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Answer by Henry for Expected winning score in a round robin tournament

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You can get a good approximation by making the correct assumption that each player's score is a binomial distribution with parameters $99$ and $\frac12$ and the slightly wrong assumption that each player's total score is independent of the other players.

You could use the distributions for order statistics for $100$ independent samples from this distribution, using the beta and binomial distributions. Using R for a form of numerical integration:

  • For $S_1$mean(qbinom(qbeta(ppoints(10^6),100,1),99,1/2)) gives about $61.91$ compared to Gabriel Romon's simulation of $61.98$
  • For $S_{10}$mean(qbinom(qbeta(ppoints(10^6),91,10),99,1/2)) gives about $55.99$ compared to Gabriel Romon's simulation of $56.03$
  • For $S_{50}$mean(qbinom(qbeta(ppoints(10^6),51,50),99,1/2)) gives about $49.56$ compared to Gabriel Romon's simulation of $49.57$

The simulation has a small amount of noise while the approximation makes a false assumption which affects the results, but they are broadly similar.


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