user3405: 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?

user1761: 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).

user167: 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


3405: <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>
 1761: <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>
 167: <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>
