# Computer Chess Club Archives

## Messages

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

Date: 09:23:00 01/07/04

Go up one level in this thread

```On January 06, 2004 at 19:03:58, Peter Fendrich wrote:

>On January 06, 2004 at 14:03:05, Dieter Buerssner wrote:
>
>>On January 05, 2004 at 18:11:25, Peter Fendrich wrote:
>>
>>>The rest is basic formulas for
>>>the standard deviation.
>>
>>Actually, I am puzzled about these basic formulas for the standard deviation.
>>Somehow I don't get it (in German, ich vermute, ich sitze auf der Leitung).

In between, I got it. But still not yet for the case below.

>>However, the function m vs. p(m) looks different, when I assume different w/d/l
>>probabilities for white and black. So, we not only have W, D, L, but wW, wD, wL,
>>bW, bD, bL and say nb, nw, mb, mw (typically nb=nw). Can you also give a formula
>>for this scenario for s?

>I'm sorry I've been quite busy today and will be for a while.

There is no need to hurry. Also, it is not that important. I am just curios.

>I must confess that I didn't study your examples carefully enough.
>Anyway I'm not sure what you're after with the variables above. The variance in
>the sample is what it is and is computed as I told before. In order to estimate
>the standard deviation of the population and to use preknown distribution of
>W/D/L you are into the Bayesian thing. Is this what you want?

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).

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. 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.

>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.

>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.
I think, the Elo just complicates things, without giving more insight. I only
used it, to be able to compare the numbers with elostat. Otherwise, your A (or
without the factor 1.96 for the 95%, as in the examples above) should be enough.

Regards,
Dieter

```

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