Answers.Body: <p>Jacob Aagaard gives <a href="http://www.qualitychess.co.uk/blog/?p=1445#comment-10550" rel="nofollow noreferrer">in a comment in his blog</a> 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.</p>
  <p>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).</p>
  
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Answers.Text: 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.
Answers.UserDisplayName: D M
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Answers.CreationDate: 2022-09-29T22:11:14.993Z
Answers.Text: @DM Thanks for catching it! The 2^ and log_2 were swapped in the formulas.
Answers.UserDisplayName: JiK
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Answers.Text: The edited formula says 10 times better is around 448 points, which is much more reasonable.
Answers.UserDisplayName: D M
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Answers.UserDisplayName: JiK
Answers.Body: <p>This answer addresses one concrete portion of the OP; namely, for an arbitrary <code>N &gt; 0</code>, what rating difference between Player A and Player B corresponds to Player A's expected score against Player B being <code>N</code> 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.</p>
  
  <p>If Player A's expected score is <code>N</code> times that of Player B when they face each other, then Player A's expected score is <code>N / (N + 1)</code> while Player B's expected score is <code>1 / (N + 1)</code> (since the two expected scores must sum to <code>1</code>). Let <code>A</code> be Player A's rating and let <code>B</code> be Player B's rating. According to the USCF rating system, these ratings <a href="http://en.wikipedia.org/wiki/Elo_rating_system#Mathematical_details">are intended to indicate</a> that Player A's expected score is then</p>
  
  <pre><code>1 / (1 + 10^((B - A) / 400))
  </code></pre>
  
  <p>Thus if we solve the equation</p>
  
  <pre><code>1 / (1 + 10^((B - A) / 400)) = N / (N + 1)
  </code></pre>
  
  <p>for <code>A - B</code>, we will have an answer as to what rating difference corresponds to Player A's expected score being <code>N</code> times greater than Player B's when they match up. What we get is that</p>
  
  <pre><code>A - B = -400 * ln(1 / N) / ln(10)
  </code></pre>
  
  <p>So we have the following approximate rating differences corresponding to certain values of <code>N</code>:</p>
  
  <pre><code>N     A - B
  -     -----
  1     0
  1.5   70
  2     120
  3     191
  4     241
  5     280
  </code></pre>
  
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Answers.Text: 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.
Answers.UserDisplayName: D M
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Answers.Text: 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
Answers.UserDisplayName: MikhailTal
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Answers.Text: It is the same for FIDE.
Answers.UserDisplayName: dfan
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Answers.Text: @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](http://en.wikipedia.org/wiki/Logistic_distribution#Cumulative_distribution_function), 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](http://en.chessbase.com/post/arpad-elo-and-the-elo-rating-system).)
Answers.UserDisplayName: ETD
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Answers.Text: Well, this is an interesting question. Maybe I can put a bounty in it, and make progress checks.
Answers.UserDisplayName: MikhailTal
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Answers.Text: @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.
Answers.UserDisplayName: dfan
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Answers.Text: @dfan, understood. I was reading more into your "the same" than you intended. With my chess player hat on, I agree that the differences aren't particularly crucial.
Answers.UserDisplayName: ETD
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Answers.Text: Indeed they aren't, but still supplying both of them would be much more interesting.
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Body: <p>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.</p>
  
  <p>What I think that being n times weaker than a player is.</p>
  
  <p>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.</p>
  
  <p>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?</p>
  
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2014-09-01T15:52:21.087Z
Comments.Text: Define "half as good."
Comments.UserDisplayName: Tony Ennis
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Comments.CreationDate: 2014-09-01T16:14:37.007Z
Comments.Text: That is what the question asks. What is the meaning of half as good, or n times worse, in terms of ELO or other solutions.
Comments.UserDisplayName: MikhailTal
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2014-09-01T16:34:04.007Z
Comments.Text: I've never heard someone say a player is half as good/n times as good as another player...
Comments.UserDisplayName: Patrick Coulombe
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Comments.CreationDate: 2014-09-01T16:35:38.777Z
Comments.Text: My point is that if you can't tell me what 'half as good' actually means, I can't tell you the ELO difference.
Comments.UserDisplayName: Tony Ennis
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Comments.CreationDate: 2014-09-01T17:06:48.437Z
Comments.Text: @Patrick It was actually "He barely knows half of what I know", but still this sprung in my mind.
Comments.UserDisplayName: MikhailTal
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Comments.CreationDate: 2014-09-01T17:07:50.170Z
Comments.Text: @TonyEnnis That is the point. Form an idea for what actually half as good means. You can read mine in the post. Maybe you have something you think is more accurate.
Comments.UserDisplayName: MikhailTal
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Comments.CreationDate: 2014-09-01T17:24:44.000Z
Comments.Text: @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.)
Comments.UserDisplayName: ETD
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Comments.CreationDate: 2014-09-01T18:48:48.683Z
Comments.Text: @MikhailTal Do you mean "the weaker player should get _x_ points, while the strong player n*x"?
Comments.UserDisplayName: dfan
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Comments.CreationDate: 2014-09-01T23:32:37.433Z
Comments.Text: I would define half as good as something like, in a hypothetical tournament of 12 games between these players the score is expected to come out like 4-8.
Comments.UserDisplayName: wim
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Comments.CreationDate: 2014-09-02T07:58:42.573Z
Comments.Text: Yep. that is my idea.
Comments.UserDisplayName: MikhailTal
Tags: <rating><elo>
Title: What is the Elo of a player who is half as good as another player?
ThreadText: # What is the Elo of a player who is half as good as another player?
  
  (Asked by: username_0 on 2014-09-01)
  
  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?
  
  **Tags:** rating, elo
  
  > **Comments:**
  > **username_1**: Define "half as good."  
  > **username_0**: That is what the question asks. What is the meaning of half as good, or n times worse, in terms of ELO or other solutions.  
  > **username_2**: I've never heard someone say a player is half as good/n times as good as another player...  
  > **username_1**: My point is that if you can't tell me what 'half as good' actually means, I can't tell you the ELO difference.  
  > **username_0**: @username_2 It was actually "He barely knows half of what I know", but still this sprung in my mind.  
  > **username_0**: @username_1 That is the point. Form an idea for what actually half as good means. You can read mine in the post. Maybe you have something you think is more accurate.  
  > **username_3**: @username_0, 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 username_1's comments are already getting at.)  
  > **username_4**: @username_0 Do you mean "the weaker player should get _x_ points, while the strong player n*x"?  
  > **username_5**: I would define half as good as something like, in a hypothetical tournament of 12 games between these players the score is expected to come out like 4-8.  
  > **username_0**: Yep. that is my idea.  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_3 on 2014-09-01. Score: 14 | Accepted answer
  
  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](http://en.wikipedia.org/wiki/Elo_rating_system#Mathematical_details) 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
  
  ```
  
  > **Comments:**
  > **username_0**: 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  
  > **username_4**: It is the same for FIDE.  
  > **username_3**: @username_4, 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](http://en.wikipedia.org/wiki/Logistic_distribution#Cumulative_distribution_function), 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](http://en.chessbase.com/post/arpad-elo-and-the-elo-rating-system).)  
  > **username_0**: Well, this is an interesting question. Maybe I can put a bounty in it, and make progress checks.  
  > **username_4**: @username_3 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.  
  > **username_3**: @username_4, understood. I was reading more into your "the same" than you intended. With my chess player hat on, I agree that the differences aren't particularly crucial.  
  > **username_0**: Indeed they aren't, but still supplying both of them would be much more interesting.  
  > **username_7**: 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.  
  
  ---
  
  ### Answer by username_6 on 2014-09-02. Score: 3
  
  Jacob Aagaard gives [in a comment in his blog](http://www.qualitychess.co.uk/blog/?p=1445#comment-10550) 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).
  
  > **Comments:**
  > **username_7**: 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.  
  > **username_6**: @DM Thanks for catching it! The 2^ and log_2 were swapped in the formulas.  
  > **username_7**: The edited formula says 10 times better is around 448 points, which is much more reasonable.  
  
  ---
  
  