# Computer Chess Club Archives

## Messages

### Subject: Re: A question about statistics...

Author: Peter Fendrich

Date: 05:53:58 01/09/04

Go up one level in this thread

```On January 07, 2004 at 12:23:00, Dieter Buerssner wrote:

>Perhaps I can myself clearer by another example. Assume again the 1000 games,
>W=510, D=0, L=490. Player a wins as white 255 games and as black 255 games. >With your forumula we get for the result 0.51 +/- 0.01586 (67% margin).

Why 67% ???

>If I simualtate this match (1 million such matches), and plot p(m) vs. m, and I
>fit a curve
>
>1/(sqrt(2*pi)*sigma*n)*exp(-0.5*(x-x0)^2/sigma^2)
>
>(I kept n constant at 1000, because the integral must be 1/1000), I get
>sigma=0.01581. As we can expect, practically the same number as calculated by
>your formula. But now assume Player A won all his games with white, and
>additionally 20 games with black. No draws. Again m=0.51. W/D/L the same as
>above.

Just to check if I got it right: You mean that A plays 490 white games and 510
black games?

>But we can already assume now (we know more about the match than just
>W/D/L), that here player a is most probably better (very similar to the W/D/L
>20/980/0 case). The game might be a game, where typically white wins, and only
>when he does a sublte fault will lose. Player A showed, that he was clearly
>better, because he allowed B not a single win with black. The distribution of m
>will be much narrower. Actually, the fit yields sigma=3.1e-3 (although now -
>again not too surprisingly - the distribution is slightly tilted from the normal
>distribution; I did not try to do the mathematics, but I guess, it is a
>binominal distribution). This also shows, that player A ist stronger with big
>confidence.
>Your formula would not give an adequate error margin in this case, because it
>ignores the more detailled knowledge about performance as white and black.

The formulas I've given is general for a stochastic variable X with 3 outcomes.
You could redefine the X to [wW, wD, wL, bW, bD, bL] all from player A's
perspective. Apply the same formulas but with the 6 outcomes instead. I'm not
sure if this is the trick you are looking for...
If not I associate to 2 different tecniques:
ANOVA: Split up you sample in groups and analyse the Variance and Mean
differences between groups. I think the 2-way ANOVA is applicable here.

Bayesian statistics: Gives you some good techniques to include preknown
knowledge in the model in a more general fasion.

Use Google! There a numerous descriptions.

Finally you can always try the method I used to evaluate a result. That was
before Rémi found a more elegant solution ignoring the draws.

I think I've sent you the text describing it otherwise give me a hint.
I suppose this method can be used to estiamte all variables. It's really a
Bayesian approach computerised.

>>Maybe we can continue by email - I will send you one when things are calmed down
>>here...
>
>No problem with me, to disuss it in email. I just thought, I post an followup
>here, and try to explain it a bit clearer. Perhaps, we are not the only 2
>persons interested.

Maybe so, I doubt there are more ...

>>I did put your examples into my formulas and got slightly different Rating-Dif's
>>than you have. Maybe it's just decimal errors.
>
>Indeed, I had a few rounding error. I wrote down some intermediate results,
>rounded them (including one error I made in the last digit). And even more
>embarrassing, I did some circular calculation (from m to Elo and back, and did
>not recognize, that what came back was not exactly the same to what was put in)
>
>>I also used another formula than
>>you did.
>
>I guess, you are referring to the Elo calculations.

Yes

>I think, the Elo just complicates things, without giving more insight.

:-)
I wouldn't put it that way. It's the other way around. The -400LG(x) formula is
a simplification. But is it exact?
I haven't tried to derivate it from Elo formulas but it seems to me as an
approximation.

/Peter

```