Author: Uri Blass
Date: 11:30:55 04/25/01
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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. Assumptions: 1)There is a match between 2 opponents J and F. 1)The results of different games independent events. 2)The results of different games when J is white have the same distribution. 3)The result of different game when F is white have the same distribution The variance of the score after n games when J is white and m games when F is white is the sum of the folowing numbers: 1)n times the variance of the score of 1 game when J is white 2)m times the variance of the score of 1 game when F is white The variance of the score of 1 game is at most 1/4 and you can calculate it by the following formula: (1/2*1/2*p(1/2)+1*1*p(1))-(1/2*p(1/2)+p(1))^2 example:if p(J wins with white) =0.5,p(draw when J is white)=0.2,p(J lose with white)=0.3 then you get for J with white the following variance: (1/4*0.2+0.5)-(1/2*0.2+0.5)^2=0.55-0.36=0.19 Uri
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