user11144: 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.

user12957: 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.

11144: <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>
 12957: <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>
