Author: Sune Fischer
Date: 03:36:46 11/25/03
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On November 25, 2003 at 05:23:31, Rémi Coulom wrote: >On November 24, 2003 at 18:38:06, Sune Fischer wrote: > >> >>While reading it I managed to convince myself you are right. >> >>Intuitively it seems obvious that if the normalization constraint >>p0+p0.5+p1=1 is the only correlation between p0 and p1, then >>p0.5 isn't going to have any say in whether p0>p1 or not, right? > >Right, but it is not that simple. What we get from the match is not p0, p0.5 and >p1, but the number of wins, losses, and draws. p0, p0.5, and p1 are unknown. Of >course, if you know p0 and p1, you don't need p0.5. That does not oviously >implies that if you know n0 and n1, you don't need n0.5. That's true, in statistics everything is done 'backwards' :) >>The only exception is if p0.5=1, but then we wouldn't have a three parameter >>distribution in the first place which is sort of the assumption. > >This is not really an exception. All the maths work the same if p0.5=1. In this >case, what you'll get is a match with only draws, so the formula says >P(p1>p0)=0.5. Right, unless you played an infinite number of games which all ended drawn. Then P(p1=p0)=1 which would answer our question. I guess I'm always assuming an infinite number of games, since I know the distribution :) -S. >>-S. >>
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