3

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.

Improve this question
6
  • 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?
    – SecretAgentMan
    Nov 2, 2022 at 12:49
  • 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.
    – SecretAgentMan
    Nov 2, 2022 at 12:51
  • 2
    Related (not dupes): Player rating distribution is not Gaussian. See Elo distribution, FIDE Rating Distribution, & Distribution of Elo Ratings.
    – SecretAgentMan
    Nov 2, 2022 at 12:56
  • I simplified the question and hopefully added all used assumptions. The question should be answerable with a simple urn model.
    – Hauke Reddmann
    Nov 2, 2022 at 18:28
  • If all players have equal strength, wouldn't p = 0.5 ?
    – SecretAgentMan
    Nov 3, 2022 at 12:25

1 Answer 1

Reset to default
1

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 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.

Improve this answer
1
  • Agrees with what my instinct said (alas, I'm no math profi :-) - and the complicated joint distributions are only interesting for a profi...
    – Hauke Reddmann
    Nov 3, 2022 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.