Should we abstain from voting? (In nondeterministic elections)
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Looking at Jason Brennan's "Polluting the polls" argument in the context of probabilistic voting
https://bobjacobs.substack.com/p/should-we-abstain-from-voting-in -
@bmjacobs Interesting article, but a couple of things. Where does it come from that in Borda you need 2/3 of the vote to ensure a win? Also Borda and winner-takes-all systems wouldn't have the step function if there are more than two candidates. E.g. having 40% of the vote under First Past the Post will still give you the win in many elections. If it's to guarantee the win then yes, but I don't think it's clear that that's what the graph is trying to say.
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@toby-pereira Yes, this was about what "ensures a win" aka what guarantees it. This post didn't really go into the details on which threshold is necessary since it was more about comparing step functions and non-step functions, and what that entails for voters. However, since you're curious I'll try to explain why you need ⅔:
Assume you control X% of the votes, want to make sure that option A wins, you do not know how the other 100–X% will vote, and there are K many options overall. No matter how you vote, the average rank of A within your X% of votes will be at least 1, and there will be some option B other than A that has an average rank of at most K/2 + 1 within your X% votes. The other 100–X% of voters might provide A with a rank of K and B with a rank of 1, in which case the overall average rank of A will be at least X% · 1 + (100 − X%)·K = K − X% · (K − 1) and that of B will be at most X% · (K/2 + 1) + (100 − X%) · 1 = 1 + X% · K/2. So, to guarantee that the latter is larger than the former, 1 + X% · K/2 > K − X% · (K − 1), you need to have X% > (K − 1)/(3K/2 − 1) ≈ K/(3K/2) = ⅔. -
@bmjacobs OK, I think I get that. But I still don't think the graph is correct - it's giving a different measurement for a deterministic and non-deterministic systems. For non-deterministic methods, it seems to be working on the probability of getting elected, which I presume is what the "controlled winning probability" on the y axis means. But for the deterministic methods, it just gives a score of 100 if they are guaranteed to be elected and 0 otherwise.
If it's a about guaranteed election then the non-deterministic methods should give 0 across the board. If it's probabilistic, then the non-deterministic methods are correct on the graph but the deterministic ones are wrong - though there isn't an exact probability you can give as it depends on voting patterns.
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@toby-pereira See the "effective power" more as something over time. So for the random ballot, with a guaranteed 51% of the vote you'd be in power 51% of the time, but with a guaranteed 51% of the vote in a majoritarian system, you'd be in power 100% of the time. I'd gestured at this by saying "if you have 51% of the vote you are in power 100% of the time", but I'll make this all a bit clearer if I use this graph again in the future.
