Author: Uri Blass
Date: 00:19:50 01/04/01
Go up one level in this thread
On January 04, 2001 at 01:22:57, Ricardo Gibert wrote: >On January 04, 2001 at 00:00:39, Robert Hyatt wrote: > >>On January 03, 2001 at 17:50:38, José Carlos wrote: >> >>>On January 03, 2001 at 16:26:19, Robert Hyatt wrote: >>> >>>>On January 03, 2001 at 09:52:06, José Carlos wrote: >>>> >>>>> Lately, people have been talking here about significant results. I'm not >>>>>really sure if probabilistic calculus is appropiate here, because chess games >>>>>are not stocastic events. >>>>> So, I suggest an experiment to mesure the probabilistic noise: >>>>> >>>>> -chose a random program and make it play itself. >>>>> -write down the result after 10 games, 50 games, 100 games... >>>>> >>>>> It should tend to be an even result, and it would be possible to know how many >>>>>games are needed to get a result with a certain degree of confidence. >>>>> If we try this for several programs, and the results are similar, we can draw >>>>>a conclusion, in comparison with pure probabilistic calculus. >>>>> >>>>> Does this idea make sense, or am I still sleeping? :) >>>>> >>>>> José C. >>>> >>>>It is statistically invalid. IE if you flip a coin 500 times do you _really_ >>>>expect to get 250 heads and 250 tails? Probability distribution says you >>>>won't get that very often at all. In fact, if you flip long enough, you will >>>>either get 500 straight heads or tails, or else prove the coin is _not_ actually >>>>perfectly random with 50-50 probability of getting a head or tail. >>> >>> But don't you think the more times you flip the coin, the closer the number of >>>head and tails (in %) will be? Maybe the coin is not the better comparison, as >>>it is a random event, and a chess game is not, but I still think it should work. >>>But I expect a different rate of "closeness" (is this word correct?) for the >>>same number of tries with the coin (random event) and the games (partially >>>random -book, pondering, ... and partially not -eval function, search algos...), >>>and that difference is what I want to measure. >>> >>> José C. >> >> >>No I don't. Suppose that 500-0 run comes _first_. How long will you have to >>flip to get back to even? You may _never_ get back to even. Remember this is >>a bell-curve shaped probability distribution. Not a single spike on the curve >>at the mid-point of the distribution. You probably need to play 40 forty-game >>matches to get the beginning of an idea of who is better. > >I have a vague recollection of a statistics theorem that guarantees you will >cross the 50% line an infinite number of times given an infinite number of >*fair* coin flips. The unlucky 500 run coming first is irrelevant. >Unfortunately, I do not have a statistics book or an infinite amount of time to >verify this. The book is right and I have a simple idea to prove it. The idea is that the bad start is not relevant and after enough trials the probability to have majority for the unlucky side is almost 50%. There is a constant C such that even when you were unlucky in the first x trials and got x tails the probability to get majority for heads after C*x^2 trials is always more than 40%. It means that the probability to get majority in the future is 1 because the probability not to get majority after C*x^2 steps and after C*(C*x^2)^2 steps and after C*(C*(C*x^2)^2)^2 ... is less than 60%*60%*60%... Uri
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