# Computer Chess Club Archives

## Messages

### Subject: Re: ELO performance?

Author: Stefan Meyer-Kahlen

Date: 04:55:59 05/23/99

Go up one level in this thread

```On May 22, 1999 at 14:22:26, Peter Fendrich wrote:

>Sorry, I forgot to answear the 95% question... :)
>
>To estimate the standard deviation of the Performance Rating there are
>lots of options, giving about the same result. You need the individual results
>to know the the error margin and that's what this is all about. There is a
>difference between a result of only wins and losses compared to a result with a
>lot of draws. The SSDF method is best described as:
>       (1) s=SQRT(W(1-m)**2 + D(0.5-m)**2 + L(0-m)**2/(n-1))
>       (2) A=1.96 * s/SQRT(n)
>
>       where s is estimated standard deviation
>             n     is number of games
>             m     is score/n
>             W,D,L is number of Wins, Draws and Losts respectivelly
>             A     is the margin of error (for score, not rating points)
>             1.96  is fethed from the Normal Distribution table to get
>                   95% reliability
>             SQRT  is the square root
>             **    is used as 'the power of'
>
>    Now we have an 95% interval of *scores* from m-A to m+A
>
>    Compute the ratings for each m-A and m+A and here we go!
>    These ratings are the end points in the interval.
>
>//Peter

Yes, that's what I was looking for, thanks.
I think (1) should be

(1) s=SQRT((W(1-m)**2 + D(0.5-m)**2 + L(0-m)**2)/(n-1))

right?

Let's assume A and B play 3 games, A wins 2, 1 is a draw.
If I use your fromula I get
S = 0.2886
and
A = 0,3266

If a calculate m-A I get 0.5067, which is > 0.5!
There must be somethin wrong, because I don't think you can say that A is better
than B with 95% prob. just after 3 games.

Stefan

```

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