You often hear players refer to others as half as good as me, while they have the same Elo. Is there a formula to measure this? It must take in account that being half as good as Carlsen, for example, is somewhere near a 2600 GM level, while half as good as a club player, is at 1000 level.

What I think that being n times weaker than a player is.

If an n times weaker player plays the player he is n times weaker, it means that if they played an infinite match, the weaker player should get n points, while the stronger player n*x.

Any other ideas? Are there any papers written on this? If my idea were to be true, is there a formula to calculate the Elo difference?

  • 11
    Define "half as good."
    – Tony Ennis
    Sep 1 '14 at 15:52
  • 4
    I've never heard someone say a player is half as good/n times as good as another player... Sep 1 '14 at 16:34
  • 2
    My point is that if you can't tell me what 'half as good' actually means, I can't tell you the ELO difference.
    – Tony Ennis
    Sep 1 '14 at 16:35
  • 3
    @MikhailTal, but then we'd just be arguing about the semantics of what people should mean by saying "half as good," and there would just be a bunch of opinions with no real question to answer. (And that point is what Tony's comments are already getting at.)
    – ETD
    Sep 1 '14 at 17:24
  • 1
    @MikhailTal Do you mean "the weaker player should get x points, while the strong player n*x"?
    – dfan
    Sep 1 '14 at 18:48

This answer addresses one concrete portion of the OP; namely, for an arbitrary N > 0, what rating difference between Player A and Player B corresponds to Player A's expected score against Player B being N times the expected score for Player B? I'll speak in terms of the USCF's Elo rating system, since I'm most familiar with it.

If Player A's expected score is N times that of Player B when they face each other, then Player A's expected score is N / (N + 1) while Player B's expected score is 1 / (N + 1) (since the two expected scores must sum to 1). Let A be Player A's rating and let B be Player B's rating. According to the USCF rating system, these ratings are intended to indicate that Player A's expected score is then

1 / (1 + 10^((B - A) / 400))

Thus if we solve the equation

1 / (1 + 10^((B - A) / 400)) = N / (N + 1)

for A - B, we will have an answer as to what rating difference corresponds to Player A's expected score being N times greater than Player B's when they match up. What we get is that

A - B = -400 * ln(1 / N) / ln(10)

So we have the following approximate rating differences corresponding to certain values of N:

N     A - B
-     -----
1     0
1.5   70
2     120
3     191
4     241
5     280
  • That was what I wanted, thank you. The Glicko System takes in account more stuff that the FIDE system. If there was also something for the FIDE, I would be grateful, but I guess it is about the same
    – MikhailTal
    Sep 1 '14 at 18:15
  • It is the same for FIDE.
    – dfan
    Sep 1 '14 at 18:49
  • 1
    @dfan, going just by Wikipedia (as I don't know any better), there are differences between the FIDE and USCF systems. My formula for Player A's expected score above, for instance, takes the form of the CDF of a logistic distribution, but according to Wikipedia's entry on Elo ratings, that differs from the FIDE system, which is still based on the normal distribution. (The source behind Wikipedia's claim on this point is a Chessbase article.)
    – ETD
    Sep 1 '14 at 19:04
  • Well, this is an interesting question. Maybe I can put a bounty in it, and make progress checks.
    – MikhailTal
    Sep 1 '14 at 19:12
  • @ETD Thanks. I will amend my comment to "It is almost the same for FIDE". Every Elo table I look at is extremely similar. Certainly there is little point (in my opinion) in obsessing whether someone who wins 2/3 of the points should be rated 120 or 121 points above his opponent.
    – dfan
    Sep 1 '14 at 19:15

Jacob Aagaard gives in a comment in his blog a rule of thumb that doubling one's strength corresponds to increasing elo by 135. He gives no citation, but it might sound plausible if you consider the amount of work needed to achieve certain level etc. This (coincidentally?) seems to be close to the value given in ETD's answer for the rating difference when the better player scores 2/3 and the worse player scores 1/3 in a match.

If you believe Aagaard's formula, then with elo difference of x, the better player is log_2(x/135) times better, and if the a player is n times better than another, their elo difference is 135*2^n.

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