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table
The probability function D800 is given by D800(P) = 4C(P - 0.5), Class interval C = 200. (Elo, 1978, p.146). Rating performance using D800 produces the same result as "algorithm of 400".

quotations
The development of the Percentage Expectancy Table (table 2.11) is described in more detail by Elo (1978, p159) as follows: The normal probabilities may be taken directly from the standard tables of the areas under the normal curve when the difference in rating is expressed as a z score. Since the standard deviation σ of individual performances is defined as 200 points, the standard deviation σ' of the differences in performances becomes √2 or 282.84. the z value of a difference then is D/282.84. This will then divide the area under the curve into two parts, the larger giving P for the higher rated player and the smaller giving P for the lower rated player.

For example, let D = 160. Then z = 160/282.84 = .566. The table gives .7143 and .2857 as the areas of the two portions under the curve. Thes probabilities are rounded to two figures in table 2.11. The table is actually build with 2000/7 as an approximation for √2.
 * author=Arad E. Elo
 * title=The Rating of Chessplayers

"Ceci n'est pas une pipe. This is Not a Pipe."

Foreign language
Zunächst sind ebenso viele Nummern zu verloosen, als Theilnehmer vorhanden sein: bei 14 Theilnehmern z. B. die Nummern 1, 2, 3 bis 14. Die Paarung geschieht alsdann in folgender Weise: Man fertigt ein Schema an, welches soviel senkrechte Reihen hat, als die Hälfte der Theilnehmer beträgt. Bei einer ungeraden Anzähl ist die Anzahl der senkrechten Reihen die Hälfte der um 1 vergrösserten Theilnemer- zahl. Bei 14 Theilnehmern sind daher, 14/2 = 7, bei 13 Theilnehmern (!3+1)/2= 14 / 2, also ebenfalls 7 senkrechte Reihen zu bilden. Je suis English. Je suis English.

Math
$$ {P(D)} = \frac 1 {1 + 10^{-D/4C}} ~.$$

Base 10 or base $$\sqrt{10}$$
In The Rating of Chessplayers, 1978, page 141 we read at 8.43:

"When the logarithms in equation (38) are taken to the base $\sqrt{10}$, then the Verhulst and the logistic take the following forms:

$ {P(D)} = \frac 1 {1 + 10^{-D/2C}} ~   $ (46)
 * undefined"

$$10^{1/2}$$

And subsequently in chapter 8.46 Percentage Expectancy Table, page 143 the table does have the header: "Logistic Probabilities to base $\sqrt{10}$"

Based on the above sources, it is $$\sqrt{10}.$$

LSM rating
Let Gm,m be bipartite with two disjoint parts A, B and m-regular. Let SA, SB be the sum of the scores si of Xi in A, B respectively.

Rating vector q equals:

qi = (2.si + T / m) / γ,  where γ = n.m,  T = SA if Xi in A, and SB if Xi in B.

Cite
Cite direct journal.

.

Edmund Landau

[Dr. W. Ahrens]

[Citeer Arxiv]

Chance

Roblox

In practice, there is little difference between the shape of the logistic and normal curve. So it does not matter whether the logistic or normal distribution is used to calculate the expected scores.

Repeat pp Repeat pp  Repeat at  Repeat loc

Diversen
Interwiki: pairwise comparison


 * (E1)      f(x) = x + 1

Deze iteratie is met pen en papier of pocket calculator uit te voeren. Het resultaat is een benadering. De relatie tussen ratingverschil (D) en winstkans (P) is niet lineair. Het rekenkundige gemiddelde van de tegenstanderratings is een benadering van van de rating waarvan de verwachting van de gespeelde partijen 50% is. Daarom zal deze procedure niet convergeren naar een oplossing, waarin de werkelijke en verwachte waarden identiek zijn.

In de beschouwing wordt uitgegaan van een binomiale verdeling, uitsluitend winst (1) en verlies (0) komen voor. In de schaakpraktijk is remise geen uitzondering. Daardoor wordt de verdeling vlakker. De binomiale variantie van een partij tussen even sterke spelers is per definitie gelijk aan: μ = (0 + 1)/2, en σ$2$ = (1 - μ)$2$/2 + (0 - μ)$2$/2 = 1/4. Inclusief remise (½) wordt dit: μ = (0 + ½ + 1)/3, en σ$2$ = (1 - μ)$2$/3 + (½ - μ)$2$/3 + (0 - μ)$2$/3 = 1/6. Dit is een factor 2/3 kleiner.

Calculating of the Rating Difference table
The standard normal distribution, with mean value μ = 0 and standard deviation σ = 1, in x = 2.17 equals to 98.500%. This is the first value in the 99% category. The corresponding rating is 620. Therefore the standard deviation employed in the table equals to 620 / 2.17 = 2000 / 7. The Elo table contains a few irregularities. A more accurate expectation of 620 equals to 98.4997%, which falls into the 98% range. The table assigns 344 to expectation 88%. However the expectation due to the normal distribution is 88.5705%, which is clearly in the 89%. Needed are references to the construction of this table in order to explain the underlying calculation.

If Player A has a rating of $$R_A$$ and Player B a rating of $$R_B$$, the exact formula (using the logistic curve). for the expected score of Player A is

Next
Elo made references to the papers of Good 1955, David 1959, Trawinski and David 1963 and Buhlman and Huber 1963.

The above schedule can also be represented by a graph, as shown below:

Designing Schedules for Leagues and Tournaments. (http://www.emba.uvm.edu/~jdinitz/preprints/design_tourney_talk.pdf)

Brewer's Dictionary of Phrase and Fable

Ramon Llull Ars Magna (cyclic pattterns), Martin Gardner, Last Recreations.

Round robin references

nl:IJmuiden IJmuiden (nnle)

Édouard Lucas, Récréations Mathématiques, four volumes, Gauthier-Villars, Paris, 1882-94 (https://archive.org/stream/rcrationsmathma01lucagoog#page/n0/mode/2up)

M. Walecki, professeur de Mathémntiques spéciales au lycée Condorcet

http://www.les-mathematiques.net/phorum/read.php?17,569123,569771

Pour Walecki: Walecki Felix Charles Louis

Né le 12/01/1844 à Moselle - Metz. Agrégé de mathématiques. A eu la légion d'honneur. A enseigné au lycée de Nancy avant de sévir à Paris, au lycée Condorcet. » C’est donc son anniversaire aujourd’hui  : 166 ans aujourd’hui. Mais il faut aller plus loin  : qui est Walecki ? Qu’a-t-il publié ? démontré ? qu’est-ce ce qui pousse à s’intéresser aujourd’hui à lui ? Comment procéder méthodologiquement pour en savoir plus sur lui ?

Schurig does not provide a proof nor a motivation for his algorithm. For more historical details, see Ahrens.

This method was discovered by Felix Walecki and reported by Lucas (1883) as a solution to the Walking schoolgirl problem (Les Promenades du Pensionnat).

Lucas, who describes the method as simple and ingenious, attributes the solution to Felix Walecki. Lucas also shows an alternative solution by means of a sliding puzzle

Other articlea are: and .

Teeds method .

Alternatively Berger tables,

In France this is called the Carousel-Berger system (Système Rutch-Berger).

See also Combinatorial design

Schurig tables.

According to the Oxford Companion, the pairing tables were first published by Richard Schurig in 1886 in Deutsche Schachzeitung v41. "With scant regard for chess history FIDE calls Schurig's creation 'Berger tables' because Johann Berger gave them, duly acknowledged, in his two Schachjahrbucher (1892-3 and 1899-1900) ,

Try to do something in Beta.

A quad is a Round-robin tournament.

Berger published the pairing tables in his two Schachjahrbucher, with due reference to its inventor Richard Schurig.

Original construction of pairing tables by Richard Schurig (1886)
For 7 or 8 players, Schurig builds a table with $$n/2$$ vertical rows and $$n-1$$ horizontal rows, as follows:

Then a second table is constructed as shown below:

By merging above tables we arrive at:

Then the first column is updated. The first or second position are alternating substituted by a bye, if $$n$$ is not even, or by $$n$$.

The pairing tables were published as an annex concerning the arrangements for the holding of master tournaments.

Relatieve ratings uit de periode Morphy
Als er toernooiresultaten bekend zijn over een langere periode, dan kunnen de relatieve ratings worden vastgesteld, ook als spelers niet tegen elkaar hebben gespeeld. Elo werkt dit uit, op basis van de onderstaande kruistabel.

De relatieve rating van een speler wordt berekend op basis de formule
 * Rp = Rc + d(p) (E1)

Hierin is Rp de eigen rating en Rc de gemiddelde rating van de tegenstanders, gewogen per gespeelde partij.

De relatieve rating wordt nu door successieve benaderingen berekend:


 * 1) Wijs aan alle spelers één initiële rating Ri toe, groot genoeg om tijdens de iteratie positief te blijven
 * 2) Vind voor iedere speler de d(p) op basis van de werkelijke score en de relatie tussen winstkans en rating verschil.
 * 3) Bereken vervolgens voor iedere speler de eerste correctie R1 op basis van regel (E1), met Rc = Ri
 * 4) Bepaal vervolgens voor iedere speler het gewogen gemiddelde van de tegenstanderratings Rc1.
 * 5) Bepaal de tweede benadering op basis van formule (E1), met Rc = Rc1
 * 6) Vervolg de berekening totdat de berekende ratings weinig veranderen.

Deze methode convergeert niet bijzonder snel.

De oplossing kan ook beschouwd worden als het nulpunt van de vergelijking:


 * We(x) = W,

waarbij We(x) de verwachte score is, afhankelijk van de relatieve ratings. Het nulpunt kan met moderne iteratieve methodes efficiënt worden bepaald. Zie ook voor een moderne benadering van dit probleem.

Irregularities in The Percentage Expectancy Table
Tables of Normal Probability Functions U. S. Department of Commerce National Bureau of Standards Applied Mathematics Series #23 Issued June 5, 1953

Fide Table Handbook Fide, https://handbook.fide.com/chapter/B022022 The Rating of Chessplayers.

Below we find the irregularities in the table: σ=2000/7 Rtg H |z-score |table x           |Tabel y 50% + 2y |Calc             | P Dif    |        |                  |                 |                 | 54 58%| 0,1890 |0,149907183209998 |0,574953591604999|0,574953591604999| 57%     88%| 1,2000 |0,769860659556583 |0,884930329778292|0,884930329778292| 343    | 1,2005 |                  |                 |0,885027364557095| 89%     88%| 1,2010 |0,770248798671795 |0,885124399335897|0,885124399335898| 344 88%| 1,2040 |0,771410421454696 |0,885705210727348|0,885705210727348| 89% *        |        |                  |                 |                 | 358 90%| 1,253  |0,789794293303286 |0,894897146651643|0,894897146651643| 89% * 392 92%| 1,372  |0,829936562186873 |0,914968281093437|0,914968281093436| 91% * 620 99%| 2,170  |0,969993154052536 |0,984996577026268|0,984996577026268| 98% *

Inflation in FIDE ratings
There is no relationship between strength and rating increase. Elo rating is differential. So if FIDE decides to lower the Elo rating of all its members in the pool by 100, the scoring probabilities between players remain the same. As more rating points are added, the average rating per player in the pool will increase.

The graph shows the expectation of the champion's rating from the FIDE rating list, compared to the ratings of the following 19 players. Robert Fischer's superiority is undeniable. The line represents the average rating development. The rating inflation since 1985 is clear.

Magnus Carlsen's rating as of February 2022 is 2865, and the average rating difference with the next 19 players is 102 (64%). Rating inflation have stabilized.

The meaning of the K factor
The K-factor determines how we weigh past and present. A low K gives more weight to past performances. (Elo, 1978, p. 25). In the following example we show the effect of the K-factor on the rating development in more detail. The K factor in this example is set to 32. The new rating (Rn) is approximately equal to the rating performance (Rp) calculated over the games played in the rating period (N = 5) supplemented by (800/K - N) = 20 fictitious draws against own rating.

Let N = 5, Ne = 20, N + Ne = 25, K = 4C / 25, R0 = 1613.

The class interval C = 200 is rooted in tradition (Elo, 1978, p. 19). In this example, the past weighs 4 times as much as the present. If K is set to 10 then the ratio between present and past becomes 1 to 15.

We note the following:

RnElo&ensp;&ensp;= 1617,33 = 1613 + 32(13,000 – 12,865) RnP800&ensp;= 1617,60 = 1613 + 32(13,000 – 12,856) RpD800&ensp;= 1617,60 = 1601,600 + 16 RpElo&emsp;= 1615,93 = 1601,600 + 14,330

As the expectation curve between -300 and +300 is almost linear we have:


 * RnElo ≈≈ RnD800 == RpD800 ≈≈ RpElo, calculated over the above 25 games.

Percentage Expectancy Curve
 * P(D) = norm.dist(D, 0, 2000 / 7, cumulative), (Elo, 1978, p.28)

Linear Approximation Formulae ( "algorithm of 400")


 * P800(D) = D / 4C + 50%, slope = 1 / 4C, intercept = 50%.
 * D800(P) = (P - 50%)4C.

kolommen
$$ \begin{array}{lll} Rn_{Elo} &=&1601.33=&1613 + 32(12.500-12.865) \\ Rn_{D800} &=&1601.60=&1613 + 32(12.500-12.856) \\ Rp_{DP800}&=&1601.60=&1601.600 + 0 \\ Rp_{Elo} &=&1601.60=&1601.600+0 \end{array} $$