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I thought this would just be an application of Benford's law, but they note in the beginning that using Benford's law doesn't work for elections (citing this pa
by blueintegral 10y ago
I thought this would just be an application of Benford's law, but they note in the beginning that using Benford's law doesn't work for elections (citing this paper: http://www.vote.caltech.edu/sites/default/files/benford_pdf_4b97cc5b5b.pdf http://www.vote.caltech.edu/sites/default/files/benford_pdf_...)
Does anyone know why Benford's law doesn't work here but does work for other made up numbers in applications like accounting?
- e2e8 10y agoBenford's law best applies to data that spans multiple orders of magnitude.
- TeMPOraL 10y agoIsn't it that Benford's law applies when parts of data are being multiplied by other parts (like in most things in real world)? I.e. two uniform distributions multiplied together give a non-uniform distribution?
- sampo 10y agoFor Benford's law to apply, the numbers (the non-fabricated numbers) need to be coming from a source/distribution covering several orders of magnitude. The fabricated voting percentages are limited to between 0 and 100.
- moptar 10y agoHuh? You can have many orders of magnitude between 0 and 100%. 0.0001 is 4 orders of magnitude smaller than 1.
- empath75 10y agoRaw vote totals then?
- oceliker 10y agoWe ran a test for one of the problem sets in a course I was TAing. Raw vote counts did follow Benford's law in 2000 US presidential elections. We did not portray it as a tool to detect or disprove fraud, however.
- sampo 10y agoPerhaps the sizes (number of eligible voters, and also number of people who actually voted) of the voting districts follows the law? Then you could take the total vote counts, and distribute them back to the candidates in a lot of artificial ways (e.g. always 50-50, a random percentage between 30 and 70, say) and probably obtain data that still follows the law.
- ethan_g 10y agoBenford's law applies when the underlying distribution is approximately exponential. Because candidates with an exponentially small chance to win are not likely to run, it wouldn't make much sense for elections to have an exponential distribution. Much more plausible distributions are (truncated) normal or uniform, neither of which satisfies Benford's law.
- Terr_ 10y agoI like to visualize it as throwing random darts at graph-paper that has logarithmic marks on both axes.
- colanderman 10y agoI have seen Benford's law applied to voter turnout, where the distribution could be expected to be closer to exponential (but still not truly, due to districting practices).