• bamboo@lemmy.blahaj.zone
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    3 months ago

    I might need an EIL5 for this. Is the fishy part that when you expand the vote percentages to 7 decimal places, both candidates are .19999 and then that gets rounded up to .2?

    • Skua@kbin.earth
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      3 months ago

      0.1% of the vote with roughly 10,000,000 voters is 10,000 votes. So for each 0.1% result, there are 10,000 possible numbers of voters that would round to that.

      For the sake of easy maths I’m going to briefly pretend that there were exactly 10,000,000 votes cast. To get 51.2% of the vote, I could get anywhere between 5,115,000 and 5,124,999 votes. What I got, though, was exactly 5,120,000 votes. The other two results look the same as well (although if the first two hit this requirement, the third necessarily must as well because it’s the “other” category)

      The seven decimal places thing is because there weren’t exactly 10,000,000 votes cast, so there’s not actually a number that produces exactly 51.2%. The author expanded the percentage out to show how close it is to 51.2% despite this - in fact, it is the closest possible whole number of votes

      • bamboo@lemmy.blahaj.zone
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        3 months ago

        I see, so the fishy part is that the vote counts basically correspond exactly to the nearest 0.1% for the top two candidates, whereas that’s unlikely. And the odds of that happening for both candidates is a 1 in roughly (10,000^2)?

        EDIT: rereading the article, I see the probability calculations at the bottom.