Rating Formula Question

[doucy]

The following equation is used in conjuction with a 2 player game. At the beginning of the game, each player has a rating that reflects his skill (usually somewhere between 100 and 2000; the higher the rating, the better the player.) This equation is used to determine, at the end of the game, the adjustment to each player's rating based on the outcome of the game (you win the game, your rating increases, you lose, your rating decreases.)

<font color="white"> ________________</font> 1
r1 + k * (w - --------------------)
<font color="white"> ___________</font> 1 + 10^((r2-r1)/400)


r1 = your pre-game rating
r2 = your opponent's pre-game rating
let w = 1 if you win, .5 if you draw, and 0 if you lose
<font color="white"> ________ </font> |d|
k = 20 + ----- where d = score difference, with a max value of 200.
<font color="white"> ________</font> 10

After observing the equation, you can see that the post-game rating adjustment is determined by 2 things: the margin of victory, and the difference in pre-game ratings. The greatest possible post-game rating adjustment is 40 points.

Question: You play a game, you lose, and your rating decreases by 40 points (the greatest possible number.) How many games will you have to play before that game will no longer have any effect on your rating? Assume that each subsequent game you play will be against a player with a rating equivalent to yours.

Another way to think of the question: The formula is such that as you play more and more games, the importance of previous games decreases. If you played a million games, the very first game likely would have no effect on your rating at all (i.e. regardless of if you won, lost, or tied that first game, your present rating would be the same.) However, the last game you played could have an effect up to 40 points. I am trying to figure out how many previously played games have an effect on a player's rating, under the assumption that the player is always competing against opponents whose ratings are identical to his own.

[bigpooch]

I don't understand why

k = 20 + abs(d)/10.

If the rating system is to accurately reflect playing
strength, is a victory with a score difference of ZERO
significant enough to include the "20 +" ? I would think
that you should use a different form of

k = a + abs(d)/b where the maximum value of k is 40, but
with a and b being smaller.

The rationale is that if someone wins by 200 points, is it
"only" worth twice as much as victory by ZERO points ?

Curiously, what is this used to rate?

[Siegmund]

This type of formula has been used for rating, well, almost everything. With a different definition of k, it was invented for chess, and is almost universally used in backgammon. It has been applied with varying degrees of success for other games. I find it only mildly inadequate for pool and cribbage (you can claim that the margin of victory doesn't matter, if you agree ahead of time to 'per-game' stakes rather than per ball or per point.) Its performance is very disappointing indeed when applied to bridge.

First word of advice for the OP: 40 points per game is way, way, way, way too big. Your entire rating system consists of nothing but random noise. In backgammon the adjustment is 2 points for a one-point match, about 5 points for standard hour-long matches, and maybe 10 points for an all-afternoon-long match -- and even under those circumstances a player's rating may move up or down by about 100 points in a month's time by chance.

As to your question about rates of decay.... applying it to your particular game will require repeating some experiments or carefully applying some scaling factors, but I think you will find Doug Zare's answer to this same question in backgammon (http://www.math.columbia.edu/~zare/rdraft.html) enlightening.