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’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.
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.
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’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)
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.