I had a feeling youâ€™d say something like this. Well, to that, Iâ€™ve prepared an answer: you could do is a favour and post a direct link yourself â€” I just donâ€™t use X and the like, and that particular channel happens to be in my bookmarks, as I use at as a reference and information source for myself. Couldnâ€™t be bothered to dig to the original source.

Thatâ€™s the whole point of it.

Objective data is objective data, whosoever posted it (even if someone like you ).

One could guess, but thereâ€™s no point trying to peek inside someoneâ€™s soul, especially online.

Iâ€™m really surprised how any of you could make anything out of the info thatâ€™s displayed on the website (maybe there are some obscure tabs that I missed, dunno): itâ€™s absolutely unclear at which ranks those games were played; for example, in a bot-infested D5-D1, scoring something like 20-0 isnâ€™t something outstanding.

Besides, since youâ€™ve made me use my head here for once, letâ€™s take a look at those numbers and partake in some maths on the level of bumpkins like Maw and Paw (since Iâ€™m not that well versed in the intricate differences between hayseeds, hicks, hillbillies and rednecks, youâ€™ll have to excuse me if I miss a proper title; besides, Maw and Paw look kindaâ€¦ zombie-green, so, I guess, the last-named option would be â€˜greennecksâ€™? ).

Letâ€™s assume a fixed probability of winning a single game `p=0.5`

â€” a little crude estimate, but itâ€™ll do. The probability to win `k`

games out of `n`

is given by the binomial distribution, and, as is well-known, the mean value is equal to `np`

, and the dispersion (or variance) is `np(1-p)`

, or, in this case, just `np^2`

, thus the standard deviation (SD) is `sqrt(n)*p`

. I canâ€™t easily tell you the factor to multiply the SD in order to estimate the stochastic error in a given confidence interval for the binomial distribution (see also this; apparently, no maths gurus here to help me out), but suffice to say that the relative stochastic error (SD/Mean) is proportional to the inverse square root of `n`

. Thus, for an order of 100 games, itâ€™s the order of 10% (thus the most expected observed win rates would be within 45%-55% for a â€˜one sigmaâ€™ confidence intervalâ€¦ which isnâ€™t even that great â€” for example, about 68% for the normal distribution, which *isnâ€™t* the case here, according to Wikipedia, and youâ€™d want â€˜three sigmaâ€™ to be rather sure â€” although itâ€™s not always quite enough), for 10 000 â€” the order of 1% and so on.

With such a small sample size `n`

, according to these estimates, itâ€™s not even that impossible or improbable to have figures like that, although Iâ€™d like to see the confidene intervals for different values of those coefficients.

Iâ€™d really ask you how exactly you estimated that, be it nitpicking or not, but is there any pointâ€¦