I am the Keeper of the Spreadsheet for the SDMB Weekly Poker Game. To calculate standings over a quarter, we use a formula that we picked up from some web site. It works like this:
Say there are 10 players in a particular game. The raw numbers are place of finish for each player (F[sub]i[/sub]), and the total number of players in the game (N). A raw number is calculated from LN((N+1)/F[sub]i[/sub]). (In the hypothetical 10-player game, the raw numbers are 2.4, 1.7, 1.3, 1.01, .79, .61, .45, .32, .2 and .1.)
The raw numbers are averaged over the course of a quarter, taking into account how many games each player played (A[sub]i[/sub]). At the end, the average is converted into a ranking by (1-EXP(-A[sub]i[/sub]))*100).
So here is my question. It appears to me that the formula places greater weight for high finishes and low finishes, but I don’t really know. What **is **the effect of the way this is calculated? How is it different than if we just averaged (N-F[sub]i[/sub])/(N)?
Well, it appears to diminish the hit you’ll take from playing short-handed. It also diminishes the hit you’ll take from finishing low (your score gives zero points to the low scorer). I don’t really see that it weights high finishes all that much.
So if you have one player who’s consistently ahead or behind, it’ll take them longer to move away from the crowd with this ranking than it would otherwise. I haven’t verified this, but I find it plausible.
Put more elementarily, this system calculates your final ranking as 100% - the kth root of (p[sub]1[/sub]/(N[sub]1[/sub] + 1) * p[sub]2[/sub]/(N[sub]2[/sub] + 1) * … * p[sub]k[/sub]/(N[sub]k[/sub] + 1)), where you’ve played k many games, finishing in place p[sub]i[/sub] out of N[sub]i[/sub] many players in the i-th game.
Basically, this system first calculates the geometric mean of (your finishing place)/(number of players + 1) over all your games, and interprets the result (a number between 0% and 100% [with neither end actually achievable]) as something like “the percentage of everyone else who is better than you” (thus, 0% is the best and 100% is the worst). Then, as a minor last step, since we’re accustomed to reading higher numbers as better, we flip the scale into “the percentage of everyone else who is worse than you” by subtracting from 100%.
Alright, but why the geometric mean of (your finishing place)/(number of players + 1)? Why look to the product of these as indicative of your average performance (instead of their sum or something else)?
Well, that’s a fine question, and something I have to think about…
I think the calculation of the raw score is a little more complicated than it has to be, if you just want to understand how it weights things. The raw score is
S = log((N+1)/F) = log(N+1) - log(F) = c - k*log[sub]2/sub ,
where c and k are constants we don’t really care about. So your score is basically the negative of the base-2 logarithm (or any base you like, up to a constant factor) of your finishing place. So a first-place finish is worth 0, a second-place finish -1, a fourth-place finish -2, eighth place -3, and so on. Basically you get penalized the same amount for every factor of 2 drop in your finishing place; so dropping from first to second is a greater penalty than from third to fourth, for example. The constants c and k are just chosen to keep all of the scores positive and use the natural logarithm.
I found something interesting. Maybe someone can explain the how’s and why’s of it.
Suppose my weekly game is always ten players. Player A always finishes either first or tenth. Player B always finishes fifth or sixth. Player C finishes all over the place, finishing first, second, third, etc. equally. Taking a nonweighted average, all would be rated .55.
But with the log-weighted average, Player A gets a .71, Player B gets a .50, and Player C gets a .59.
So, high finishes are somehow weighted more than low finishes with this system? Is there a reason this result is desirable?