I really like this argument. I think there’s another way of framing it that occurred to me when reading it, that I also found insightful (though it may already be obvious):
Suppose the value of your candidate winning is X, and their probability of winning if you don’t do anything is p.
If you could buy all the votes, you would pay X(1-p) to do so (value of your candidate winning minus a correction because they could have won anyway). This works out at X(1-p)/N per vote on average.
If p>1/2, then buying votes probably has diminishing returns (certainly this is implied by the unimodal assumption).
Therefore, if p>1/2, the amount you would pay for a single vote must be bounded below by X(1-p)/N.
If p<1/2, I think you can just suppose that you are in a zero-sum game with the opposition party(ies), and take their perspective instead to get the same bound reflected about p=1/2.
The lower bound this gives seems less strict (1/2 X/N in the case that p=1/2, instead of X/N), but it helps me understand intuitively why the answer has to come out this way, and why the value of contributing to voting is directly analogous to the value of contributing to, say, Parfit’s water tank for the injured soldiers, even though there are no probabilities involved there.
If as a group you do something with value O(1), then the value of individual contributions should usually be O(1/N), since value (even in expectation) is additive.
I really like this argument. I think there’s another way of framing it that occurred to me when reading it, that I also found insightful (though it may already be obvious):
Suppose the value of your candidate winning is X, and their probability of winning if you don’t do anything is p.
If you could buy all the votes, you would pay X(1-p) to do so (value of your candidate winning minus a correction because they could have won anyway). This works out at X(1-p)/N per vote on average.
If p>1/2, then buying votes probably has diminishing returns (certainly this is implied by the unimodal assumption).
Therefore, if p>1/2, the amount you would pay for a single vote must be bounded below by X(1-p)/N.
If p<1/2, I think you can just suppose that you are in a zero-sum game with the opposition party(ies), and take their perspective instead to get the same bound reflected about p=1/2.
The lower bound this gives seems less strict (1/2 X/N in the case that p=1/2, instead of X/N), but it helps me understand intuitively why the answer has to come out this way, and why the value of contributing to voting is directly analogous to the value of contributing to, say, Parfit’s water tank for the injured soldiers, even though there are no probabilities involved there.
If as a group you do something with value O(1), then the value of individual contributions should usually be O(1/N), since value (even in expectation) is additive.