5 ms·

If I'm in charge of forging a presidential election, how difficult is it for me to use realistic, "ugly" numbers to sell it more effectively?

anvuong 2y ago
You just need to run some Monte Carlo sims where the priors are your desired rates, then use this results to alter the real numbers. It's as random as it gets.
verbify 2y ago
This kind of mistake is easy to avoid. The problem is that there are a lot of potential mistakes that could be made, this is just one of them.
- energy123 2y ago
  And the people doing the fraud aren't going to be computer scientists or statisticians. They were chosen for their loyalty to the dear leader.
  - AnimalMuppet 2y ago
    The dear leader wasn't chosen for his ability, either, but for his loyalty to the previous dear leader.
aaplok 2y ago
In light of recent analyses of suspicious elections (Iran, Russia, Venezuela), it seems harder than it sounds to avoid discernible patterns. On the other hand, the goal is to get away with fraud, not to convince an international community who will likely look for any confirmation of their suspicion. It would be interesting to look for patterns in a (presumably) fair election like the recent British one for comparison. Disclaimer: I have never tried to rig elections myself so I don't really know how hard it is.
- colimbarna 2y ago
  > On the other hand, the goal is to get away with fraud, not to convince an international community who will likely look for any confirmation of their suspicion. Despite my joke in a sibling comment, this is key. When you're a politician everything is power relations. Sometimes it's necessary to show that you have the power to semi-obviously rig an election. Your bargaining position is different if it requires military force to remove you vs just an unhappy electorate. You can achieve different things.
  - lazide 2y ago
    Yup. It’s a special kind of power that can flat out rig an election and have opponents ‘fall out of windows’ with no repercussions. The type no one wants to even be seen trying to challenge.
- IAmGraydon 2y ago
  >It would be interesting to look for patterns in a (presumably) fair election like the recent British one for comparison. Someone should run this same analysis on all the election data they can get their hands on. Who knows what might be found.
  - nick238 2y ago
    There were some pretty shit analysises of the 2020 US elections that Matt Parker covered with videos like "Why do Biden's votes not follow Benford's law?"[1] and "Why was Biden's win calculated to be 1 in 1,000,000,000,000,000?"[2]. I don't know about other countries, but the amount of data that every county in every US state produces makes systematic fraud pretty much impossible. If there's literally only 3 numbers produced by the Venezuelan government, you need to be seriously incompetent to have detectable fraud because techniques like Benford's or Zipf's law need lots of individual numbers. [1]: https://youtu.be/etx0k1nLn78 https://youtu.be/etx0k1nLn78 [2]: https://youtu.be/ua5aOFi-DKs https://youtu.be/ua5aOFi-DKs
    - IAmGraydon 2y ago
DonsDiscountGas 2y ago
Not difficult at all. Just pick the approximate numbers you want and then introduce a random error of a few percent. (Normal, uniform, doesn't really matter). This is also not hard for statistics experts to detect, but it's much harder to prove (aka you've got plausible deniability). One wonders why they didn't even bother to do fraud slightly better.
- stouset 2y ago
  Unless done carefully this will almost certainly fail Benford’s Law. Manipulating statistics is harder than you think.
  - dllthomas 2y ago
    It's my understanding that legitimate vote totals aren't likely to conform to Benford's law in the first place. Even if that's the case, though, there might very well be other applicable tests this would run afoul of.
    - stouset 2y ago
      I’m not a statistician so I may be confusing it with Zipf’s law. But IIRC tallies from individual precincts should roughly conform to Benford’s law.
      - dllthomas 2y ago
        I think the concern is that precinct size tends to cluster in ways that mean results can cluster in ways that - for a large portion of the data - does not span a full order of magnitude.
        dllthomas 2y ago
        To elaborate, if we imagine a polity with precincts that turn out 10,000 people each election, with two major party candidates that each get between 20% and 80% of the vote, we'd see precisely 0% precincts with a leading digit of 1, much less the ~30% predicted by Benford's law. Of course that doesn't exactly describe any real polity, but it doesn't seem surprising that real elections would be enough like that to screw with the pattern.
a0123 2y ago
No you see, they had the perfect plan. But they got foiled by that one thing they always forget at every single election (that never gets brought up when your government agrees with the result, because in that case it's just a "statistical anomaly" or "shit happens sometimes")
onlyrealcuzzo 2y ago
Votes are not random. So it would be non-trivial to make results look real. Additionally, if even a few polling places release real data - that can really complicate things. It will look very, very suspicious when some polling places display wildly different behaviors (especially if they match expectations) then the rest of the polling places.
tptacek 2y ago
I don't know, but the numbers they provided here are impossible.
forinti 2y ago
Use the real numbers but change the owners of each count. Except, of course, if the winning party has a huge advantage which you don't think people would buy.
wrsh07 2y ago
Hard. Naively, run the election, use real vote counts, claim that you're the one that got more. Ok but now we need to fake this on a local level. But everybody knows people in this district are more Party A and in that are more Party B. Well.... let's correct for that... If you want to give top line numbers, fine. Credible local numbers would get really hairy
- zaik 2y ago
  > Naively, run the election, use real vote counts, claim that you're the one that got more. I think this would be even more obvious, even to the general public, especially when there is a landslide victory. Oh, your result is 70%? That's exactly what the exit polls said about our candidate.
ordu 2y ago
Generate 10,058,744 random floating numbers from [0,1), and count how many will fall into each of the intervals: [0, 0.522), [0.522, 0.522+0.442), or [0.522+0.442, 1). You can do it with less calculations if you know how to generate random numbers from binomial distribution. Then you should calculate percentages from these three numbers just to be sure they are right. It is not difficult at all, but it needs some basic programming skills and some basic knowledge of statistics.
glenstein 2y ago
Actually think this is a fascinating question, although perhaps for different reasons within whatever reasons might have led you to ask it. I think you raise a legitimate point that it's not that hard to create ugly numbers. I also think that authoritarian social dynamics come to these questions with a kind of brutal simplicity, lack of intellectual curiosity or creativity, and a lot of the traits that would entail a value for democracy are mutually exclusive with the brute simplicity of authoritarian mindset. And so the story they choose to tell of how they won is going to have similar hallmarks of brute simplicity and absence of nuance.
BurningFrog 2y ago
It's not hard for you or me. But for the brutal thugs running Venezuela, it is a very advanced conept.
- fragmede 2y ago
  it's like Russia killing someone with polonium. Hitting 60% exactly sends a message to anyone with the bright idea of running against the brutal thug's preferred candidate that it might not be a good idea, because there are brutal thugs involved.
worewood 2y ago
Depends on if you're in on it with the forgers or if you're against it but being forced to do it. It's the perfect clue to leave in, as its intention is plausibly deniable and you can tip statisticians to uncover the fraud.

anvuong 2y ago

You just need to run some Monte Carlo sims where the priors are your desired rates, then use this results to alter the real numbers. It's as random as it gets.

verbify 2y ago

This kind of mistake is easy to avoid. The problem is that there are a lot of potential mistakes that could be made, this is just one of them.

energy123 2y ago
And the people doing the fraud aren't going to be computer scientists or statisticians. They were chosen for their loyalty to the dear leader.
- AnimalMuppet 2y ago
  The dear leader wasn't chosen for his ability, either, but for his loyalty to the previous dear leader.

aaplok 2y ago

In light of recent analyses of suspicious elections (Iran, Russia, Venezuela), it seems harder than it sounds to avoid discernible patterns. On the other hand, the goal is to get away with fraud, not to convince an international community who will likely look for any confirmation of their suspicion. It would be interesting to look for patterns in a (presumably) fair election like the recent British one for comparison. Disclaimer: I have never tried to rig elections myself so I don't really know how hard it is.

colimbarna 2y ago
> On the other hand, the goal is to get away with fraud, not to convince an international community who will likely look for any confirmation of their suspicion. Despite my joke in a sibling comment, this is key. When you're a politician everything is power relations. Sometimes it's necessary to show that you have the power to semi-obviously rig an election. Your bargaining position is different if it requires military force to remove you vs just an unhappy electorate. You can achieve different things.
- lazide 2y ago
  Yup. It’s a special kind of power that can flat out rig an election and have opponents ‘fall out of windows’ with no repercussions. The type no one wants to even be seen trying to challenge.
IAmGraydon 2y ago
>It would be interesting to look for patterns in a (presumably) fair election like the recent British one for comparison. Someone should run this same analysis on all the election data they can get their hands on. Who knows what might be found.
- nick238 2y ago
  There were some pretty shit analysises of the 2020 US elections that Matt Parker covered with videos like "Why do Biden's votes not follow Benford's law?"[1] and "Why was Biden's win calculated to be 1 in 1,000,000,000,000,000?"[2]. I don't know about other countries, but the amount of data that every county in every US state produces makes systematic fraud pretty much impossible. If there's literally only 3 numbers produced by the Venezuelan government, you need to be seriously incompetent to have detectable fraud because techniques like Benford's or Zipf's law need lots of individual numbers. [1]: https://youtu.be/etx0k1nLn78 https://youtu.be/etx0k1nLn78 [2]: https://youtu.be/ua5aOFi-DKs https://youtu.be/ua5aOFi-DKs
  - IAmGraydon 2y ago

DonsDiscountGas 2y ago

Not difficult at all. Just pick the approximate numbers you want and then introduce a random error of a few percent. (Normal, uniform, doesn't really matter). This is also not hard for statistics experts to detect, but it's much harder to prove (aka you've got plausible deniability). One wonders why they didn't even bother to do fraud slightly better.

stouset 2y ago
Unless done carefully this will almost certainly fail Benford’s Law. Manipulating statistics is harder than you think.
- dllthomas 2y ago
  It's my understanding that legitimate vote totals aren't likely to conform to Benford's law in the first place. Even if that's the case, though, there might very well be other applicable tests this would run afoul of.
  - stouset 2y ago
    I’m not a statistician so I may be confusing it with Zipf’s law. But IIRC tallies from individual precincts should roughly conform to Benford’s law.
    - dllthomas 2y ago
      I think the concern is that precinct size tends to cluster in ways that mean results can cluster in ways that - for a large portion of the data - does not span a full order of magnitude.
      - dllthomas 2y ago
        To elaborate, if we imagine a polity with precincts that turn out 10,000 people each election, with two major party candidates that each get between 20% and 80% of the vote, we'd see precisely 0% precincts with a leading digit of 1, much less the ~30% predicted by Benford's law. Of course that doesn't exactly describe any real polity, but it doesn't seem surprising that real elections would be enough like that to screw with the pattern.

a0123 2y ago

No you see, they had the perfect plan. But they got foiled by that one thing they always forget at every single election (that never gets brought up when your government agrees with the result, because in that case it's just a "statistical anomaly" or "shit happens sometimes")

onlyrealcuzzo 2y ago

Votes are not random. So it would be non-trivial to make results look real. Additionally, if even a few polling places release real data - that can really complicate things. It will look very, very suspicious when some polling places display wildly different behaviors (especially if they match expectations) then the rest of the polling places.

tptacek 2y ago

I don't know, but the numbers they provided here are impossible.

forinti 2y ago

Use the real numbers but change the owners of each count. Except, of course, if the winning party has a huge advantage which you don't think people would buy.

wrsh07 2y ago

Hard. Naively, run the election, use real vote counts, claim that you're the one that got more. Ok but now we need to fake this on a local level. But everybody knows people in this district are more Party A and in that are more Party B. Well.... let's correct for that... If you want to give top line numbers, fine. Credible local numbers would get really hairy

zaik 2y ago
> Naively, run the election, use real vote counts, claim that you're the one that got more. I think this would be even more obvious, even to the general public, especially when there is a landslide victory. Oh, your result is 70%? That's exactly what the exit polls said about our candidate.

ordu 2y ago

Generate 10,058,744 random floating numbers from [0,1), and count how many will fall into each of the intervals: [0, 0.522), [0.522, 0.522+0.442), or [0.522+0.442, 1). You can do it with less calculations if you know how to generate random numbers from binomial distribution. Then you should calculate percentages from these three numbers just to be sure they are right. It is not difficult at all, but it needs some basic programming skills and some basic knowledge of statistics.

glenstein 2y ago

Actually think this is a fascinating question, although perhaps for different reasons within whatever reasons might have led you to ask it. I think you raise a legitimate point that it's not that hard to create ugly numbers. I also think that authoritarian social dynamics come to these questions with a kind of brutal simplicity, lack of intellectual curiosity or creativity, and a lot of the traits that would entail a value for democracy are mutually exclusive with the brute simplicity of authoritarian mindset. And so the story they choose to tell of how they won is going to have similar hallmarks of brute simplicity and absence of nuance.

BurningFrog 2y ago

It's not hard for you or me. But for the brutal thugs running Venezuela, it is a very advanced conept.

fragmede 2y ago
it's like Russia killing someone with polonium. Hitting 60% exactly sends a message to anyone with the bright idea of running against the brutal thug's preferred candidate that it might not be a good idea, because there are brutal thugs involved.

worewood 2y ago

Depends on if you're in on it with the forgers or if you're against it but being forced to do it. It's the perfect clue to leave in, as its intention is plausibly deniable and you can tip statisticians to uncover the fraud.