= 27d567bc2d48d9f40e9ccc20ea370b15)
DanielLC comments on [SEQ RERUN] Stop Voting For Nincompoops - Less Wrong Discussion
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (54)
Your vote has a small chance of deciding the President. This is magnified by voting for a nincompoop. It also has a small effect on the party platforms. This is magnified by voting for a third party. If we know which of these makes a bigger difference, it will be clear whether or not we should vote for a third party. Does anyone know which makes a bigger difference?
Additional factors to consider, especially for legislative seats rather than presidential:
1) Voting for a third party can increase its recognition and power even without giving it a chance to win
2) If you are in a 'safe district' and would be more aligned with the winning lizard, voting for a 3rd party drags down your lizard's vote margin, making it feel the threat of accountability more credibly.
3) If you are in a 'safe district' and would be more aligned with the losing lizard, voting for a 3rd party may actually be more likely to change the election, as the losing lizard probably has a negative aura effect going on
4) If you are in a 'safe district', whichever side, voting for a 3rd party can limit the effectiveness of Gerrymandering, by limiting the information about true lizard preference (in the event you ever want to pick between the lizards rather than going for a 3rd party)
Will it? If my vote has X influence on the election if I vote for a nincompoop, and Y influence on party platforms if I vote for a third party, what relationship between X and Y makes it clear that I should vote for a nincompoop?
X > Y
I suppose the problem is defining influence.
I suppose it would be something like the probability of a change times the magnitude of the change times the number of bits of control you have over the direction.
If you vote for a main party candidate, you only have one bit of control over the direction. If you vote for any of eight parties, you have three bits.
If consider influencing the candidates for the next election, you have a high probability of a low magnitude change. If consider deciding the winner for this election, you have a low probability of a high magnitude change.
Number of bits is a property of the choice, not a property of the chosen option. So it is the same for both candidates. Probability of change is what precisely? Some change is pretty certain. The most charitable and still reasonably literal interpretation is to take the probability distribution for each magnitude of change and integrate over, which means that you suggest to vote for the candidate whose victory causes higher expected change. It may be sensible (if you are actually trying to maximise change), but I don't see what it has to do with your influence.
If, on the other hand, by "probability of change" you mean "probability of victory of that candidate", it makes more sense. But it's still imperfect.
Compare with utilitarian answer: Suppose U(third party candidate X wins) > U(lizard Y wins) > U(lizard Z wins). Then I am deciding between voting for X or Y (there is clearly no sense in voting for Z and let's assume I must vote). We compare
p(X | vote for X) is equal to p(X | I don't vote) + p(X has exactly as many votes as Y and more votes than Z | I don't vote) + p(X has exactly as many votes as Z and more votes than Y | I don't vote). Similarly, p(X | vote for Y) is equal to p(X | I don't vote) - p(X has one more vote than Y and more votes than Z | I don't vote) - p(X has one more vote than Z and more votes than Y | I don't vote). The p(X | don't vote) cancel out (similarly for Y and Z) and we are left to estimate the probability that my vote is the decisive one.
To find out this probability we must have a distribution over exact vote counts, not only over winning candidates. It is not clear that p(draw between X and some other candidate) is proportional to p(X wins).
Basically.
No. It's which vote has the highest expected change. Some of the expected change comes from the candidate winning. Some comes from secondary effects, such as changing the party platform.
Here seems a good place to start. Essentially, if you live in a large country, and the split of support is not extremely close to 50/50 then your vote has an exponentially small chance of determining the winner. Thus the signalling effect of voting third party is probably much bigger.