Answers.Body: <p>It depends which quantities you are interested in; the whole distribution is a complex concept to describe.</p>
  <p>Statistics that look at the score of a single player (mean, variance, etc.) are easy to compute, because they are the sum of independent game outcomes. It is essentially (up to rescaling) a <a href="https://en.wikipedia.org/wiki/Multinomial_distribution" rel="nofollow noreferrer">multinomial distribution</a> with k=3. So average scores, standard deviations, etc. are easy to compute. And the central limit theorem predicts that the limit for n to infinity is a Gaussian distribution with mean and variance that are easy to compute.</p>
  <p>The trickier part is finding out quantities that involve the joint distribution of 2 or more player scores, since these scores are not independent: for instance, no two players can both score 0 in a round-robin tournament. So probabilities like &quot;the probability that 5 different players score at least m points&quot; are more complicated.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2022-11-03T08:21:26.347Z
Answers.Text: Agrees with what my instinct said (alas, I'm no math profi :-) - and the complicated joint distributions are only interesting for a profi...
Answers.UserDisplayName: Hauke Reddmann
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Answers.CreationDate: 2022-11-03T08:14:16.780Z
Answers.LastActivityDate: 2022-11-03T08:14:16.780Z
Answers.UserDisplayName: Federico Poloni
Body: <p>Maybe this is a bit too much towards math, but we have Noam :-)</p>
  <p>If you view the final point standings of any tournament, you'll probably see many players with an average number of points, but few with a very high or low score. For Swiss this is almost by construction, but for a round robin (easier to analyze) it also resembles a Gaussian distribution.</p>
  <p>My question is whether it is actually one, or something only similar, assuming some simplifications:</p>
  <ul>
  <li>n players play a round robin with n-1 rounds. n is assumed to be a very large number (ideally infinite).</li>
  <li>Each player is assumed to have equal strength.</li>
  <li>Thus the result of each game is (independently) 1, 1/2, 0 with probability p,1-2*p,p with p in [0,1/2].</li>
  </ul>
  <p>This should be homework for a statistics student...and I can immediately claim the mean of the score distribution is (n-1)/2 in any case.</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-02T12:49:45.067Z
Comments.Text: This is a very worthy and interesting question, but is lacking in the precise details to make it answerable mathematically.  What "resulting table"?  (what is in the table?)  From your final paragraph, are you asking for the distribution of the final scores for each of the _m_ players?
Comments.UserDisplayName: SecretAgentMan
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Comments.CreationDate: 2022-11-02T12:51:04.383Z
Comments.Text: Finally, are you assuming each game is independent of the others?  OP should [edit] the question to include all mathematical assumptions and precise definitions to make this answerable. There appear to be multiple methods to answer this question, but it is unclear what is appropriate at this point.
Comments.UserDisplayName: SecretAgentMan
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-02T12:56:05.170Z
Comments.Text: Related (not dupes):  Player rating distribution is not Gaussian.  See [Elo distribution](https://chess.stackexchange.com/a/33443/26335), [FIDE Rating Distribution](https://chess.stackexchange.com/a/39079/26335), & [Distribution of Elo Ratings](https://chess.stackexchange.com/q/37549/26335).
Comments.UserDisplayName: SecretAgentMan
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-02T18:28:02.870Z
Comments.Text: I simplified the question and hopefully added all used assumptions. The question should be answerable with a simple urn model.
Comments.UserDisplayName: Hauke Reddmann
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-03T12:25:03.560Z
Comments.Text: If all players have equal strength, wouldn't _p_ = 0.5 ?
Comments.UserDisplayName: SecretAgentMan
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-03T14:17:33.143Z
Comments.Text: @SecretAgentMan: Go on, play a tournament with n equally rated players and wait for n(n-1)/2 draws. Happened...but... ;-) (And if you play a tournament with n clones of me, be sure that the drawing rate is even lower than usual!)
Comments.UserDisplayName: Hauke Reddmann
Tags: <statistics><round-robin>
Title: Point distribution of an "average" tournament
ThreadText: # Point distribution of an "average" tournament
  
  (Asked by: username_0 on 2022-11-02)
  
  Maybe this is a bit too much towards math, but we have Noam :-)
  
  If you view the final point standings of any tournament, you'll probably see many players with an average number of points, but few with a very high or low score. For Swiss this is almost by construction, but for a round robin (easier to analyze) it also resembles a Gaussian distribution.
  
  My question is whether it is actually one, or something only similar, assuming some simplifications:
  
  * n players play a round robin with n-1 rounds. n is assumed to be a very large number (ideally infinite).
  * Each player is assumed to have equal strength.
  * Thus the result of each game is (independently) 1, 1/2, 0 with probability p,1-2\*p,p with p in [0,1/2].
  
  This should be homework for a statistics student...and I can immediately claim the mean of the score distribution is (n-1)/2 in any case.
  
  **Tags:** statistics, round-robin
  
  > **Comments:**
  > **username_1**: This is a very worthy and interesting question, but is lacking in the precise details to make it answerable mathematically.  What "resulting table"?  (what is in the table?)  From your final paragraph, are you asking for the distribution of the final scores for each of the _m_ players?  
  > **username_1**: Finally, are you assuming each game is independent of the others?  OP should [edit] the question to include all mathematical assumptions and precise definitions to make this answerable. There appear to be multiple methods to answer this question, but it is unclear what is appropriate at this point.  
  > **username_1**: Related (not dupes):  Player rating distribution is not Gaussian.  See [Elo distribution](https://chess.stackexchange.com/a/33443/26335), [FIDE Rating Distribution](https://chess.stackexchange.com/a/39079/26335), & [Distribution of Elo Ratings](https://chess.stackexchange.com/q/37549/26335).  
  > **username_0**: I simplified the question and hopefully added all used assumptions. The question should be answerable with a simple urn model.  
  > **username_1**: If all players have equal strength, wouldn't _p_ = 0.5 ?  
  > **username_0**: @username_1: Go on, play a tournament with n equally rated players and wait for n(n-1)/2 draws. Happened...but... ;-) (And if you play a tournament with n clones of me, be sure that the drawing rate is even lower than usual!)  
  
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  ## 1 answers
  
  ### Answer by username_2 on 2022-11-03. Score: 1 | Accepted answer
  
  It depends which quantities you are interested in; the whole distribution is a complex concept to describe.
  
  Statistics that look at the score of a single player (mean, variance, etc.) are easy to compute, because they are the sum of independent game outcomes. It is essentially (up to rescaling) a [multinomial distribution](https://en.wikipedia.org/wiki/Multinomial_distribution) with k=3. So average scores, standard deviations, etc. are easy to compute. And the central limit theorem predicts that the limit for n to infinity is a Gaussian distribution with mean and variance that are easy to compute.
  
  The trickier part is finding out quantities that involve the joint distribution of 2 or more player scores, since these scores are not independent: for instance, no two players can both score 0 in a round-robin tournament. So probabilities like "the probability that 5 different players score at least m points" are more complicated.
  
  > **Comments:**
  > **username_0**: Agrees with what my instinct said (alas, I'm no math profi :-) - and the complicated joint distributions are only interesting for a profi...  
  
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