Answers.Body: <p>Indeed the limit of this scenario would be infinity for the perpetually winning player. This is because FIDE has a rule saying that if one player is rated more than 400 points higher than his opponent, their difference should be set to 400 when calculating rating gain/loss. That means no matter how weak the opposition is, our winning player will receive minimum 0.8 rating points for winning. </p>
  
  <p>See this discussion as to why this rule exists: <a href="https://chess.stackexchange.com/questions/22328/why-is-there-a-minimum-rating-gain-when-you-win">Why is there a minimum rating gain when you win?</a></p>
  
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Answers.Text: Thanks. By excluding cut offs I also meant to exclude the 400 point rule. Is my understanding correct, that in this case, the rating limit would be 3000?
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Answers.Body: <p>If we work solely from the standard mathematical formula for Elo ratings, it's possible to prove that the winning player can get an arbitrarily high rating, but that it'll probably take exponentially long (i.e., the winning player's rating increases logarithmically).</p>
  
  <p>I'll suppose that the initial Elo rating of the two players is 0, not 1500, because it makes the math simpler. (If this bugs you, just add 1500 to every number below.) Let r(n) be the rating of the winning player after n games, so the rating of the losing player is -r(n). Then in the (n+1)st game, the rating difference is 2r(n).
  By the calculation given in Brian's answer, r(n) obeys the formula</p>
  
  <p>r(n+1)-r(n)=K/(10^(r(n)/200)+1)</p>
  
  <p>with r(0)=1.</p>
  
  <p>This recursion is painful to solve explicitly, but we can use a differential equation to approximate the solution. Explicitly, let f(r)=10^(r/200)+1; then the above recursion can be approximated as</p>
  
  <p>r'(t)=K/f(r(t)).</p>
  
  <p>Intuitively, passing to the differential equation is considering the limit of the process where we repeatedly play twice as many games, but with a K-factor that's half as large. Every time we do this, we slow down the rating growth a little bit, because the winning player has a higher rating as they play the second game in each pair than they do in the first game, so the second game leads to slightly less Elo growth than the first game.</p>
  
  <p>The differential equation r'(t)=K/f(r(t)) is a separable differential equation, which can be solved. I won't bore you with the details on a platform that doesn't let me write formulas, but the solution turns out to obey the equation</p>
  
  <p>t=(200e^(r/200)-200+r ln 10)/(2K ln 10) .</p>
  
  <p>Notice that this is an equation for t (the number of games played) in terms of r (the rating of the better player). There's essentially no nice way to invert this equation to write r in terms of t. However, we can notice that t is well-defined for any positive r, and is an increasing function of r. So, for any given rating, we can find the amount of time required to achieve that rating, which means it's possible to achieve arbitrarily high ratings.</p>
  
  <p>But since t is an exponential function of r, the amount of time taken may be exponentially long; that is, r will grow logarithmically over time.</p>
  
  <p>This argument shows that the winning player's rating can grow arbitrarily large in the limit where we played many more games, but with a correspondingly smaller K-factor. But as we argued above, the winning player's rating increases faster in the original setup than it does in this limiting setup. Since ratings can grow arbitrarily large in the limiting setup, this means that ratings can also grow arbitrarily large in the original setup.</p>
  
  <p>I'm reasonably certain (but have not proved) that this growth is still logarithmic.</p>
  
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Answers.Text: (If you ask this question on Math.SE, I'd be happy to provide a version of this answer where the intermediate algebraic steps are included. But I don't really want to write them out in plain text...)
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Answers.Text: Thanks. Great answer. I asked at Math.SE: https://math.stackexchange.com/questions/3669086/how-does-elo-rating-scale-with-games-played
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Answers.Body: <p>It kind of depends exactly what you mean by artificial cutoffs. There is one explicitly stated artificial cutoff in the <a href="https://old.fide.com/fide/handbook.html?id=197&amp;view=article" rel="noreferrer">FIDE rating regulations</a>. That says that if the rating difference is greater than 400 it will be treated as 400. But there is another hidden cutoff. The actual elo algorithm for calculating expected score or scoring probability involves power functions but FIDE provide a look-up table with approximations. This bottoms out when the rating difference is 735 with the expected score of the stronger player at 1.</p>
  
  <p>So, if the 400 point cutoff is excluded but the exact FIDE algorithm with its table of expected scores is used then the maximum rating difference is 735 points and once the two players' ratings reach 2235 and 765 further wins for the stronger player will not change ratings. That is part of the reason the 400 point rule was introduced.</p>
  
  <p>If, however, you are not talking about FIDE's implementation of the elo algorithms then the problem becomes a purely mathematical one. I look at the algorithms below.</p>
  
  <p>The formulae for working out rating change for elo ratings work like this:</p>
  
  <p>Suppose the initial ratings of the two players are r(1) and r(2)  and their recalculated ratings (which we want to work out) are r'(1) and r'(2).</p>
  
  <p>First we need to calculate a power function for each initial rating to use to calculate the expected result of the match. So, call these power functions for the two players R(1) and R(2) and their expected scores E(1) and E(2).</p>
  
  <p>So -<br>
  R(1) = 10^(r(1)/400) and R(2) = 10^(r(2)/400)</p>
  
  <p>And -<br>
  E(1) = R(1) / (R(1) + R(2)) and E(2) = (1 - E(1)) = R(2) / (R(1) + R(2))</p>
  
  <p>Now let S(1) = player 1's score in the game (going to be 1 in this case) and<br>
  let S(2) = player 2's score in the game (going to be 0 in this case) then</p>
  
  <p>r'(1) = r(1) + k * (S(1) – E(1)) and<br>
  r'(2) = r(2) + K * (S(2) – E(2)) where k is the k factor</p>
  
  <p>If there are no artificial cutoffs then the two ratings will eventually reach 3000 and 0 (with the same k factor for both, one player's rating gain is the other player's rating loss).</p>
  
  <p>What happens in the next game?</p>
  
  <p>Well r(1) = 3000 and r(2) = 0 so -<br>
  R(1) = 10^7.5 = 3162276.6 (to 8 sf) and<br>
  R(2) = 1<br>
  E(1) = 0.999999968<br>
  E(2) = 0.000000032, so<br>
  r'(2) = -0.000000032*k</p>
  
  <p>The second player's rating continues to decrease and goes negative.</p>
  
  <p>I'm fairly sure that as these algorithms are power function based the limits will be power function limits, i.e. the first player's rating will tend to infinity and the second's to -infinity</p>
  
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Answers.UserDisplayName: Brian Towers
Body: <p>I only take two players starting out at 1500 Elo and have them play against each other with one player winning all games. Also I leave out any artificially introduced cut-offs.</p>
  
  <p>So the winner is constantly gaining rating while the other player is losing Elo rating. The gain is decreasing because the rating difference of the players increases.</p>
  
  <p>How does (mathematically) the raiting gain scale with the number of games (n) played for large n?</p>
  
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Comments.Text: Just a guess but I think that it would approach to a finite limit as n tends to infinity.
Comments.UserDisplayName: AKP2002
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Comments.Text: @AKP2002 Just to clarify. Do you mean the rating goes to a finite limite or the rating gain per game?
Comments.UserDisplayName: user1583209
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Comments.CreationDate: 2020-05-10T09:47:29.237Z
Comments.Text: Iwas taking about rating, rating gain would become infinitesimally small .
Comments.UserDisplayName: AKP2002
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Comments.CreationDate: 2022-01-24T02:05:18.673Z
Comments.Text: Does this answer your question? [Why is there a minimum rating gain when you win?](https://chess.stackexchange.com/questions/22328/why-is-there-a-minimum-rating-gain-when-you-win)
Comments.UserDisplayName: BCLC
Tags: <rating><elo><mathematics>
Title: How does Elo rating gain scale with games played?
ThreadText: # How does Elo rating gain scale with games played?
  
  (Asked by: username_0 on 2020-05-10)
  
  I only take two players starting out at 1500 Elo and have them play against each other with one player winning all games. Also I leave out any artificially introduced cut-offs.
  
  So the winner is constantly gaining rating while the other player is losing Elo rating. The gain is decreasing because the rating difference of the players increases.
  
  How does (mathematically) the raiting gain scale with the number of games (n) played for large n?
  
  **Tags:** rating, elo, mathematics
  
  > **Comments:**
  > **username_1**: Just a guess but I think that it would approach to a finite limit as n tends to infinity.  
  > **username_0**: @username_1 Just to clarify. Do you mean the rating goes to a finite limite or the rating gain per game?  
  > **username_1**: Iwas taking about rating, rating gain would become infinitesimally small .  
  > **username_2**: Does this answer your question? [Why is there a minimum rating gain when you win?](https://chess.stackexchange.com/questions/22328/why-is-there-a-minimum-rating-gain-when-you-win)  
  
  ---
  
  ## 3 answers
  
  ### Answer by username_5 on 2020-05-10. Score: 7 | Accepted answer
  
  It kind of depends exactly what you mean by artificial cutoffs. There is one explicitly stated artificial cutoff in the [FIDE rating regulations](https://old.fide.com/fide/handbook.html?id=197&view=article). That says that if the rating difference is greater than 400 it will be treated as 400. But there is another hidden cutoff. The actual elo algorithm for calculating expected score or scoring probability involves power functions but FIDE provide a look-up table with approximations. This bottoms out when the rating difference is 735 with the expected score of the stronger player at 1.
  
  So, if the 400 point cutoff is excluded but the exact FIDE algorithm with its table of expected scores is used then the maximum rating difference is 735 points and once the two players' ratings reach 2235 and 765 further wins for the stronger player will not change ratings. That is part of the reason the 400 point rule was introduced.
  
  If, however, you are not talking about FIDE's implementation of the elo algorithms then the problem becomes a purely mathematical one. I look at the algorithms below.
  
  The formulae for working out rating change for elo ratings work like this:
  
  Suppose the initial ratings of the two players are r(1) and r(2) and their recalculated ratings (which we want to work out) are r'(1) and r'(2).
  
  First we need to calculate a power function for each initial rating to use to calculate the expected result of the match. So, call these power functions for the two players R(1) and R(2) and their expected scores E(1) and E(2).
  
  So -  
  R(1) = 10^(r(1)/400) and R(2) = 10^(r(2)/400)
  
  And -  
  E(1) = R(1) / (R(1) + R(2)) and E(2) = (1 - E(1)) = R(2) / (R(1) + R(2))
  
  Now let S(1) = player 1's score in the game (going to be 1 in this case) and  
  let S(2) = player 2's score in the game (going to be 0 in this case) then
  
  r'(1) = r(1) + k \* (S(1) – E(1)) and  
  r'(2) = r(2) + K \* (S(2) – E(2)) where k is the k factor
  
  If there are no artificial cutoffs then the two ratings will eventually reach 3000 and 0 (with the same k factor for both, one player's rating gain is the other player's rating loss).
  
  What happens in the next game?
  
  Well r(1) = 3000 and r(2) = 0 so -  
  R(1) = 10^7.5 = 3162276.6 (to 8 sf) and  
  R(2) = 1  
  E(1) = 0.999999968  
  E(2) = 0.000000032, so  
  r'(2) = -0.000000032\*k
  
  The second player's rating continues to decrease and goes negative.
  
  I'm fairly sure that as these algorithms are power function based the limits will be power function limits, i.e. the first player's rating will tend to infinity and the second's to -infinity
  
  
  ---
  
  ### Answer by username_3 on 2020-05-10. Score: 4
  
  Indeed the limit of this scenario would be infinity for the perpetually winning player. This is because FIDE has a rule saying that if one player is rated more than 400 points higher than his opponent, their difference should be set to 400 when calculating rating gain/loss. That means no matter how weak the opposition is, our winning player will receive minimum 0.8 rating points for winning.
  
  See this discussion as to why this rule exists: [Why is there a minimum rating gain when you win?](https://chess.stackexchange.com/questions/22328/why-is-there-a-minimum-rating-gain-when-you-win)
  
  > **Comments:**
  > **username_0**: Thanks. By excluding cut offs I also meant to exclude the 400 point rule. Is my understanding correct, that in this case, the rating limit would be 3000?  
  
  ---
  
  ### Answer by username_4 on 2020-05-10. Score: 3
  
  If we work solely from the standard mathematical formula for Elo ratings, it's possible to prove that the winning player can get an arbitrarily high rating, but that it'll probably take exponentially long (i.e., the winning player's rating increases logarithmically).
  
  I'll suppose that the initial Elo rating of the two players is 0, not 1500, because it makes the math simpler. (If this bugs you, just add 1500 to every number below.) Let r(n) be the rating of the winning player after n games, so the rating of the losing player is -r(n). Then in the (n+1)st game, the rating difference is 2r(n).
  By the calculation given in username_5's answer, r(n) obeys the formula
  
  r(n+1)-r(n)=K/(10^(r(n)/200)+1)
  
  with r(0)=1.
  
  This recursion is painful to solve explicitly, but we can use a differential equation to approximate the solution. Explicitly, let f(r)=10^(r/200)+1; then the above recursion can be approximated as
  
  r'(t)=K/f(r(t)).
  
  Intuitively, passing to the differential equation is considering the limit of the process where we repeatedly play twice as many games, but with a K-factor that's half as large. Every time we do this, we slow down the rating growth a little bit, because the winning player has a higher rating as they play the second game in each pair than they do in the first game, so the second game leads to slightly less Elo growth than the first game.
  
  The differential equation r'(t)=K/f(r(t)) is a separable differential equation, which can be solved. I won't bore you with the details on a platform that doesn't let me write formulas, but the solution turns out to obey the equation
  
  t=(200e^(r/200)-200+r ln 10)/(2K ln 10) .
  
  Notice that this is an equation for t (the number of games played) in terms of r (the rating of the better player). There's essentially no nice way to invert this equation to write r in terms of t. However, we can notice that t is well-defined for any positive r, and is an increasing function of r. So, for any given rating, we can find the amount of time required to achieve that rating, which means it's possible to achieve arbitrarily high ratings.
  
  But since t is an exponential function of r, the amount of time taken may be exponentially long; that is, r will grow logarithmically over time.
  
  This argument shows that the winning player's rating can grow arbitrarily large in the limit where we played many more games, but with a correspondingly smaller K-factor. But as we argued above, the winning player's rating increases faster in the original setup than it does in this limiting setup. Since ratings can grow arbitrarily large in the limiting setup, this means that ratings can also grow arbitrarily large in the original setup.
  
  I'm reasonably certain (but have not proved) that this growth is still logarithmic.
  
  > **Comments:**
  > **username_4**: (If you ask this question on Math.SE, I'd be happy to provide a version of this answer where the intermediate algebraic steps are included. But I don't really want to write them out in plain text...)  
  > **username_0**: Thanks. Great answer. I asked at Math.SE: https://math.stackexchange.com/questions/3669086/how-does-elo-rating-scale-with-games-played  
  
  ---
  
  