3 ms·

I'd at least want some idea of what to compare it to. That's certainly closer than if we assigned percentages at random. On the other hand, my intuition is they

by bbctol 10y ago

I'd at least want some idea of what to compare it to. That's certainly closer than if we assigned percentages at random. On the other hand, my intuition is they're less close than if every voter was picking at random, though I'm not sure if either of those counterworlds is relevant to the argument.

mikeash 10y ago
Each voter picking at random will get you almost exactly 50/50 every time, because the sample size is so large. The expected variance goes with the square root of the sample size, so with 100 million voters, you'd expect the outcome of each voter choosing randomly to be about 10,000 off from exactly 50/50. Here's a quick Python snippet to test it empirically: print sum(random.randint(0, 1) for x in xrange(0, 100000000)) (I strongly recommend using pypy to run it. CPython takes forever to complete.) With 8 runs, I got these results: 49995906 50003341 50009287 49996048 49996424 49993962 49989590 50002171 All of which require at least two decimal places in the percentage to be distinguishable from 50% each way.
- foota 10y ago
  Parent is saying if we randomly chose the outcome, not the votes.
  - AnimalMuppet 10y ago
    Sure. But why should we expect random percentages to have any relationship to reality? (I would argue that random votes isn't related to reality either...)
  - mikeash 10y ago
    Huh? "than if every voter was picking at random" Sounds like what I showed to me.
- ryanmonroe 10y ago
  Pretty amazing. If the outcome is determined by 100 million independent coin flips, the probability of a margin greater than one tenth of a percent is 1.5e-23. In R n <- 100000000 diff <- 0.1/100 a_wins <- n*(1-diff)/2 b_wins <- n*(1+diff)/2 p1 <- pbinom(a_wins, size = n, prob = 0.5) p2 <- pbinom(b_wins, size = n, prob = 0.5, lower.tail = F) p1 + p2 # = 1.523959e-23

by bbctol 10y ago

mikeash 10y ago

Each voter picking at random will get you almost exactly 50/50 every time, because the sample size is so large. The expected variance goes with the square root of the sample size, so with 100 million voters, you'd expect the outcome of each voter choosing randomly to be about 10,000 off from exactly 50/50. Here's a quick Python snippet to test it empirically: print sum(random.randint(0, 1) for x in xrange(0, 100000000)) (I strongly recommend using pypy to run it. CPython takes forever to complete.) With 8 runs, I got these results: 49995906 50003341 50009287 49996048 49996424 49993962 49989590 50002171 All of which require at least two decimal places in the percentage to be distinguishable from 50% each way.

foota 10y ago
Parent is saying if we randomly chose the outcome, not the votes.
- AnimalMuppet 10y ago
  Sure. But why should we expect random percentages to have any relationship to reality? (I would argue that random votes isn't related to reality either...)
- mikeash 10y ago
  Huh? "than if every voter was picking at random" Sounds like what I showed to me.
ryanmonroe 10y ago
Pretty amazing. If the outcome is determined by 100 million independent coin flips, the probability of a margin greater than one tenth of a percent is 1.5e-23. In R n <- 100000000 diff <- 0.1/100 a_wins <- n*(1-diff)/2 b_wins <- n*(1+diff)/2 p1 <- pbinom(a_wins, size = n, prob = 0.5) p2 <- pbinom(b_wins, size = n, prob = 0.5, lower.tail = F) p1 + p2 # = 1.523959e-23