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>>> ((Mean + Magic1*Sigma) + Magic2*Sigma*(math.factorial(Count_M)/math.pow(math.factorial((Count_M-Rank)),Rank))*(math.log(Count_M)-(round(Gamma+math.log(Rank-1)))))-((Mean + Magic1*Sigma) + Magic2*Sigma*(math.factorial(Count_F)/math.pow(math.factorial((Count_F-Rank)),Rank))*(math.log(Count_F)-(round(Gamma+math.log(Rank-1)))))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
OverflowError: long int too large to convert to float

>>> ((Mean + Magic1*Sigma) + Magic2*Sigma*math.exp(0.5*(math.log(2)+math.log(math.pi)+math.log(Count_M))+Count_M*math.log(Count_M)-1 - 0.5*(math.log(2)+math.log(math.pi)+math.log((Count_M-Rank))+(Count_M-Rank)*math.log((Count_M-Rank))-1 + Rank*math.log(Count_M)))*(math.log(Count_M)-(round(Gamma+math.log(Rank-1)))))-((Mean + Magic1*Sigma) + Magic2*Sigma*math.exp(0.5*(math.log(2)+math.log(math.pi)+math.log(Count_F))+Count_F*math.log(Count_F)-1 - 0.5*(math.log(2)+math.log(math.pi)+math.log((Count_F-Rank))+(Count_F-Rank)*math.log((Count_F-Rank))-1 + Rank*math.log(Count_F)))*(math.log(Count_F)-(round(Gamma+math.log(Rank-1)))))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
OverflowError: math range error

math.factorial(Count_M)/math.pow(math.factorial((Count_M-Rank)),Rank))

Posted by funkyass
you don't need to put the entire formula in a single line.

math.factorial(Count_M)/math.pow(math.factorial((Count_M-Rank)),Rank))

that seems off. is it not n!/((n-k)! * n^k)?

thats also probably where the overflows are coming from. you can reduce the size of the number that produces with a bit of work, but not easily expressed in a single line.

Posted by wertigon
Seems like you still do not understand how the math works. It is pretty simple, really:

https://en.wikipedia.org/wiki/Order_statistic#Probability_distributions_of_order_statistics

This is, in other words, a standard function.

Sympy could help with the bigger calculations, but more specifically try out this:

https://www.sympygamma.com/input/

Of course, it is a bit of manual repetitive labor to input 100 different equations, but maybe you could simplify the equation if you plug in some values.

Posted by wertigon
Not sure how you get your formula for the model so messed up. To be clear, the formula should be:

µ + sigma (c₁ + (c₂ n! (ln n - H(k-1))) / ((n-k)!n^k))

Where H(k-1) = 1 + 1/2 + 1/3 + ... + 1/(k-1).

Incidentally, H(n) can be written as follows:

H(n) = log₁₀ n + 33841/58628 + 1/2n - 1/12n²

With this it should be possible to reach a good enough accuracy with a symbolic calculator (e.g. a calculator that understands the notion of 100! / 101!), at least for the purpose of determining the difference in ELO.

1500 + 350 (1.25 + (0.287 60000! (ln 60000 - (log((30000-1))/log(10) + 33841/58628 + 1/2(30000-1) - 1/12((30000-1)^2)))) / ((60000-30000)!(60000^30000)))

1500 + 350 (1.25 + (0.287 6000! (ln 6000 - (log((3000-1))/log(10) + 33841/58628 + 1/2(3000-1) - 1/12((3000-1)^2)))) / ((6000-3000)!(6000^3000)))

Posted by funkyass
have you looked into how the chess scoring system works?

Posted by wertigon
Not quite an expert at this particular math and model, but if your calculations show up wrong it is probably because you made a mistake along the way.

When I cannot solve a piece like this I usually turn towards a university and send an email to one of my old professors. For them, it doesn't take long to solve those particular models. Perhaps you should ask them?

Posted by sureanem
I wouldn't think there's anything wrong with the calculations. The replication study also said it gave very high results, like estimating the top German player above the current world champion. Plugging in k = 7, it says that the gap according to the model would be 228-298 points, depending on your values for n. This is about what the study finds too.

If you're saying the model is horribly broken and does not accurately reflect reality, then that makes sense. I've yet to understand what was wrong with the far simpler model that can be calculated on a better pocket calculator.

Posted by wertigon
The hypothesis is that the skill difference between the k:th persons of the two pools correlate to their respective pool size, yes?

Suppose we have a total population size (Z) of 1000 individuals. In this pool size, we have 900 people from one group (X), and 100 people from a different group (Y).

For a simplistic model, if a correlation exists between skill difference and pool size, and it is linear, the model could be something like: D_n = c₁ * n + c₂

Where c₁ and c₂ are calculated from the ratio between X and Y.

If no difference exist, X and Y should have the same mean. But since they do, X and Y should have different means, especially compared to Z. This is why I think you made a mistake in your model calculations.

Either way a correlation has been proven. Perhaps a ML method could shed further light upon this. But, yes. I think we have reached the end for now.