"New EA cause area: voting"; or, "what's wrong with this calculation?"

post by Optimization Process · 2021-02-26T16:50:54.275Z · LW · GW · 14 comments

This is a question post.

Contents

  Answers
    17 CarlShulman
    12 shadonra
None
14 comments

Consider my home county:

Expected GDP increase from my voting
... = (fraction of GDP at stake) * P(I swing election) * (P(I'm good) - P(I'm bad))
... = ($400B * 0.1%) * (1/800) * (60% - 40%)
... = $100k

...which seems absurdly large! And it just gets crazier as you look at larger areas, since GDP goes up like while P(swing) only goes down like . For the United States, the same calculation yields a benefit of $300k.

What's going wrong here? (Or, is nothing going wrong? In which case, I guess I'll stop donating to charity and devote that time and energy to Getting Out The Vote instead.)

Answers

answer by CarlShulman · 2021-02-27T21:21:58.035Z · LW(p) · GW(p)

The error is a result of assuming the coin is exactly 50%, in fact polling uncertainties mean your probability distribution over its 'weighting' is smeared over at least several percentage points. E.g. if your credence from polls/538/prediction markets is smeared uniformly from 49% to 54%, then the chance of the election being decided by a single vote is one divided by 5% of the # of voters.

You can see your assumption is wrong because it predicts that tied elections should be many orders of magnitude more common than they are. There is a symmetric error where people assume that the coin has a weighting away from 50%, so the chances of your vote mattering approach zero. Once you have a reasonable empirical distribution over voting propensities fit to reproduce actual election margins both these errors go away.

See Andrew Gelman's papers on this.

answer by shadonra · 2021-02-26T20:20:41.445Z · LW(p) · GW(p)

Accepting your calculation at face value implies that if you convince 800 (times some constant independent of n) nonvoters to vote with you, you will... (expect to) always win the election?

Your probability of swinging the election comes from modelling each voter as a fair coin flip, I believe, but this is not really a good model - if each voter is a 51% weighted coin flip then the calculus changes significantly.

↑ comment by Optimization Process · 2021-02-26T22:25:30.682Z · LW(p) · GW(p)

Ah! You're saying: if my "500k coin flips" model were accurate, then most elections would be very tight (with the winner winning by a margin of around 1/800, i.e. 0.125%), which empirically isn't what happens. So, in reality, if you don't know how an election is going to turn out, it's not that there are 500k fair coins, it's that there are either 500k 51% coins or 500k 49% coins, and the uncertainty in the election outcome comes from not knowing which of those worlds you're in. But, in either case, your chance of swinging the election is vanishingly small, because both of those worlds put extremely little probability-mass on the outcome being a one-vote margin.

(see also: johnwentworth's comment below [LW(p) · GW(p)])

Replies from: CarlShulman
↑ comment by CarlShulman · 2021-02-27T21:25:06.341Z · LW(p) · GW(p)

That is the opposite error, where one cuts off the close election cases. The joint probability density function over vote totals is smooth because of uncertainty (which you can see from polling errors), so your chance of being decisive scales proportionally with the size of the electorate and the margin of error in polling estimation.

14 comments

Comments sorted by top scores.

comment by Stanisław Barzowski (sbarzowski) · 2021-02-26T17:39:11.025Z · LW(p) · GW(p)

Could you explain where P(I swing election) = 1/sqrt(nVoters) is coming from?

Replies from: Optimization Process
↑ comment by Optimization Process · 2021-02-26T21:56:47.761Z · LW(p) · GW(p)

Sure! I'm modeling the election as being coin flips: if there are more Heads than Tails, then candidate H wins, else candidate T wins.

If you flip coins, each coin coming up Heads with probability , then the number of Heads is binomially distributed with standard deviation , which I lazily rounded to .

The probability of being at a particular value near the peak of that distribution is approximately 1 / [that standard deviation]. ("Proof": numerical simulation of flipping 500k coins 1M times, getting 250k Heads about 1/800 of the time.)

comment by johnswentworth · 2021-02-26T20:23:09.742Z · LW(p) · GW(p)

Wait... your county has a GDP of over half a million dollars per capita? That is insanely high!

Also, note that your probability of swinging the election is only  if the population is split exactly 50/50; it drops off superexponentially as the distribution shifts to one side or the other by voters or more. On the other hand, if you're actively pushing an election, not just voting yourself, then that plausibly has a much bigger impact than just your one vote.

Replies from: Optimization Process, Optimization Process
↑ comment by Optimization Process · 2021-02-26T22:24:01.042Z · LW(p) · GW(p)

Also, note that your probability of swinging the election is only 1/√n if the population is split exactly 50/50; it drops off superexponentially as the distribution shifts to one side or the other by √n voters or more.

Yesss, this seems related to shadonra's answer [LW(p) · GW(p)]. If my "500k coin flips" model were accurate, then most elections would be very tight (with the winner winning by a margin of 1/800, i.e. 0.125%), which empirically isn't what happens. So, in reality, if you don't know how an election is going to turn out, it's not that there are 500k fair coins, it's that there are either 500k 51% coins or 500k 49% coins, and the uncertainty in the election outcome comes from not knowing which of those worlds you're in. But, in either case, your chance of swinging the election is vanishingly small, because both of those worlds put extremely little probability-mass on the outcome being a one-vote margin.

if you're actively pushing an election, not just voting yourself, then that plausibly has a much bigger impact than just your one vote

That... is... a very interesting corollary. Although... you only get the "superexponential" benefit in the case where you're far out on the tail of the PDF -- in the "500k 49% coins" world, throwing 100 votes for Heads instead of 1 would increase your chances of swinging the election by a factor of much, much more than 100x, but your probability of swinging the election is still negligible, since the 50% mark is, uh, 14 standard deviations out from the mean. Right?

↑ comment by Optimization Process · 2021-02-26T21:59:51.519Z · LW(p) · GW(p)

Wait... your county has a GDP of over half a million dollars per capita? That is insanely high!

I agree! (Well, actually more like $1-200k/capita, because there are more people than voters, but still.) Sources: population, GDP, turnout.

Replies from: Dagon
↑ comment by Dagon · 2021-02-26T22:14:16.227Z · LW(p) · GW(p)

Umm, that's a very misleading statistic - King County, WA has a very uneven distribution of contribution to GDP, as it includes a number of international powerhouses that local politics don't affect very much (nonzero, but nowhere near the few percent you're estimating).  per-capita averages are just about useless for any planning or valuation of action.

Replies from: Optimization Process
↑ comment by Optimization Process · 2021-02-26T22:52:38.934Z · LW(p) · GW(p)

Yeah, a fair point!

comment by Optimization Process · 2021-02-26T17:04:26.514Z · LW(p) · GW(p)

  • Possible answer: "Sure, it's individually rational for you to devote your energy to Getting Out The Vote instead of donating to charity, but the group-level rational thing for people to do is to donate to charity, rather than playing tug-o'-war against each other."

    Ugh, yeah, maybe. I see the point of this sort of... double-think... but I've never been fully comfortable with it. It sounds like this argument is saying "Hey, you put yourself at a 60% probability of being right, but actually, Outside View, it should be much smaller, like 51%." But, buddy, the 60% is already me trying to take the outside view! My inside view is that it's more like 95%!

    It sounds like down this path lies a discussion around how overconfident I should expect my brain to be (and therefore how hard I should correct for that). Which is important, sure, but also, ugh.

comment by ChristianKl · 2021-02-27T08:59:09.239Z · LW(p) · GW(p)

Which country has only 700 thousand voters but $400 billion GDP? That seems like an order of magnitude is wrong somewhere. 

I would be very surprised if the probability of swinging an election with 700k voters is 1/800. That would suggest that most elections are won with a margin of a lot less then 2000 votes which seems unlikely to me. 

Replies from: AnthonyC
↑ comment by AnthonyC · 2021-02-27T18:14:28.990Z · LW(p) · GW(p)

Possibly the UAE or another country where voting is limited to a smaller subset of the population?

comment by Optimization Process · 2021-02-26T17:03:02.924Z · LW(p) · GW(p)

  • Possible answer: "No election is decided by a single vote; if it's that close, it'll be decided by lawyers."

    Rebuttal: yeah, it's a little fuzzy, but, without having cranked through the math, I don't think it matters: my null hypothesis is that my vote shifts the probability distribution for who wins the legal battle in my desired direction, with an effect size around the same as in the naive lawyer-free model.

comment by Optimization Process · 2021-02-26T17:03:19.028Z · LW(p) · GW(p)

  • Possible answer: "You're doing a causal-decision-theory calculation here (assuming that your vote might swing the election while everything else stays constant); but in reality, we need to break out [functional decision theory or whatever the new hotness is], on account of politicians predicting and "pricing in" your vote as they design their platforms."

    Hmm, yeah, maybe. In which case, the model shouldn't be "my vote might swing the election," but instead "my vote will acausally incrementally change candidates' platforms," which I don't have very good models for.