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koube | 1 year ago
For example the probability of getting [1,2,3,4,5,6] as the winning numbers in the lottery is the same as any random set of numbers.
koube | 1 year ago
For example the probability of getting [1,2,3,4,5,6] as the winning numbers in the lottery is the same as any random set of numbers.
jvanderbot|1 year ago
So for a given number of votes, which determines a split, how many times does the split come out so nice? Answer: Effectively none - there are always ugly numbers with lots of decimal places.
Now that analysis comes after they conjecture that the percentages were fixed apriori. The first comment "That seems fishy" basically says this. "How can it be that we're so close to even 1/10 percentages. How can it be that we're exactly one vote off from nice 1/10 percentages"? Fishy indeed - must be rounding.
And they tell you: it's very unlikely to be 1 vote off from nice 0.1% percentage splits.
tumult|1 year ago
How likely is it that you'd get these votes distributions
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.
enoch_r|1 year ago
But in reality, we can say that:
- her probability of winning in the world where her father is cheating is very high
- her probability of winning in the world where her father isn't cheating is very low
Together these two facts give us evidence about which world we're actually inhabiting - though of course we can never be completely certain!
In the same way, yes, it's equally (im)probable that the winning percent will be 51.211643879% or 51.200000000%. But the latter is more likely to occur in a world where Maduro said "get me 51.2% of the votes" and someone just did that mechanically with a pocket calculator, which is good evidence about which world we live in.
ChadNauseam|1 year ago
60% is a nice, round percentage. In an honest election, this is just as likely to be reported as any nearby percentage, like 59.7% or 60.3%. As you mention, any particular percentage is equally (and extremely) unlikely. SUppose this you estimate the chance of this occurring, given an honest election, is 1/1000.
60% however is a much more likely outcome if the election results were faked sloppily. A sloppy fake is reasonably likely to say "Well, why not just say we won 60%". Suppose you estimate the chance of this occurring, given a sloppily faked election, are 1/100.
Bayes' theorem tells us that we can use this information to "update our beliefs" in favor of the election being faked sloppily and away from the election being honest. Say we previously (before seeing this evidence) thought the honest:faked odds were 5:1. That is, we felt it was 5 times more likely that it was honest than that it was sloppily faked. We can then multiply the "honest" by 1/1000 (chance of seeing this if it was honest), and the "faked" by 1/100 (chance of seeing this if it was faked), to get new odds of (5 * 1/1000):(1 * 1/100), which simplifies to 1:2. So in light of the new evidence, and assuming these numbers that I made up, it seems twice as likely that the election was faked.
This exact analysis of course relies on numbers I made up, but the critical thing to see here is that as long as we're more likely to see this result given the election being faked than given it being honest, it is evidence of it being faked.
a0123|1 year ago
To think they could have got away with it if only they hadn't forgotten.
That´s what you get when you defer the dirty work to interns on their first day, I guess. Which you always rely on to stay in power. Wouldn't want to rely on competent advisers who would have reminded you to come up with a non-round number with 8 or 9 decimals.
paxys|1 year ago
Here's an example – if I generate 10 random numbers between 1 and 100, what is more likely: all ten are multiples of 10, or at least one is not a multiple of 10?
Waterluvian|1 year ago
The issue here is that a bunch of the percentages imply super round numbers.
The signal isn’t that there’s a round number. It’s that they’re all round numbers.
neura|1 year ago
SoftTalker|1 year ago
Yes, but the comparison is not to "any random set of numbers" it's "all other random sets of numbers"
The candidate got 52.200000% of the vote instead of any other percentage, not another specific percentage.
happyopossum|1 year ago
No, he got 51.1999971%. It’s right there in the second table of the article
neura|1 year ago
1. It's very easy to arrive at the provided values, if you make up some percentages that only go to a single decimal value (1/10th). Though doing so would result in vote counts that are decimal, as well. Then if you just remove the decimal from those values, the given percentages don't change enough to be incorrect, but even when taken to 7 decimal places, the new values are pretty clearly due to the rounding (44.2%: 44.1999989%, 4.6%: 4.6000039%).
2. While yes, the chance of these vote counts coming up in this kind of pattern is similar to the example you provided, even if you were using 0-9 for your example of 6 values, the total combinations is about an order of magnitude less than the total vote count provided here.
3. The finer point made is that there's a very small chance for one of the vote counts to show up as a number that so nicely fits the single decimal percentage, but in this case, all 3 vote counts fit this pattern. The calculations are shown for just 2 of the candidates (so not including the "other") resulting only a 1 in 100 million chance.
tumult|1 year ago
shadowgovt|1 year ago
Having read the article, I came away with similar questions and I appreciate the sibling comments clarifying.
unknown|1 year ago
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