(no title)
tumult | 1 year ago
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.
contravariant|1 year ago
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.
neura|1 year ago
The chances of those numbers occurring normally for all 3 vote counts together is just ridiculously tiny.
culi|1 year ago
achempion|1 year ago
kadoban|1 year ago
jvanderbot|1 year ago
But clean numbers? Much more likely.
unknown|1 year ago
[deleted]
happyopossum|1 year ago
jvanderbot|1 year ago
So the argument is once removed, but still compelling.
tumult|1 year ago
noitpmeder|1 year ago
melenaboija|1 year ago
staunton|1 year ago