z_2\right),$$ while the total change after three games is $$Z'_1+Z'_2+Z'_3\approx \left(1-\frac{K}{695}\right)^2Z_1+\left(1-\frac{K}{695}\right)Z_2+Z_3.$$ After repeating this algorithm till the $n$th game, we get the total change $Z'_1+Z'_2+...+Z'_n$ which is \begin{equation*} \left(1-\frac{K}{695}\right)^{n-1} Z_1+ \left(1-\frac{K}{695}\right)^{n-2}Z_2+ ...+ \left(1-\frac{K}{695}\right)Z_{n-1}+Z_n. \end{equation*}

This whould roughly be the evaluation of the rating (without rating lag) on the Interent chess servers, but in the spirit of the FIDE there should not be a difference in the order of games played. Going through all the possible permutations of games can be shown in the simplest way through aproximated using of the mean value of changes and thus \begin{equation*} \left(\left(1-\frac{K}{695}\right)^{n-1} + \left(1-\frac{K}{695}\right)^{n-2}+ ...+ \left(1-\frac{K}{695}\right)+1\right)\frac{Z_1+Z_2+...+Z_n}{n}. \end{equation*}

Since $$\sum_{i=0}^{n-1} \left(1-\frac{K}{695}\right)^{i}= \frac{1-\left(1-\frac{K}{695}\right)^n}{1-\left(1-\frac{K}{695}\right)}= \frac{1-\left(1-\frac{K}{695}\right)^n}{\frac{K}{695}},$$ for the total change holds \begin{equation*} \frac{1-\left(1-\frac{K}{695}\right)^n}{\frac{K}{695}}\cdot \frac{Z}{n}= \left(\frac{695}{nK} \left(1-\left(1-\frac{K}{695}\right)^n\right) \right)Z. \end{equation*} This result shows that the total change is $M(n,K)\cdot Z$, where $Z$ is a classical calculated change,

## The proposal of a solution for the rating lag

As we saw in the previous calculation, we can eliminate the rating lag by multiplying the classical total change calculation with the number $M(n,K)$ and in that way reduce the total change of rating. Let us look at the values of the function $M(n,K)$ for small $n$.

 n=1 n=2 n=3 n=4 n=5 n=6 n=7 n=8 n=9 n=10 n=11 n=12 n=13 n=14 n=15 K=10 1.000 0.993 0.986 0.979 0.972 0.965 0.958 0.951 0.944 0.938 0.931 0.925 0.918 0.912 0.905 K=15 1.000 0.989 0.979 0.968 0.958 0.948 0.938 0.928 0.918 0.908 0.899 0.889 0.880 0.871 0.862 K=20 1.000 0.986 0.971 0.958 0.944 0.931 0.918 0.905 0.892 0.880 0.868 0.856 0.844 0.833 0.822 K=30 1.000 0.978 0.957 0.937 0.917 0.898 0.879 0.861 0.844 0.827 0.810 0.794 0.778 0.763 0.748

If we assume that we have $K$ such that it gives an excelent rating system without unwanted anomalies (i.e. for small number $n$ of played games), and if we want to adopt the new way of calculations, we should raise $K$ to $K_1$ so that $K\approx K_1\cdot M(n,K_1)$ holds. For example, we can assume that the most regular thing for a player is to play one tournament with usual $n=9$ rounds within the rating period. The rating change for a classical $K=10$ player will be $M(9,10)=94.4\%$ of old change, so we get the better coefficient is $K_1\approx 10.6$. Of course, we should not exaggerate, so it is probably better to keep the old solution where the round $K=10$. In a case of $K=15$ player, it whould be better to use $K_1=16$.

However, the problem here is not classical chess, because one cannot play too many games in one month, and therefore, in that case we can keep the old system. Let us concentrate on blitz rating. The current FIDE blitz coefficient $K=20$ does not make much sense (blitz coefficient should not be higher than classical), so we should just pick any decent coefficient in our calculation. Let us look at tables of $M(n,K)$ for the large number of games $n$.

 n=10 n=20 n=30 n=40 n=50 n=60 n=70 n=80 n=90 n=100 n=110 n=120 n=130 n=140 n=150 K=10 0.938 0.874 0.817 0.764 0.717 0.673 0.633 0.596 0.563 0.532 0.504 0.477 0.453 0.431 0.411 K=15 0.908 0.819 0.742 0.674 0.615 0.564 0.518 0.478 0.443 0.411 0.383 0.358 0.336 0.315 0.297 K=20 0.880 0.769 0.676 0.599 0.534 0.479 0.432 0.392 0.358 0.329 0.303 0.281 0.261 0.244 0.229 K=30 0.827 0.679 0.567 0.480 0.412 0.359 0.316 0.281 0.253 0.229 0.209 0.192 0.178 0.165 0.154

From the table we clearly see that if we use $K=20$, as the FIDE does, and someone play 100 games, we should multiply the change with 0.329, i.e. the real change is worth less than one third of the FIDE value. Enough is enough...