sxae answers Voting-like mechanisms which address size of preferences?

[sxae]

Another solution to “Objection 1: Quadratic voting discourages voting on losing propositions” is the idea that 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. Everyone who voted for A gets a full refund.

Consider in your example of selecting a head of state. If a political party overextend their voting to elect their guy—that is, that they overvalue the position—then they will be punished by a lack of voting tokens compared to the opposition for other cabinet positions.

[abramdemski]

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: ${\sum }_i{P}_i{U}_i$

Expected value if you do: ${\sum }_i{P}_i{U}_i-{\sum }_i\frac{{ϵ}_x}{n}{U}_i+{ϵ}_x{U}_x-c\cdot (P_x+{ϵ}_x-\frac{{ϵ}_x}{n})$

Here, ${ϵ}_x$ 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: ${ϵ}_x{U}_x-{\sum }_i\frac{{ϵ}_x}{n}{U}_i-c\cdot (P_x+{ϵ}_x-\frac{{ϵ}_x}{n})$

Break-even point: $c={ϵ}_x\left(U_x-{\sum }_i\frac{1}{n}{U}_i\right)/\left(P_x+{ϵ}_x-\frac{{ϵ}_x}{n}\right)$

That’s pretty similar to before, but now comes the critical point: it seems like ${ϵ}_x$ is proportional to $P_x$. 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 ${ϵ}_x=\alpha P_x$:

$c=\alpha P_x\left(U_x-{\sum }_i\frac{1}{n}{U}_i\right)/\left(P_x+\alpha P_x(1-\frac{1}{n})\right)$

$=\alpha \left(U_x-{\sum }_i\frac{1}{n}{U}_i\right)/\left(1+\alpha (1-\frac{1}{n})\right)$

$\propto U_x-{\sum }_i\frac{1}{n}{U}_i$

A happy ending! Not only does $P_x$ disappear, but all $P_i$ 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.

[sxae]

I’d be lying if I claimed to fully grok the maths, but I’m glad it was a useful suggestion!

[abramdemski]

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 v${}^2$ 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?

[sxae]

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 :)

[abramdemski]

Oooh, is it really that simple?

Expected state of affairs if I don’t buy one more vote (ignoring what else I could have done with the money):

${\sum }_i{P}_i{U}_i$, where $P_i$ is the probability of candidate $i$ winning if I do nothing, and $U_i$ is the utility of candidate $i$ winning.

Expectation if I do buy one more vote:

$(1-ϵ)\left({\sum }_{i\ne x}\right(P_i{U}_i\left)\right)+(ϵ+P_x)(U_x-c)$, where $x$ is the candidate under consideration, $ϵ$ is the probability there would have been a tie w/​o this one extra vote, and $c$ is the utility adjustment for losing your money in the case of a win. (I’m going to go ahead and pretend $c$ is precisely the monetary cost, for convenience.)

So this is worth it when the expected benefits outweigh the expected cost:

$c(P_x+ϵ)<ϵ(U_x-{\sum }_{i\ne x}(P_i{U}_i\left)\right)$

So the price at which we’re indifferent would be:

$c=\frac{ϵ(U_x-{\sum }_{i\ne x}(P_i{U}_i\left)\right)}{P_x+ϵ}$

Hm, this is weird.

• We were looking for $P_x$ to disappear, indicating that the probability of a candidate no longer factors into our considerations. It didn’t disappear. In fact, improbable candidates now look better, because we know we get to avoid the cost of paying for votes (guaranteed refund).

• We were looking for $c$ to be proportional to $U_x$. So $c=ϵU_x$ would be fine.$c=ϵ(U_x-{\sum }_i{P}_i{U}_i)$ would also be fine; that just means $c$ is proportional to a normalized utility, where we normalize compared to the baseline of expected election outcomes. But instead we’re normalizing compared to the expectation minus the component from candidate $x$, which is weird—it means we’re not really normalizing at all (because this adjustment will be different for different values of $x$).

[sxae]

We were looking for $P_x$ to disappear, indicating that the probability of a candidate no longer factors into our considerations.

I’m surprised that we are looking for $P_x$ to disappear entirely, I’m not sure I understand that. Quadratic voting shines when you have lots of votes with the same voting token pool, because you force people to allocate resources to decisions they really care about. It’s absolutely not meant to decide one decision—it’s meant to force people to allocate limited resources over a long period, and by doing so reveal their true valuation of those decisions. I would therefore fully expect $P_x$ to play a part in every agent’s considerations, as they must consider the probability of success in each vote in order to plan allocation of voting tokens for every other vote.

[abramdemski]

Interesting. But then how do you argue that it gives approximately correct results? As I understand it, Weyl sees the argument as just: votes end up being roughly proportional to utility (under a lot of differenc scenarios/​assumptions). When this condition holds, the quadratic vote is a good representation of the utilitarian value of the different options.

So, the reason I think we’re looking for $P_x$ to disappear entirely is because $P_x$ is a function of $x$! It’s fine if $c=\alpha +\gamma \frac{U_x-\sqrt{\zeta }}{BB\left(7\right)}$ or whatever, so long as none of those extra terms are a function of $x$, so that in the end $c$ is still proportional to (some normalization of) $U_x$.

You’re effectively arguing that it’s OK to divide by $P_x+ϵ$, because this just represents voters rationally investing less in cases where $P_x$ will probably win anyway. But this means the quadratic vote systematically undervalues the candidates who are seen as the probable winners!

Quadratic voting shines when you have lots of votes with the same voting token pool, because you force people to allocate resources to decisions they really care about. It’s absolutely not meant to decide one decision—it’s meant to force people to allocate limited resources over a long period, and by doing so reveal their true valuation of those decisions.

OK, but this should mean the quantity we are willing to spend on an election is overall adjusted up or down based on properties of that particular election (IE, how much the issue matters to us). This should not mean a dependence on $P_x$; a dependence on $P_x$ distorts the vote, compromising it as a representation of collective utility.

[sxae]

Interesting, you make some great points here and I don’t think I have any good refutations to any of them. Perhaps if we play around with the auction structure by which we take away and refund these tokens?