[EM] round robin tournaments

[Jameson Quinn]

I like this approach. But I think you've multiplied by a factor which should
be dividing:

2011/6/23 <fsimmons at pcc.edu>
...

> Here's a way to resolve it.
>
> Let N=W+L.  Let p=W/N.  Let q=L/N.  Let sigma = sqrt(N*p*q).
>
> Measure defeat strength by  S=(p-q)*sigma.
>

Let's say one team happens to consist of penguins. They score 0. The chances
of them winning a rematch are 0. Their defeat strength should be...
infinite. But with your formula, it comes out to 0.

 ....

> What is the heuristic behind S=(p-q)*sigma?
>
> The Binomial distribution has standard deviation sigma = sqrt(N*p*q).
>
> From the point of view of the winning team the fraction p is the proportion
> of favorable outcomes, while
> the fraction q is the proportion of unfavorable outcomes.  So (p-q) is the
> difference in estimates of the
> underlying fractions of favorable and unfavorable outcomes.  We multiply
> this estimate by the standard
> deviation to take into account "sample size."
>

You should be dividing by the standard deviation. Higher standard deviation,
more uncertainty about the repeated result.

And even that would only give you number of standard deviations. To convert
this to a probability, you'd need to use a bell curve distribution.

Of course, if the game were soccer(UK football) instead of basketball, where
a score of 0 is common and nothing to be particularly ashamed of, this
wouldn't work. I think you'd have to add some epsilon to both scores to
avoid the infinities.

>
> If a real statistician, like Jobst were reading this, he (or she) could
> refine this estimate, or at least give a
> better explanation.
>

Disclaimer: I am not a real statistician by academic credentials, though
actually I've been gainfully employed as one in Guatemala.

JQ
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