Author: Shep
Date: 06:12:25 02/15/00
Hi all,
I was trying to write this post earlier, but found a flaw in my calculations, so
here goes the fixed one:
Suppose you want to play a match on two machines with PB and only have two
machines of different speed. (Obviously, just adjusting the time controls to
reflect the speed difference won't make it fair.)
Assume that machine A is the faster one and that machine B is slower by a factor
of k>1. The natural impulse would be to give machine B k-times as much
reflection time.
For simplicity, let's assume that both machines predict precisely 50% of their
opponent's moves. Then the effective reflection times for both side are:
R_a(k) = 1+k/2
R_b(k) = k+1/2
Both reflection time functions are linear in k and intersect at k=1, thus it's
easy to see that
R_a(k) < R_b(k) for all k>1
Thus
l := R_b(k)/R_a(k) > 1 for all k>1
is the factor by which B's reflection time is too big to be fair.
Consequently, to have a fair match, the multiplier for B's reflection time needs
to be k/l instead of k.
A quick calculation yields the final formula:
m := k/l = (k^2+2k)/(2k+1)
Thus, if machine B is given m-times the reflection time of machine A, the match
is fair (under the assumption that both sides predict 50% which is not precise,
but rather realistic - the final formula is slightly more complex for uneven
prediction rates).
Example table:
k m
--------
2 ~1.6
3 ~2.1
4 ~2.7
5 ~3.1
10 ~5.8
---
Shep
This page took 0.01 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.