Elo theory

The theory Elo is a theory which makes it possible to develop a model of evaluation of the force of a player of failures, which makes it possible to establish its Classement Elo.

Intuitive approach

; 1st step

Let us consider very large open with for example 20 rounds and approximately 300 participants (Swiss system or other by removing the null parts).

At the end of the tournament, the distribution of the points would take the following form roughly: August 1st

A theoretical competition made up of an infinity of rounds and participants would produce results summarized by a smoothed curve: the normal Law.

; 2nd step

Let us take a player who should play against the 300 participants, it is certain that its result would correspond to the same classification that if it had taken part in the great tournament with 20 rounds.

; 3rd step

Now, let us imagine that we do not want to make play the 300 parts, one takes a sample of ten players divided in the various categories. The result would be again similar, all things considered.

; 4th step

Let us adapt our scales by refining measurement, i.e. while replacing from 0 to 20 points by 1 to 1000. The curve resembles then that published for the calculation of points of classification.

; 5th step

One can imagine that the curve expressing the results of a given player opposed to a whole of players whose mean level corresponds to that of their adversary would be also out of bell, with comparable characteristics (but not identical, because each player has a different answer).

The FIDE proposed to unify the value of the statistics of the players so that all the countries have the same base of calculation.

Theoretical approach

The formula of Elo rests on the solution of a differential equation with an assumption which is checked relatively well. What explains the small differences that one can note between the Elo system and the normal Loi.

Let us take 3 players has, B, C.

The notations $P\; (a/b)$, $P\; (a/c)$, $P\; (b/c)$ express the probability respectively that has gains against B; the probability that has gains against C and to finish the probability that B gains against C.

One can write: $\backslash \; frac\; \{P\; (a/c)\}\; \{1-P\; (a/c)\}\; =\; \backslash \; frac\; \{P\; (a/b)\}\; \{1-P\; (a/b)\}\; \backslash \; times\; \backslash \; frac\; \{P\; (b/c)\}\; \{1-P\; (B:\; c)\}$ {1}
By taking the logarithm:
$\backslash \; ln\; \backslash \; left\; (\backslash \; frac\; \{P\; (a/c)\}\; \{1-P\; (a/c)\}\; \backslash \; right)\; =\; \backslash \; ln\; \backslash \; left\; (\backslash \; frac\; \{P\; (a/b)\}\; \{1-P\; (a/b)\}\; \backslash \; right)\; +\; \backslash \; ln\; \backslash \; left\; (\backslash \; frac\; \{P\; (b/c)\}\; \{1-P\; (b/c)\}\; \backslash \; right)$ {2}
Let us pose: $D\_\; \{ab\}\; =Z\; \backslash \; times\; \backslash \; ln\; \backslash \; left\; (\backslash \; frac\; \{P\; (a/b)\}\; \{1-P\; (a/b)\}\; \backslash \; right)$; in the same way for $D\_\; \{ac\}$ and $D\_\; \{bc\}$ {3}

We obtain: $D\_\; \{ac\}\; =D\_\; \{ab\}\; +\; D\_\; \{bc\}$ {4}

One can take the function reverses: $\backslash \; frac\; \{P\; (a/b)\}\; \{1-P\; (a/b)\}\; =e^\; \{\backslash \; frac\; \{D\_\; \{ab\}\}\; \{Z\}\}$

Let us pose: $\backslash \; mathcal\; \{E\}\; =\; e^\; \{\backslash \; frac\; \{D\_\; \{ab\}\}\; \{Z\}\}$

While solving, one finds: $P=\; \backslash \; mathcal\; \{E\}\; \backslash \; times\; (1-P)$;

$\backslash \; mathcal\; \{E\}\; =\; \backslash \; mathcal\; \{E\}\; \backslash \; times\; P+P$;
$\backslash \; mathcal\; \{E\}\; =P\; \backslash \; times\; (\backslash \; mathcal\; \{E\}\; +1)$;

$P=\; \backslash \; frac\; \{\backslash \; mathcal\; \{E\}\}\; \{\backslash \; mathcal\; \{E\}\; +1\}$; or $P=\; \backslash \; frac\; \{\backslash \; mathcal\; \{E\}\}\; \{\backslash \; mathcal\; \{E\}\; \backslash \; times\; \backslash \; left\; (\backslash \; frac\; \{1\}\; \{\backslash \; mathcal\; \{E\}\}\; +1\; \backslash \; right)\}=\; \backslash \; frac\; \{1\}\; \{1+\; \backslash \; frac\; \{1\}\; \{\backslash \; mathcal\; \{E\}\}\}$;

$P=\; \backslash \; frac\; \{1\}\; \{1+e^\; \{-\; \backslash \; frac\; \{D\}\; \{Z\}\}\}$ {6};

While differentiating, one finds: $\backslash \; frac\; \{dP\}\; \{dD\}\; =\; \backslash \; frac\; \{e^\; \{-\; \backslash \; frac\; \{D\}\; \{Z\}\}\}\; \{\backslash \; left\; (1+e^\; \{-\; \backslash \; frac\; \{D\}\; \{Z\}\}\; \backslash \; right)\; ^2\}$ {7}, which is the distribution of Verhulst

All the demonstration rests on the formula 1 which remains to be checked, the transposition of the natural logarithm in a logarithm bases 10 of them being direct:

One will take D equal to the difference in classification and Z equal to $\backslash \; frac\; \{400\}\; \{\backslash \; ln10\}$.

Results

• : 1000 Elo is the level of a beginner who knows the rules of the game.

• : 1200 Elo is the level of an occasional player.
• : 1600 Elo corresponds to a level of player of average club.
• : 2000 Elo corresponds to a strong player of club.
• : 2300 Elo corresponds to a level of Maître Fide.
• : 2400 Elo corresponds to a level of international Maître.
• : 2500 Elo corresponds to a level of international Large-Master.
• : 2800 Elo corresponds to the level of the world champion.

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