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Background:

The result of a chess game between two players is a win ,a loss or a draw which are (usually) scored respectively $1$ point, $0$ point or $0.5$ point for the appropriate player. Team matches, between two teams, are played in which each team's score is the sum of the scores of their players.

National and international chess rating systems periodically (typically monthly, six monthly or annually) assign to individual players a rating based on their (recent) games. The difference in the ratings between two players gives a prediction of the expected score of each player in a single game. For example in the English Chess Federation (ECF) system a rating point difference of $20$ points predicts the expected score of the weaker player is $0.3$ and that of the stronger player is $0.7$. Similarly for a $40$ point rating difference the expected scores are $0.1$ and $0.9$. In the ECF system when rating point differences are more than $40$ points special rules apply.

For (typical) illustration in an individual game with a $20$ point rating difference the probabilities of loss , draw or win for the weaker player are respectively: $0.7-x$, $2x$, $0.3-x$ where $0 \le x \le 0.3$, $x$ is unknowable and the application of a current $20$ point difference to a specific game is questionable. However $x = 0$ corresponds to the maximum variance case and $x=0.3$ corresponds to the minimum variance case. This problem has its origins in the London (UK) Chess League ,Division 1 where teams have $12$ players each. In that case , if teams differ by $20$ points on average the expected score is $8.4:3.6$. In the London League there is often more than one team from the same club in the same Division.

Thus assuming normality for the match score, bounds for the probability of extreme events, such as the weaker side at least drawing the match, can be made in the obvious way . Simulation can be used to check normality or directly to obtain probabilities. Chebyshev's theorem may also be useful.

My question

My question concerns the non-normal case and asks for pathological examples (if any) with rating point differences less than $40$ points where the probability of an extreme team event (say the probability of the weaker side winning) is not bounded by those obtained from the minimum and maximum variance cases defined in the obvious way. Generalisations might also be interesting.

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The average score difference does not suffice to predict the probability of the outcome. Suppose all players in team A are of equal strength, while all but one player in team B are somewhat stronger then the players of team A, but one player is a lot weaker. Then team B will win most of the first 11 games, and lose the last one. So you can arrange the scores in such a way that team B will most likely win close to 11:1, yet the average score of team A is higher then the average score of team B.

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  • $\begingroup$ As this is the only response, I am reluctant to criticiseit. However it does not address the question. $\endgroup$ – Ian Calvert Feb 15 '15 at 12:36
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For teams of 1, n=1, the results are trivial.

For n=2, and each match having an expected score for the weaker side of 0.3, elementary probability and contour maps can be used to illustrate that :

(I) the probability of the weaker side at least drawing is maximised if both x's are 0, (II)the probability of the weaker side scoring at least 1.5 is maximised if one x is 0 and the other is 0.3.

I know nothing about higher n, except n=12, when I suspect, in all practical cases, x(1)...x(12) equal zero: the "go for it" solution. In this seasons London League there are three "Drunken Knights" teams

In any event, I still believe and hope that the general n behaviour is an appropriate question for this site.

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