Computer Chess Club Archives




Subject: Re: Proving something is better

Author: Uri Blass

Date: 04:04:45 12/24/02

Go up one level in this thread

On December 23, 2002 at 12:14:57, Peter Fendrich wrote:

>On December 23, 2002 at 09:10:59, Rémi Coulom wrote:
>>On December 23, 2002 at 06:31:57, Peter Fendrich wrote:
>>>I don't get the same results as you:
>>>(all the sample values of Pw, Pd, Trin and cumTrin are the same)
>>>          Mine         Yours
>>>4 6 0 :   0.2745363   0.2745533
>>>4 6 1 :   0.274543    0.2747703
>>>4 6 10:   0.2740947   0.2773552
>>>I'm not sure what's happening here but as you say, the Monte Carlo method
>>>doesn't give exact values.
>>>I thought that the program could be reliable to 3 decimals but maybe not...
>>>However, if I'm right about draws, the prob would slowly move towards 0.5 when
>>>the number of draws increases. I continued up to 500 draws and got the
>>>4 6 100:  0.2758096
>>>4 6 200:  0.2820387
>>>4 6 300:  0.2906555
>>>4 6 400:  0.3076096
>>>4 6 500:  0.3273436
>>That is because the Monte Carlo method is inaccurate: think about the x^n
>>function, x varying between 0 and 1: when n grows large, the function has an
>>extremely thin peak, that is very difficult to integrate accurately with a Monte
>>Carlo method.
>>I am totally certain about this, because I first started by implementing a
>>program that calculated the big trinomial integrals. I wrote a small polynom
>>library based on the GNU multiprecision library so that the _exact_ value was
>>found. That's how I noticed that the result does not depend on draws, and went
>>further in the calculations.
>Yes, that's quite possible. What's funny is that I repeatably get these kind of
>results for other combinations of win/lose. I have a sligth hunch about what
>might happen. Let's come back to that later.
>>>The probability for A neatly grows with more draws.
>>>OTOH I can't argue against your formulas. Give me some more time.
>>>Could you please elaborate the first formula in section "3 Draws do nout Count".
>>>How do explain 1 - p0 - p0.5 = 1 - u + up0.5 - p0.5?
>>>I'm a bit interested in the term up0.5 that seems to be superfluous.
>>u is defined by p0 = u * (1 - p0.5). If you replace p0 by u(1 - p0.5) in
>>1 - p0 - p0.5, the you get 1 - u + up0.5 - p0.5
>Ok, u is a variable and nothing else...
>Then it's prefectly clear that your math is allright. The connection to the
>rating system is not as close as I, from the beginning, took for granted. It's
>not the probability that is changing it is the interval describing the bell
>curve, surronding 0.5 that is shrinking when the number of draws increases. I
>had to get a short discussion with another person in order to get really
>convinced about this strange fact.
>That gave rise to a new question.
>Maybe he will post something himself otherwise I will come back with an better
>explanantion of what's folling ...
>The new question was about the problem formulation. The issue is not right
>formulated from the beginning ("is A *better* than B" given a certain match
>result). Draws contains some information and when we have a question formulated
>in a way that makes draws not to be included, we will not get full quality of
>the conclusion. The area of interpreting match results is interesting and not as
>evident one might think.
>I'll return later on...

If you want to give an interval for the difference in rating that you want to be
sure in 95% that the real difference in rating is in the interval then the
number of draws is relevant.


This page took 0.01 seconds to execute

Last modified: Thu, 07 Jul 11 08:48:38 -0700

Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.