Wait, so, my previous analysis doesn’t make that much sense. I now think your claim is pretty plausible.
Expected value if you don’t buy one more vote: ∑iPiUi
Expected value if you do: ∑iPiUi−∑iϵxnUi+ϵxUx−c⋅(Px+ϵx−ϵxn)
Here, ϵx is specifically the probability that candidate x ends up in a tie without our one additional vole. I’m assuming the vote uses the (rather dumb) tie-break procedure of choosing randomly from all the candidates, for simplicity. Hence, breaking the tie steals probability equally from each candidate (including candidate x). I’m also neglecting the possibility of creating a new tie (hopefully that doesn’t matter too much).
Difference in expected values: ϵxUx−∑iϵxnUi−c⋅(Px+ϵx−ϵxn)
Break-even point: c=ϵx(Ux−∑i1nUi)/(Px+ϵx−ϵxn)
That’s pretty similar to before, but now comes the critical point: it seems like ϵx is proportional to Px. I’m sure this isn’t exactly true, but it roughly makes sense—you have to have a good chance of being on top in order to have a good chance of tying for top. If we assume it’s true, then everything cancels nicely. Suppose ϵx=αPx:
c=αPx(Ux−∑i1nUi)/(Px+αPx(1−1n))
=α(Ux−∑i1nUi)/(1+α(1−1n))
∝Ux−∑i1nUi
A happy ending! Not only does Px disappear, but all Pi disappear. This is good, because it means voters don’t even have to think about the probabilities of the different outcomes when voting; they should just think about how much they like the different outcomes.
The analysis is still rough, but at least now things are looking good.
Hm, I just noticed that I didn’t really get your whole proposal in the first place—I latched onto “full refund for losing positions”, but ignored the rest.
[...] only the winners of a quadratic vote actually pay an average of the tokens, and everyone else gets a refund—sort of like a blind Dutch auction of the decision.
For example, a quadratic vote is taken between two binary options A and B. A receives 400 votes, B receives 500. B wins the vote, so an average of 450 is taken from the voting token pool of B and 50 tokens are redistributed equally amongst B.
What’s supposed to be going on here? You seem to be ignoring the quadratic formula—an individual must spend v2 tokens to purchase v votes on a candidate. Thus the number of tokens total spend on a candidate isn’t a simple function of the number of votes on that candidate; it also depends on the distribution of votes among voters. So the concept of “redistributing equally” gets kind of complicated.
If it’s supposed to be like a Dutch auction, maybe the idea is that we look at the total number of votes needed to win, and we refund votes (starting with most expensive and going down) until we reach the number needed to win, at which point we stop refunding. This way people don’t need to worry too much about overpaying for success. I don’t know if that’s good or bad, though.
Anyway, I’m curious where you got the idea. What made this suggestion natural to you? I’m personally not that familiar with the literature on Quadratic Voting, so as far as I know, this is a standard suggestion. Or maybe it seemed natural to you based on familiarity with auctions?
So the concept of “redistributing equally” gets kind of complicated.
Ah yes, you’re right in redistributing the 50 tokens when refunding the winners in the same proportion is tricky. Probably necessitates being able to have fractional tokens so you can refund someone 0.1 token or something like that. I imagine it will be very simple for the losing choices.
Also, I don’t mean a regular Dutch auction, I mean a blind one where all bidders submit their bid at once (like an election). My understanding of a blind Dutch auction is that it resolves this “people don’t bid because they don’t think they could win” result in general auctions.
This was absolutely an intuitive suggestion from reading about voting theory and auctions, you’ve got a much deeper understanding of the VT maths than I do. I do think that thinking about elections like an auction for a decision can be a useful way of thinking about it, but I don’t have professional experience with this beyond helping to design some videogame economies. Don’t take this as any kind of standard suggestion—just mine :)
Wait, so, my previous analysis doesn’t make that much sense. I now think your claim is pretty plausible.
Expected value if you don’t buy one more vote: ∑iPiUi
Expected value if you do: ∑iPiUi−∑iϵxnUi+ϵxUx−c⋅(Px+ϵx−ϵxn)
Here, ϵx is specifically the probability that candidate x ends up in a tie without our one additional vole. I’m assuming the vote uses the (rather dumb) tie-break procedure of choosing randomly from all the candidates, for simplicity. Hence, breaking the tie steals probability equally from each candidate (including candidate x). I’m also neglecting the possibility of creating a new tie (hopefully that doesn’t matter too much).
Difference in expected values: ϵxUx−∑iϵxnUi−c⋅(Px+ϵx−ϵxn)
Break-even point: c=ϵx(Ux−∑i1nUi)/(Px+ϵx−ϵxn)
That’s pretty similar to before, but now comes the critical point: it seems like ϵx is proportional to Px. I’m sure this isn’t exactly true, but it roughly makes sense—you have to have a good chance of being on top in order to have a good chance of tying for top. If we assume it’s true, then everything cancels nicely. Suppose ϵx=αPx:
c=αPx(Ux−∑i1nUi)/(Px+αPx(1−1n))
=α(Ux−∑i1nUi)/(1+α(1−1n))
∝Ux−∑i1nUi
A happy ending! Not only does Px disappear, but all Pi disappear. This is good, because it means voters don’t even have to think about the probabilities of the different outcomes when voting; they should just think about how much they like the different outcomes.
The analysis is still rough, but at least now things are looking good.
I’d be lying if I claimed to fully grok the maths, but I’m glad it was a useful suggestion!
Hm, I just noticed that I didn’t really get your whole proposal in the first place—I latched onto “full refund for losing positions”, but ignored the rest.
What’s supposed to be going on here? You seem to be ignoring the quadratic formula—an individual must spend v2 tokens to purchase v votes on a candidate. Thus the number of tokens total spend on a candidate isn’t a simple function of the number of votes on that candidate; it also depends on the distribution of votes among voters. So the concept of “redistributing equally” gets kind of complicated.
If it’s supposed to be like a Dutch auction, maybe the idea is that we look at the total number of votes needed to win, and we refund votes (starting with most expensive and going down) until we reach the number needed to win, at which point we stop refunding. This way people don’t need to worry too much about overpaying for success. I don’t know if that’s good or bad, though.
Anyway, I’m curious where you got the idea. What made this suggestion natural to you? I’m personally not that familiar with the literature on Quadratic Voting, so as far as I know, this is a standard suggestion. Or maybe it seemed natural to you based on familiarity with auctions?
Ah yes, you’re right in redistributing the 50 tokens when refunding the winners in the same proportion is tricky. Probably necessitates being able to have fractional tokens so you can refund someone 0.1 token or something like that. I imagine it will be very simple for the losing choices.
Also, I don’t mean a regular Dutch auction, I mean a blind one where all bidders submit their bid at once (like an election). My understanding of a blind Dutch auction is that it resolves this “people don’t bid because they don’t think they could win” result in general auctions.
This was absolutely an intuitive suggestion from reading about voting theory and auctions, you’ve got a much deeper understanding of the VT maths than I do. I do think that thinking about elections like an auction for a decision can be a useful way of thinking about it, but I don’t have professional experience with this beyond helping to design some videogame economies. Don’t take this as any kind of standard suggestion—just mine :)