This guide gives the probability of player $A$ winning a game against player $B$, using their Elo ratings $R_A$, $R_B$:

$$P(A) = \frac{1}{1+10^m}$$



What is the expected number of points obtained by player $A$ in an $X$ games match, if a win is worth $1$ point, a draw is worth $0.5$ points and a loss is worth $0$ points?

Note: I am not sure if the games should be considered independent or not. One could also factor in the Elo rating changes after one player wins the first game, and the second etc. I'm not sure how much this would complicate computations, so ignoring this is also fine if considering it would be too complex.

  • $\begingroup$ $P(A)X$ and this is explained in the thread you link to. The $P(A)$ is not litterally the probability of winning, but the winning prob, that is the expected [fraction] of points per game. For an actual real match there will of course be other factors like a player that is behind might have to play more agressive in later stages, or one that is ahead by some marging might go for quick draws rather than the win but the influence of that is obviously not obtainable from the ELO number and mathematical considerations. $\endgroup$ – quid Sep 1 '16 at 14:46
  • $\begingroup$ @quid it's not explained in terms of $X$ games, just a single game. The author even highlights this. I am interested in the expected points obtained during an $X$ games match. Of course, there are other non-mathematical considerations, but ignoring those. $\endgroup$ – IVlad Sep 1 '16 at 14:53
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    $\begingroup$ It is mentioned there: "This is for single games, not matches. Carlsen's .61 does not mean he has a 61% chance of winning a match of say 10 games against Anand. It means that he is expected to get 61% of the points, or 6.1 points." The answer is $P(A)X$ as long as the rating is assumed constant. And, but I did not double check this, things are calibrated in such a way that it should stay rather constant. I mean that's the point of it all. The rating predicts the outcome, if you let things play out purely along the lines of what the rating predicts it should not change. $\endgroup$ – quid Sep 1 '16 at 15:02
  • $\begingroup$ Ah, that's true. Would it be a lot harder to compute if also taking rating changes between games into consideration? $\endgroup$ – IVlad Sep 1 '16 at 15:10
  • $\begingroup$ @IVlad as a rool ratings calculates after the tournament. Another way leads to a calculating paradoxes $\endgroup$ – Yuri Negometyanov Sep 1 '16 at 15:13

The probabilities of the results in each game of $A$ against $B$ are $$P(W)=\dfrac1{10^m+1}$$ for a win, $$P(L)=\dfrac1{10^{-m}+1}$$ for a loss, $$P(D)=1-P(W)-P(L)$$ for a draw, so expected number of points of $A$ vs $B$ in X games is $$p=X(P(W)+\dfrac12P(D)),$$ or $$p(A)=\dfrac{X}2\left(1+\dfrac1{10^{(R_B-R_A)/400}+1}-\dfrac1{10^{(R_A-R_B)/400}+1}\right)$$

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    $\begingroup$ The $P$ in the question is not the probability for a win. $\endgroup$ – quid Sep 1 '16 at 15:03

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