Wlodzimierz Holsztynski

1. The Ockham function. Predictions
============================

Given a game between two players X Y with
the pre-game ratings A B, their after-game
ratings respectively a

A' := A + r*(a*B - b*A)
B' := B + r*(b*A - a*B)

hence A'+B' = A+B -- their total rating is constant;
as in the earlier posting: a:b is the result of the
game, i.e. 0 \ a \ 1 and a+b = 1 (hence 0 \ a \ 1);
also, r is the relevance factor of the game (0 r 1).

Their next game will have relevance factor s; consider
the case when the result is the same a:b -- then

A" = A' + s*(a*B - b*A)
B" = B' + s*(b*A - a*B)

A direct computation gives:

A" = A + t*(a*B - b*A)
B" = B + t*(b*A - a*B)

where

t := 1 - (1-r)*(1-s)

(an elegant linear algebra proof is given
at the end of this post). Observe the pattern:

r = 1 - (1-r)
t = 1 - (1-r)*(1-s)

Now, a simple induction extends this pattern:

***

THEOREM 1 Let r_1 ... r_n be the relevance
========= factors of the consecutive games
of a match of players X Y, with the pre-match
ratings A B. Let's assume that the result of each
game is the same a:b. Then the post-match ratings
A_n B_n are as follows:

A_n = A + (1 - Prod(1 - r_k : k=1...n)) * (a*B - b*A)
B_n = B + (1 - Prod(1 - r_k : k=1...n)) * (b*A - a*B)

***

If the relevant game factor were constant, r_k = r,
then, for R := 1-r, we would have:

A_n = A + (1 - R^n) * (a*B - b*A)
B_n = B + (1 - R^n) * (b*A - a*B)

Then, in the case of an infinite match, we would have:

lim A_n = a*(A+B)
lim B_n = b*(A+B)

for n -- oo. The same would be true if for all n,
with possible finitely many exceptions only, we
would have r_k 1/k. The full statement is:

***

THEOREM 2 If Sum(r_k : k=1 2 ...) = oo, then
=========

lim A_n = a*(A+B)
lim B_n = b*(A+B)

***

We see that even when the relevance of the match
games approaches zero, the limit quotient of rating
is still going top be a:b, granted that the relevance
does not converge to zero too fast.


2. The meaning of the relevance constant
================================

To get a feel for the Ockham rating let's first
make a simplifying assumption that the relevance
factor is constant, say p (where 0 p 1).

It's convenient to introduce also q := 1-p
(like in probability theory).

Let's answer the question: after how many
straight loses of the higher rated player, the
ratings of the players will get equal?

Thus let's assume that A B are the two pre-match
ratings of players X Y. Now let player X keep winning.

After n games the ratings are going to be:

A_n = A + (1 - q^n) * B
B_n = B - (1 - q^n) * B

The two are going to be equal if and only if

A + (1 - q^n) * B = B - (1 - q^n) * B

(1 - q^n)*B = (B-A) / 2

q^n = 1 - (B-A) / (2*B)

q^n = (A+B)/(2*B)

n = log ((A+B)/(2*B)) / log (q)

(the numerator and the denominator are both
negative, hence the solution is positive). The
solution, as a rule, is not an integer. After floor(n)
games the player X is still going to be rated lower
than player Y, but after the ceiling(n) games the
player X will overcome player Y.

EXAMPLE An extremely strong newcomer X
======= plays a match against player Y
rated 2000. The initial rating of X is 1000.
Thus it will take

n = log(3/4) / log(q)

for the two ratings to get equal. For instance, for
q = 3/4, i.e. for p = 1/4, it would take exactly
one win by X. In general, the equalization will
happen after n wins by X when:

q = (3/4)^(1/n)
i.e. for
p = 1 - (3/4)^(1/n)

If you are in charge of selecting the parameters
of the Ockham rating function, and if you feel that
equalization of God and 2000 rated player should
happen after 6 straight wins then you'd set

p := 0.0468...

or just below 1/20.

Actually, already after 4 wins it is clear that
player X is (almost certainly) at least as strong
as player Y. Thus perhaps

p := 1 - (3/4)^(1/4) .=. 0.0694

is even better. However, it is important, that
a new player plays different players, so that
s/he will not "punish" just one--it's more fair
this way. Also, when players mix well then
the relevance constant may be lower because,
for instance, the new player will soon play
strong players instead of the same one, whose
rating would go down. Let's study this issue in
the section below. Now let me mention that
with p = 0.0694 our super-strong novice X
will go from 1000 to 1500 ratinig by playing
four games against a 2000 player (who will
go down to 1500). Then X may win 4 games
against a 3000 player and by doing so X
would gain extra half of its new rating, etc.
By playing this way 4 game matches against
stronger and stronger opponents, whose pre-match
rating each time would be the double of the
rating of X, her/his rating after n matches
would be 1000 * (1.5)^n; e.g. after 10 matches
(40 won games in a row) it'd be 57665.
Now you see that the additive representation
may be psychologically easier on your eyes.

***

Regards,

Wlod (Wlodzimierz Holsztynski)