Author: Robert Pawlak
Date: 18:50:09 04/25/01
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On April 25, 2001 at 14:30:55, Uri Blass wrote: >On April 25, 2001 at 13:46:22, guy haworth wrote: > >>A game of chess is a 'Bernoulli trial' in a sequence of say 'n' trials. >> >>One side's score after 'n' trials is a random-variable X. >> >>If chess was a win/lose 2-outcome trial, one could dial straight into the theory >>of binomial and normal distributions and pick up info about the variance of the >>score after 'n' trials. >> >>Unfortunately, it is a three-outcome trial ... win/draw/lose ... >> >>Question is ... is there an equivalent neat formula about the variance of the >>score after 'n' trials ... assuming 1 for win, 0.5 for draw, 0 for loss of >>course. To the best of my knowledge, the answer is no. Although one good source of info is the two volume Kendall stats reference. > >Assumptions: >1)There is a match between 2 opponents J and F. >1)The results of different games independent events. Not valid with book learning. >2)The results of different games when J is white have the same distribution. Not true, once again, there is book learning and probably other factors to deal with also. >3)The result of different game when F is white have the same distribution > > See above. Now, I do realize that you are not saying these are true. I'm just pointing out a few things. If you want to find the resulting multinomial distribution after N trial, you just have to perform N-1 convolutions. Any computer will do this very quickly. The problem is that is doesn't mean anything for the commercial programs. Bob
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