Okay here we go. This is gonna be fun, isn't it? I started this last night, so consider it a direct response to your post from 11:49pm:
"If you use the distribution of percentages rather than the distribution of incidences, there is no multiplier, but to compare distributions of different sample sizes (number of games) you have to *multiply* them by their standard deviations. This is all explained on my Wall."

Right. On your three year old wall of technobabble which I have trouble following as a math major. Lemme take a look at that.

"The Test Against Pure Luck (TAPL) is a measure of skill that eliminates the random fluctuations that are characteristic of the game-by-game scoring system and rewards players who play more games. "
We've all agreed at this point that it's a measure of deviation by its pure nature.

"In math jargon, the TAPL is a Pearson?s chi-square goodness of fit test against a uniform distribution."
No it isn't. You don't actually run the test. You just find the chi-square statistic. Not quite the same.

"The TAPL is the normalized sum of seven numbers. The first number is found by taking the difference between the expected number of first-place finishes and the actual number of first-place finishes, given the number of games played, squaring this difference and dividing it by the expected number of first place finishes. The second number is obtained by doing the same calculation for the second-place finishes. The third number is for the third-place finishes, and so on for all seven levels of finishes. The sum is then divided by the square root of the number of games played to get a result expressed in standard deviations."
A chi-square gof is simpler than that- there's no division or multiplication by standard deviation. It's just sum(((observed-expected)^2)/expected). that's it. that's all there is.

"The mean of the chi square goes as the number of games played, and its variance goes as the number of games played"
mean goes by k and variance goes by 2k, where k is the number of degrees of freedom- which is NOT determined by number of games played. It's 6. Number of categories minus one.

So if you decide to convert the percentages to expected COUNTS instead, your formula would be something like this:
http://i.imgur.com/mJyGe.png

... I think that's a good stopping place for now. There'll be more later, kids, don't you worry.

great! I was about to play a game after reading this but I will have to wait ten minutes. Reading your post short circuited my mind.

jk, I understand the most part of it

and well done

oh man, so much animosity! verm-chloe really hates skrum, and vice versa

It is the central limit theorem that says that as the number of samples increases, the Pearson's chi square goes to a normal distribution whose mean goes as the number of samples and whose variance goes as the number of samples.

The TAZD is not testing whether a player's distribution is significantly different than the null hypothesis; it is a way of comparing players with different numbers of games. So that is why the totals have to be scaled by the square root of the number of games.

To recapture the actual chi-square statistic of a player, divide the player's TAZD by 1000 and multiply by the square root of the number of games. Dottir, for example, has a TAZD of 25137 and has played 6652 games. Her chi-square statistic is 2050.16. For six degrees of freedom, the chi-square test is 12.59 for p = 0.05, 16.81 for p = .01, and 22.40 for p = .001. So Dottir's profile rejects the null hypothesis. If she had had the same profile but with only 66 games, her chi-square would be 204.2 which would still be significant. For a reasonable minimum number of games, we don't have to worry about whether an individual player's profile is statistically significant.

"The TAZD is not testing whether a player's distribution is significantly different than the null hypothesis"
-> the TAZD isn't a Pearson's chi square goodness of fit test. Which is in fact precisely what I said.

The same size isn't about significance- that's not even being brought into question here.

Central limit theorem states that as k approaches infinity for a chi-square distribution, the distribution of (X-k)/sqrt(2k) tends toward the normal. where k is DEGREES OF FREEDOM which is NOT determined by sample size for a pearson's chi square.

The central limit theorem says that as the number of **samples** from a distribution increases, the mean of the samples approaches a normal distribution even if the original distribution is not normal. Look at the third line in the first paragraph under the heading Classical CLT from this site:

http://en.wikipedia.org/wiki/Central_limit_theorem

and observe the variance of the mean in the expression.

In that article, k is the number of observations, not the number of cells. In some of the other Wikipedia articles, the number of observations is referred to as n. So, for a large number of observations, the central limit theorem kicks in, as it is supposed to.

Nice try, though.

In that article k is degrees of freedom.

which is not relevant here because we both agree that the Pearson's chi square used in the TAZD has six degrees of freedom.

We both should agree that the number of observations is equal to the number of games, and what triggers the central limit thorem is the number of observations.

So the central limit theorem holds, and you have to deal with it.