Author: Sune Fischer
Date: 15:31:01 02/27/02
Go up one level in this thread
On February 27, 2002 at 16:09:28, Robert Hyatt wrote: >On February 27, 2002 at 09:41:16, Sune Fischer wrote: > >>On February 27, 2002 at 09:09:05, Uri Blass wrote: >> >>>On February 27, 2002 at 06:33:17, Sune Fischer wrote: >>> >>>>On February 26, 2002 at 14:35:32, Uri Blass wrote: >>>> >>>>>The difference in elo in order to win a match 2*10^1000-1 is certainly finite >>>>>and I believe that choosing a random move is going to be enough for better score >>>>>because I believe that it is possible to get at least a draw in less than 500 >>>>>moves and the probability to be lucky and choose every one of them is more than >>>>>1/100 in every move because I believe that the number of moves in every ply is >>>>>going to be less than 100 when the opponent choose the perfect strategy. >>>> >>>>Yes I agree, but much depends on what the *chess-tree* really lookes like. >>>>Maybe black has a forced draw in 30 moves? Maybe the forced draw is really 2000 >>>>moves? As you have previously pointet out yourself, the longer the game, the >>>>greater the chance that the weaker player will make a mistake. This will >>>>probably correlate directly to the rating of the perfect player, can he drag the >>>>game on forever his rating will be much higher. >>>> >>>>>It suggest the following question >>>>>suppose that A has rating 0(I believe that the player who choose random move >>>>>will have rating that is lower than 0). >>>>> >>>>>suppose B wins against A 2*10^1000-1 >>>>> >>>>>What is going to be the rating of B based on the elo formula? >>>>>This rating is probably an upper bound for the rating of the perfect player >>>>>if you assume that the perfect player plays only against A. >>>>> >>>>>Uri >>>> >>>>Why should it be an upper bound? >>>>Your rating should be a constant no matter who you play, if your opponent is >>>>weak you will win more games, but your expected score will also be that much >>>>higher. >>>> >>>>-S. >>> >>>The rating is dependent in the opponet that the perfect player chooses to play. >> >>No it is not, look at the formula, it is a normal distribution. >> >>>The perfect player may get 100% against my program on p800 because my program is >>>a deteministic program that always does the same mistake so if you assume the >>>perfect player plays only against my program then the perfect player is going to >>>get infinite rating. >> >>Your program is deterministic by your own words, so must score even worse than >>one doing random moves. >> >>>The perfect player may get 100% against a player with a rating of 2000 when the >>>same player is going to fail to get 100% against a player that is clearly weaker >>>but not deterministic. >> >> >>Please do not ignore the small differences in probability, they are important. >>A 2000 elo player may be beaten by 10^30:1 and a 1000 elo player by 10^35:1, it >>should all add up to the same rating for the perfect player, that is how the elo >>table works. >> > >That can't possibly be right. When two players play, they have a probability >of winning and the Elo formula will increase one player's rating and decrease >the others until the difference between their ratings reflects the probability >of each winning or losing. Yes, and your point would be..? My point was, that a person will keep his rating whether he's playing 1500 opponents or 2500 opponents (if he plays enough games). There is no way to increase you rating by playing a lot of stronger or weaker players, it should remain constant (assuming you won't learn and improve from the games, naturally). >But if I play a brick, and I win _every_ game, >my rating will climb for _every_ victory. And it doesn't climb by some >amount that has a limit of zero at the upper end. It climbs by some positive >amount for every win. That spells +infinity... Again this is one of those absurd examples, a brick has -infinity rating or what? What is the elo of the brick, we need to know that to find your elo. -S.
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