Consider my home county:

Number of voters

(Corollary: probability of swinging an election: )

GDP: $400B

Amount by which GDP might go up or down depending on a single election: 0.1%

Probability I support the better side: 60%

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.)

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.

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.

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’snotthat there are 500k fair coins, it’s that there areeither500k 51% coinsor500k 49% coins, and the uncertainty in the election outcome comes fromnot knowing which of those worlds you’re in. But, in either case, your chance of swinging the election is vanishingly small, becausebothof those worlds put extremely little probability-mass on the outcome being a one-vote margin.(see also: johnwentworth’s comment below)

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.