user4543: I read somewhere that if you're around 80 points better than another player, statistically you should get around twice as many points against the same opposition. This impressed me because it's not how we generally think about ratings. I worked out that Magnus Carlsen is about 64 times the player I am, which sounds about right.
My question is, can anyone verify this using the mathematics of the Elo system?

user12873: It depends on the FIDE ratings of the players. If Person A is rated 1300 and Person B is 1220, A is not twice as good as B.  Conversly, Carlsen is rated 2843 FIDE, and I would argue he is at least twice as good as a 2763 player (if they played 10 matches, each match 10 games long, Carlsen would almost definitely win 9/10 to 10/10 matches).
The reason for this is that it becomes harder to keep increasing as you get a higher rating. 99% of chess players are below 2200, even though there's an extra 600 points above. The trend on a graph of rating vs #players is not linear; it is closer to an exponential decay function. There are vast number of players under 1400, but only an extremely select few over 2800 at a time (usually 5 players max).
Chess is a game where most people can go to roughly 1800ish with hard and dedicated work. However, only people with true talent can continue past that point.  Then, once hitting 2000, an even smaller number of people can continue forward.  This phenomena becomes stronger as you climb up the rating latter, which explains why such a small percent can ever make it to GM level.
As a result, if Person A and Person B are in a high rating bracket, Person A being 80 points higher indicates he truly has an additional "special something". Meanwhile, if A and B were in a low rating bracket, A being 80 points higher could be attributed to something like playing in a few more tournaments.
EDIT - Carlsen example fixed.

user18101: 120 points of rating difference, are expected to produce 67% of game points for the stronger player.
This apply for all ratings, so is true for a game of a 2800 vs 2680, as for a game of a 1600 vs 1480.
Anyway, expected game points are useful to calculate rating variation, not directly the players comparision.
If this really means "strong twice" or not, it is a subjective point of view.

user12523: Given the equation that models the expected score (already posted by Glorfindel), it is mathematically impossible for player A, rated X+80 to score twice as much as player B (rated X) against the same opposition (rated Y). The closest you can get is if Y is very high, in which case the odds of winning are vanishingly small but A has 58% higher chances than B.
If you change the rating difference to 120 points, then the assertion can be true, again given that Y is much higher. For example, if X=1000 and Y=2000, player A is expected to score 0.006 and player B 0.003. Also if A plays against B, the expected scores are 0.67 to 0.33, so in that sense you can say that a 120-point rating difference makes a player "twice as good". If we consider a less extreme value for Y, say 1400, then the expected scores are 0.166 for A and 0.091 for B, where A scores almost but not quite twice as much as B.

user7297: This can't be true. I (rated ~1900) expect to score 100% against a 1000 player. I don't think a 1980 rated user is able to score 200% against the same opposition.
In table 8.1b in the FIDE rating regulations, you'll find the expected scores corresponding to a rating difference. A rating difference of 80 corresponds to an expected score of 0.61, which is much lower than twice the 0.5 expected score for equally rated players. 
The only point in the table where a rating difference of 80 corresponds to a doubled expected score is at the very end. Against opposition rated 500 points higher than you, your expected score is 0.04; against opposition rated 580 points higher it's only 0.02. The expected score formula is explained on Wikipedia:

If Player A has a rating of RA and Player B a rating of RB, the exact formula (using the logistic curve) for the expected score of Player A is 
EA = 1 / (1 + 10 (RB - RA) / 400)


4543: <p>I read somewhere that if you're around 80 points better than another player, statistically you should get around twice as many points against the same opposition. This impressed me because it's not how we generally think about ratings. I worked out that Magnus Carlsen is about 64 times the player I am, which sounds about right.</p>

<p>My question is, can anyone verify this using the mathematics of the Elo system?</p>
 12873: <p>It depends on the FIDE ratings of the players. If Person A is rated 1300 and Person B is 1220, A is not twice as good as B.  Conversly, Carlsen is rated 2843 FIDE, and I would argue he is at least twice as good as a 2763 player (if they played 10 matches, each match 10 games long, Carlsen would almost definitely win 9/10 to 10/10 matches).</p>

<p>The reason for this is that it becomes harder to keep increasing as you get a higher rating. 99% of chess players are below 2200, even though there's an extra 600 points above. The trend on a graph of rating vs #players is not linear; it is closer to an exponential decay function. There are vast number of players under 1400, but only an extremely select few over 2800 at a time (usually 5 players max).</p>

<p>Chess is a game where most people can go to roughly 1800ish with hard and dedicated work. However, only people with true talent can continue past that point.  Then, once hitting 2000, an even smaller number of people can continue forward.  This phenomena becomes stronger as you climb up the rating latter, which explains why such a small percent can ever make it to GM level.</p>

<p>As a result, if Person A and Person B are in a high rating bracket, Person A being 80 points higher indicates he truly has an additional "special something". Meanwhile, if A and B were in a low rating bracket, A being 80 points higher could be attributed to something like playing in a few more tournaments.</p>

<p>EDIT - Carlsen example fixed.</p>
 18101: <p>120 points of rating difference, are expected to produce 67% of game points for the stronger player.
This apply for all ratings, so is true for a game of a 2800 vs 2680, as for a game of a 1600 vs 1480.
Anyway, expected game points are useful to calculate rating variation, not directly the players comparision.
If this really means "strong twice" or not, it is a subjective point of view.</p>
 12523: <p>Given the equation that models the expected score (already posted by Glorfindel), it is mathematically impossible for player <strong>A</strong>, rated <em>X</em>+80 to score twice as much as player <strong>B</strong> (rated <em>X</em>) against the same opposition (rated <em>Y</em>). The closest you can get is if <em>Y</em> is very high, in which case the odds of winning are vanishingly small but <strong>A</strong> has 58% higher chances than <strong>B</strong>.</p>

<p>If you change the rating difference to 120 points, then the assertion can be true, again given that <em>Y</em> is much higher. For example, if <em>X</em>=1000 and <em>Y</em>=2000, player <strong>A</strong> is expected to score 0.006 and player <strong>B</strong> 0.003. Also if <strong>A</strong> plays against <strong>B</strong>, the expected scores are 0.67 to 0.33, so in that sense you can say that a 120-point rating difference makes a player "twice as good". If we consider a less extreme value for <em>Y</em>, say 1400, then the expected scores are 0.166 for <strong>A</strong> and 0.091 for <strong>B</strong>, where <strong>A</strong> scores almost but not quite twice as much as <strong>B</strong>.</p>
 7297: <p>This can't be true. I (rated ~1900) expect to score 100% against a 1000 player. I don't think a 1980 rated user is able to score 200% against the same opposition.</p>

<p>In table 8.1b in the <a href="http://www.fide.com/component/handbook/?id=197&amp;view=article" rel="noreferrer">FIDE rating regulations</a>, you'll find the expected scores corresponding to a rating difference. A rating difference of 80 corresponds to an expected score of 0.61, which is much lower than twice the 0.5 expected score for equally rated players. </p>

<p>The only point in the table where a rating difference of 80 corresponds to a doubled expected score is at the very end. Against opposition rated 500 points <em>higher</em> than you, your expected score is 0.04; against opposition rated 580 points higher it's only 0.02. The expected score formula is explained <a href="https://en.wikipedia.org/wiki/Elo_rating_system#Mathematical_details" rel="noreferrer">on Wikipedia</a>:</p>

<blockquote>
  <p>If Player A has a rating of R<sub>A</sub> and Player B a rating of R<sub>B</sub>, the exact formula <a href="https://en.wikipedia.org/wiki/Logistic_curve" rel="noreferrer">(using the logistic curve)</a> for the expected score of Player A is </p>
  
  <p>E<sub>A</sub> = 1 / (1 + 10 <sup>(R<sub>B</sub> - R<sub>A</sub>) / 400</sup>)</p>
</blockquote>
