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Another way of writing it out: How likely is it that you'd get these votes distributions 51.2000000% 44.2000000% 04.6000000% exactly? With all of
by tumult 2y ago
Another way of writing it out:
How likely is it that you'd get these votes distributions
51.2000000%
44.2000000%
04.6000000%
exactly? With all of those clean 0s? Very low.
But it's also possible that there was sloppy reporting and the vote counts were re-processed at some point in the chain and rounded to one decimal place.
- neura 2y agoIt's more that if you start with those clean, single decimal percentages and a total number of votes, you'd end up with decimals for number of votes, which isn't possible. So if you then remove the decimal from the votes, you get slightly different percentage values when taken to 7 decimal places, but the original decimals would still be the same. The chances of those numbers occurring normally for all 3 vote counts together is just ridiculously tiny.
- deleted 2y ago[deleted]
- achempion 2y agoThe numbers are definitely sus, but what if they were 52.2543689% 44.2689426% 04.6345625% How likely is it that you'd get these votes distributions exactly with these exact tails? Compared to all other possibilities?
- kadoban 2y agoBasically the same 1/verymany chance, but that doesn't matter. The difference is that there's no particular reason to choose these numbers to start with. There _is_ a reason to choose nice round, but not too round numbers: that's what humans do.
- jvanderbot 2y agoThe question is "How likely is it that humans would begin with those numbers when fudging?" answer is low. But clean numbers? Much more likely.
- noitpmeder 2y agoThis comment perfectly distills this post.
- melenaboija 2y agoThe same probability as any combination of three results with 7 decimals and adding up to 100.
- staunton 2y agoThe question isn't "probability you get those exact numbers". It's "probability that you get numbers which all have at most N decimal places".
- deleted 2y ago[deleted]
- contravariant 2y agoWell there weren't zeros but within rounding error it was exact. That actually gives a way to estimate the probability. There's 1002 choose 2 ways to divide 1000 permils over the 3 options. While there's 10 058 776 choose 2 ways to divide the 10 058 774 votes. That works out to about 1e-8 of the possible results being an exact multiple of 0.1% up to rounding error. Of course an actual election doesn't simply pick one of the possible results at random (heck even if everyone voted randomly that wouldn't be the case). However these 'suspicious' results are distributed in a very uniform stratified fashion, any probability distribution that's much wider than 0.1% would approximately result in the same 1e-8 probability. And pretty much no reasonable person would expect a priori that the vote would result in such a suspicious number with such a high accuracy, so this should be considered strong evidence of fraud to most people.
- happyopossum 2y agoBut that’s not what happened, if you read the article it actually expands everything out to the seventh decimal, and they’re not all zeros
- tumult 2y agoIt is equivalent to all zeroes with the numeric precision used.
- jvanderbot 2y agoYes, they are not all zeros, but they are exactly what you'd expect if someone picked percentages that were all zeros, then added +1/-1 to get integer votes. So the argument is once removed, but still compelling.
- culi 2y agoActually if you're going that many decimals its 51.1999971% 44.1999989% 04.6000039%