Author: Peter Fendrich
Date: 12:41:05 06/04/99
Go up one level in this thread
On June 04, 1999 at 08:22:41, Arda Renkver wrote:
>Could someone refer to a source as to how to get these +61 -53 etc margin of
>error ratings points depending on the number of games played.Like in the new
>Swedish list; 182 games for CM6000 has higher margin than 900 games played with
>some other program.For example is just one game's margin of error in rating 800
>points? What is the maximum number of games required to be in the range of + and
>- 0.000...1 etc? What statistical formula or concept used other than ELO
>calculation as usual.
To estimate the standard deviation of the ratings there are
several options, giving about the same result, the SSDF method is:
s=SQRT( (W(1-m)**2 + D(0.5-m)**2 + L(0-m)**2/(n-1) )
A=1.96 * s/SQRT(n)
where s is estimated standard deviation
n is the number of games
m is score/n
W,D,L is the number of Wins, Draws and Losts respectivelly
A is the margin of error (for score, not rating points)
1.96 is fetched from the Normal Distribution table to get
95% reliability
SQRT is the square root
** is the power of
Now we have an 95% interval of scores from m-A to m+A
Compute the ratings for m-A and m+A and here we go!
This is done for each chess program.
//Peter Fendrich
This page took 0 seconds to execute
Last modified: Thu, 15 Apr 21 08:11:13 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.