Why can’t I think of myself as a randomly sampled voter?
Same reason you can’t ignore other relevant pieces of information—doing so makes your probability assignments less accurate. For example, if you know that John is a vocal supporter of the less-popular party, you’re not going to ignore that information and assign a high probabiity to the proposition that he votes for the winner.
If you’re looking at this ex ante, your probability of voting for the winner is ~50% because your vote is uncorrelated with everyone else’s. For every possible arrangement of their votes, there are two equally-likely possible worlds, one where you vote for the winner and one where you don’t. (Like tailcalled above, I’m assuming for the sake of simplicity that this is a large, realistic election where the chance of a 1-vote margin is negligible. I’ll drop this assumption later on.)
Your use of a coinflip is relevant because it guarantees this absence of correlation between your vote and others’. If you were voting on your preferences, these preferences would provide some evidence about the likely preferences of others; if we didn’t have any information to the contrary, our best guess would be that you are a fairly typical person with fairly typical preferences.
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If you’re looking at it after the result is known, I think it’s easiest to examine two cases: one where coinflip voting is rare, and one where it is extremely common.
In the case where it is rare, things are pretty simple. When guessing at the vote of a random voter after a 60-40 election result, we’re basically drawing from a giant urn, in which 60% of the balls are Red, 40% are blue, and a handful have an asterisk painted on them to indicate that they are coinflip voters. If we don’t know whether our random voter used a coin, we’re simply drawing from the full urn, and we have a 60% chance of picking a red ball. But if we know that they did use a coin, we can eliminate all of the un-asterisked balls. Of the remainder, we expect about half to be red and half to be blue, because most sets of coinflips are about 50:50 heads:tails, and the election result doesn’t provide much evidence about this particular set of flips. So the coinflip voter is about 50% likely to have voted for the winner.
In the case where coinflip voting is extremely common, things are more interesting. Here, we can’t assume that the coinflippers voted 50:50 Red:Blue, because the election result provides non-negligible evidence that the set of coinflips happened to skew in favour of the Red candidate. (In worlds where this happened, a Red victory was significantly more likely, and vice-versa.) So, once we know the result—and the fact that coinflip voting was extremely common—we should indeed update in favour of the proposition that random coinflipper X voted for the winner.
This might seem paradoxical—why this conflict between ex ante and ex post probabilities? Why shouldn’t I assume, ex ante, that I am more likely to vote for the winner? After all, I’m expecting to update in that direction once I hear the result, so why shouldn’t I do it now?
Well, I should—but only because of the cases where my single vote determines the outcome. For simplicity let’s suppose that everyone, including me, votes by coinflip. Suppose there are three voters, and I’m the one listed first in each possible election:
{H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T}
I only vote for the loser in two of the eight possible elections, because in a three-voter coinflip election my vote is often decisive. But in the cases where my vote is not decisive (i.e. the other two vote the same way as each other), I’m only on the winning team half the time. This holds for larger elections too.
Still, what about the cases where the coinflips do happen to heavily favour one side—like the HHH outcome above, or the (extremely unlikely, but possible) case where 60% of voters out of a million happened to flip Heads? A randomly selected voter from those elections would be certain (in the HHH case) or likely (in the 60% case) to have voted for the winner. And, supposing you managed to vote blind, there’s no reason to treat yourself differently from the random voter here: you probably voted for the winner too.
Ex ante, these skewed vote tallies are unlikely, but we know they’re possible, so why don’t we update at least a little bit in favour of you voting for the winner?
The answer is that, excluding the cases where your vote is decisive, these big margins are balanced out by the cases where your candidate loses narrowly. This is because, in a coinflip election, close outcomes are much more likely than less close ones—and when we exclude the cases where your vote is decisive, a significant number of cases with a 1-vote margin remain, in all of which you vote for the loser.
You can see this in the three-vote case above, where there are two 3-0 outcomes and six 2-1 outcomes; excluding the four cases where your vote is decisive, we’re left with two cases where you win 3-0 and two cases where you lose 2-1. Overall your probability of voting for the winner is 6⁄8, but that’s only because in 4⁄8 cases your vote is decisive.
It’s the same in a five-vote case, where there are 32 equiprobable outcomes: in two cases you win 5-0; in 8 cases you win 4-1 and in 2 cases you lose 4-1; and in twelve cases you win 3-2, while in 8 cases you lose 3-2. In all twelve 3-2 wins your vote is decisive, so excluding them, your win-loss record is 10-10. Including them, it is 22-10, because of the 12 cases where your vote is decisive.
This generalises: you always have a >50% probability of voting for the winner, but this is always fully accounted for by the chance that your vote causally affects the outcome in the usual way.
Same reason you can’t ignore other relevant pieces of information—doing so makes your probability assignments less accurate. For example, if you know that John is a vocal supporter of the less-popular party, you’re not going to ignore that information and assign a high probabiity to the proposition that he votes for the winner.
If you’re looking at this ex ante, your probability of voting for the winner is ~50% because your vote is uncorrelated with everyone else’s. For every possible arrangement of their votes, there are two equally-likely possible worlds, one where you vote for the winner and one where you don’t. (Like tailcalled above, I’m assuming for the sake of simplicity that this is a large, realistic election where the chance of a 1-vote margin is negligible. I’ll drop this assumption later on.)
Your use of a coinflip is relevant because it guarantees this absence of correlation between your vote and others’. If you were voting on your preferences, these preferences would provide some evidence about the likely preferences of others; if we didn’t have any information to the contrary, our best guess would be that you are a fairly typical person with fairly typical preferences.
*
If you’re looking at it after the result is known, I think it’s easiest to examine two cases: one where coinflip voting is rare, and one where it is extremely common.
In the case where it is rare, things are pretty simple. When guessing at the vote of a random voter after a 60-40 election result, we’re basically drawing from a giant urn, in which 60% of the balls are Red, 40% are blue, and a handful have an asterisk painted on them to indicate that they are coinflip voters. If we don’t know whether our random voter used a coin, we’re simply drawing from the full urn, and we have a 60% chance of picking a red ball. But if we know that they did use a coin, we can eliminate all of the un-asterisked balls. Of the remainder, we expect about half to be red and half to be blue, because most sets of coinflips are about 50:50 heads:tails, and the election result doesn’t provide much evidence about this particular set of flips. So the coinflip voter is about 50% likely to have voted for the winner.
In the case where coinflip voting is extremely common, things are more interesting. Here, we can’t assume that the coinflippers voted 50:50 Red:Blue, because the election result provides non-negligible evidence that the set of coinflips happened to skew in favour of the Red candidate. (In worlds where this happened, a Red victory was significantly more likely, and vice-versa.) So, once we know the result—and the fact that coinflip voting was extremely common—we should indeed update in favour of the proposition that random coinflipper X voted for the winner.
This might seem paradoxical—why this conflict between ex ante and ex post probabilities? Why shouldn’t I assume, ex ante, that I am more likely to vote for the winner? After all, I’m expecting to update in that direction once I hear the result, so why shouldn’t I do it now?
Well, I should—but only because of the cases where my single vote determines the outcome. For simplicity let’s suppose that everyone, including me, votes by coinflip. Suppose there are three voters, and I’m the one listed first in each possible election:
{H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T}
I only vote for the loser in two of the eight possible elections, because in a three-voter coinflip election my vote is often decisive. But in the cases where my vote is not decisive (i.e. the other two vote the same way as each other), I’m only on the winning team half the time. This holds for larger elections too.
Still, what about the cases where the coinflips do happen to heavily favour one side—like the HHH outcome above, or the (extremely unlikely, but possible) case where 60% of voters out of a million happened to flip Heads? A randomly selected voter from those elections would be certain (in the HHH case) or likely (in the 60% case) to have voted for the winner. And, supposing you managed to vote blind, there’s no reason to treat yourself differently from the random voter here: you probably voted for the winner too.
Ex ante, these skewed vote tallies are unlikely, but we know they’re possible, so why don’t we update at least a little bit in favour of you voting for the winner?
The answer is that, excluding the cases where your vote is decisive, these big margins are balanced out by the cases where your candidate loses narrowly. This is because, in a coinflip election, close outcomes are much more likely than less close ones—and when we exclude the cases where your vote is decisive, a significant number of cases with a 1-vote margin remain, in all of which you vote for the loser.
You can see this in the three-vote case above, where there are two 3-0 outcomes and six 2-1 outcomes; excluding the four cases where your vote is decisive, we’re left with two cases where you win 3-0 and two cases where you lose 2-1. Overall your probability of voting for the winner is 6⁄8, but that’s only because in 4⁄8 cases your vote is decisive.
It’s the same in a five-vote case, where there are 32 equiprobable outcomes: in two cases you win 5-0; in 8 cases you win 4-1 and in 2 cases you lose 4-1; and in twelve cases you win 3-2, while in 8 cases you lose 3-2. In all twelve 3-2 wins your vote is decisive, so excluding them, your win-loss record is 10-10. Including them, it is 22-10, because of the 12 cases where your vote is decisive.
This generalises: you always have a >50% probability of voting for the winner, but this is always fully accounted for by the chance that your vote causally affects the outcome in the usual way.