First you should read the blog post I use as a reference.
Elections in Donets - anatomy of a fraud.
I think this needs more publicity. I try to explain how this kind of results of elections in East Ukraine are impossible.
First the official results:
Zaharchenko : 765 340 / 969 644
Kofman : 111 024 / 969 644
Sivokonenko : 93 280 / 969 644
Discarded votes : 43 039 / (969 644+43 039)
Everything seems to be ok?
Then we calculate the percentages:
Zaharchenko : 765 340 / 969 644 = 78,92999905119820000%
Kofman : 111 024 / 969 644 = 11,44997545490920000%
Sivokonenko : 93 280 / 969 644 = 9,62002549389260000%
Discarded votes : 43 039 / (969 644+43039) = 4,24999728444143000%
The problem with these figures is that the first two decimals are ok, but then at least the next two decimals are either 00 or 99 which means they are very close to two-decimal numbers. All of them.
Lets take Zaharchenko's votes for further examination:
Zaharchenko: 765 340 / 969 644 = 78,92999905119820000%
That is really close to 78,93%
In fact it is not possible to get percentage to 78,93% because you would not get whole votes:
969 644 * 78,93% = 765340,0092000 votes
So if you first "invented" the two decimal percentage as voting result and then calculate the vote count, you would have to truncate to 765340.
There were 969 644 accepted votes. That means there are 969 644 possible results for Zaharchenko and probability of any result is 1/969644.
There are 10000 possible percentages with two decimals. Because the accurate two decimals results are not possible and it's almost always between two votes, lets take the both upper and lower vote counts. So possible vote counts for two decimal percentage are ~20000.
The probability that Zaharchenko's vote count is one vote near the two decimal percentage is 20000/969644 ~ 1/50. That is 2% probability. That is not impossible, that could happen. It means that if you held elections anywhere in the world and you pick 50 candidates, at least one of them is propable to have two decimal percentage.
Let's then get the two decimal percentages for all the candidates:
Zaharchenko : 765 340 / 969 644 = 78,93%
Kofman : 111 024 / 969 644 = 11,45%
Sivokonenko : 93 280 / 969 644 = 9,62%
Discarded votes : 43 039 / (969 644+43 039) = 4,25%
It's here when things get highly improbable.
Zaharchenko's probability 1/50
Kofman's probability 1/50
Sivokonenko probability 1/50
Discarded votes probability 1/50.
Because sum of candidates votes is 100% it is not possible that two candidates have two decimal percentages and the third to have more decimals. So we only use probabilities of the candidates because the third one's probability is "tied" to two first ones probability and therefore his probability is 1/1 meaning it always happens. The probability that the three candidates all have two decimal percentages is therefore 1/50 * 1/50 * 1 = 1/2500.
Then we add the discarded votes and the probability that all the percentages have two decimals is 1/2500 * 1/50 = 1/125 000 or 0,0008% probability . That's the smoking gun. That means DNR should have 125 000 elections and get one result like this. Of course, it is still possible. At least in Donetsk People's Republic.
PS. The elections held in May 2014 in DNR and LNR had only one decimal percentages.
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