# Point distribution of an "average" tournament

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.

• 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? Nov 2, 2022 at 12:49
• 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. Nov 2, 2022 at 12:51
• Related (not dupes): Player rating distribution is not Gaussian. See Elo distribution, FIDE Rating Distribution, & Distribution of Elo Ratings. 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. Nov 2, 2022 at 18:28
• If all players have equal strength, wouldn't p = 0.5 ? Nov 3, 2022 at 12:25