Geschrieben von:/Posted by: Uri Blass at 23 May 2004 12:26:19:
Als Antwort auf:/In reply to: Re: WBEC Ridderkerk new results. geschrieben von:/posted by: Antonio Dieguez at 23 May 2004 03:58:04:
Premier Division:
Round 16:
Quark 2.35 Paderborn ½111 3.5/4 = need 1 point vs Goliath!
GreenLightChess 3.00 ½000 0.5/4 = GLC in big problems now! (needs 2,5 point vs Ruffian)
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LittleGoliath 3.9po 0111 3.0/4 = need 1 point vs Quark!
Nejmet 3.07 1000 1.0/4 = Finished but not save yet!?
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Round 17:
Amyan 1.593b 0101 2.0/4 = "Little" Amyan is safe Antonio!!
hi!
very good perfomance..
now delete it, i never release anymore and amyan 1.593b remains a myth..
btw, my email zodiamoon@yah... is very bad because spam and viruses, it gets full everyday, in any case i also have a.dieguez at uandresbello.edu.
For comparison result of previous Amyan:
ATHLON-MP2200
1)http://wbec-ridderkerk.nl/html/his4thedition.html
Amyan1.59 scored worse than knightdreamer and almost go down to a lower division.
Amyan is place 15 and place 16 go down.
Amyan1.59 38.5/84
Knightdreamer3.2 41.5/84
gap of 3/84 for knightdreamer.
2)Amyan1.59(same amyan) scores 38/68
Knightdreamer3.2(same knightdreamer scores 23.5/68
gap of 14.5/68 for Amyan.
Hardware Mp-2400 almost the same.
What is the reason for the difference in the result:
possible explanations:
explanation1:Luck(but I think that the probability for that big difference to happen is small even if we assume constant probabilities of 1/3 for loss,draw,win)
You can try to run tournament when the result of every game is win,draw,loss with probability 1/3 when tournament 1 has 84 games and the tournament 2 has 68 games.
Now let A1 be the result of program a in tournament 1.
Let B1 be the result of program b in tournament 1.
A2 and B2 are defined in the same way for tournament 2.
Now you can calculate (A1-B1)/84 for the first tournament and (B2-A2)/68 for the second tournament
I wonder in how many cases you will get S=(A1-B1)/84+(B2-A2)/68>3/84+14.5/68
We have nearly
A1~N(0,sqrt(14))(some of independent 84 varaibles with varaiety 1/6 has varaiety 84/6).
B1~N(0,sqrt(14))
(A1-B1)~N(0,sqrt(28))
(B2-A2)~N(0,sqrt(68/3)
S~N(0,sqrt(28/(84*84)+(68/3)/(68*68)))
I get that the result is different by more than 2.5 standard deviation from the expected result by luck and the probability for it to happen is very small.
In the real case part of the programs are stronger and part of the programs are weaker and the colour also have influence on the result so the varaince is smaller relative to the varaince in case that you have 1/3 probability for everything.
Explanation 2:Amyan had a better book in the second tournament.
Explanation 3:Amyan has learning that helped it to learn better lines for the second tournament.
Uri
