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?

  • 18
    Define "half as good."
    – Tony Ennis
    Commented Sep 1, 2014 at 15:52
  • 4
    I've never heard someone say a player is half as good/n times as good as another player... Commented Sep 1, 2014 at 16:34
  • 7
    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
    Commented Sep 1, 2014 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
    Commented Sep 1, 2014 at 17:24
  • 1
    @MikhailTal Do you mean "the weaker player should get x points, while the strong player n*x"?
    – dfan
    Commented Sep 1, 2014 at 18:48

2 Answers 2


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
    Commented Sep 1, 2014 at 18:15
  • It is the same for FIDE.
    – dfan
    Commented Sep 1, 2014 at 18:49
  • 2
    @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
    Commented Sep 1, 2014 at 19:04
  • Well, this is an interesting question. Maybe I can put a bounty in it, and make progress checks.
    – MikhailTal
    Commented Sep 1, 2014 at 19:12
  • 1
    On the FIDE chart, there's no indication of exactly where 2/3 is, but a difference of 114-121 points gives an expected score of 0.66, and 122-129 gives an expected score of 0.67. So 120 is right in there.
    – D M
    Commented Oct 2, 2018 at 11:12

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 2^(x/135) times better, and if a player is n times better than another, their elo difference is 135*log_2(n).

  • 2
    That last formula can't be right. If someone is 10 times better, you're saying that the Elo difference would be 135*2^10, which would be 138240. That's plainly ridiculous.
    – D M
    Commented Sep 4, 2022 at 17:25
  • 1
    @DM Thanks for catching it! The 2^ and log_2 were swapped in the formulas.
    – JiK
    Commented Sep 29, 2022 at 22:11
  • The edited formula says 10 times better is around 448 points, which is much more reasonable.
    – D M
    Commented Sep 29, 2022 at 22:24

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