# Pareto efficiency in proportional/multi-winner systems

What do you think Pareto efficiency in a multi-system means/should mean? And is it desirable?

According to the voting wiki:

If every voter prefers alternative X over alternative Y, then the system prefers X over Y

With a multi-winner election, I would take that to mean that if Y is elected, then X must be as well. However, I don’t think that definition is strong enough. I would say that if every voter rates X at least equal to Y and at least one voter rates X ahead of Y, then the system prefers X over Y. And I think it’s desirable for a voting system to meet this criterion.

But arguably you can go further with cardinal multi-winner systems. You can define how much someone likes a set of candidates by totalling the scores (or approvals) they gave to the candidates in the set. So for a system to obey this criterion, if every voter rates set X at least equal to set Y and at least one voter prefers X to Y, then Y should not be elected. And in a an optimal system capable of giving scores to every set of candidates, X should have a higher score than Y. That last might not be so important as long as it never affects the winning set.

Is this desirable? Let’s look at an example:

2 to elect (approval voting)

5 voters: AC
4 voters: BC
1 voter: BD

I think that most people would agree that BC is the best set to elect. But compare AB with CD. In both cases every voter has one elected candidate each, so from a “Pareto” point of view they are equal. But under AB, 5 have approved A and 5 have approved B, whereas under CD, it’s 9 for C and 1 for D, which doesn’t seem very proportional. So then if you have this example:

500 voters: AC
400 voters: BC
100 voters: BD
1 voter: CD

CD now Pareto dominates AB. But is it actually the better set? You might argue that it doesn’t matter because BC is better than either, but it does at least raise the possibility that something similar could arise in the competition for best set. In such a case, which set would be better? Let’s say for some reason, it has to be AB or CD. Which set wins in your opinion?

In any case if you think this is too contrived, I will see if I can come up with an example for the winning set.

The following ballots would require AB or CD to be elected in a 2-candidate proportional election:

n voters: AC
n voters: BC
n voters: BD

Every candidate is approved by half the voters, but no-one votes for both A and B or C and D. You can see it as a two-axis voter space with A and B at opposite ends of one axis and C and D at the opposite ends of the other axis. If you make n high enough, you can then stick some other ballots on the end of this, and it would still have to be AB or CD. So we can combine it with the other example I gave:

10,000 voters: AC
10,000 voters: BC
10,000 voters: BD
500 voters: AC
400 voters: BC
100 voters: BD
1 voter: CD

which makes:

10,500 voters: AC
10,400 voters: BC
10,100 voters: BD
1 voter: CD

Which two candidates do you elect?

Well, I’m going to say that I would elect AB over CD. This result would fail Pareto efficiency under the definition I gave above:

I’m not sure this is a necessary criterion though. I think there’s more to it than just how many candidates a voter has elected. If every voter has voted for one elected candidate, then I think it’s clearly preferable for half to have approved each than for 90% to have approved one and 10% the other.

And as an aside, obviously Thiele purely looks at number of approved candidates each voter has in the set of elected candidates and bases satisfaction from that, but perhaps it could be modified to take into account that it’s preferable for a voter to “share” their candidate with fewer other voters. We wouldn’t want to go as far as dividing the worth of a candidate by the number of voters because that would lead to indifference between AB and CD in the following case:

1 voter: ABC
1 voter: ABD

Instead of saying a candidate shared among 2 voters is worth 0.5, we could instead square (1-0.5) and take that away from 1.

So if v voters approve a candidate, then the worth of that candidate to each voter is 1-((1-1/v)^2).

For 1 it would be 1.
For 2 it would be 3/4.
For 3 it would be 5/9.
For 4 it would be 7/16.
Etc.

Also the total worth of the candidate to all voters who approve it would approach 2.

To get a voter’s satisfaction score you would still have to do the harmonic function on their total. I’m not sure how sensible this is or what results it would give though.

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Just to finish on the Thiele aside, voters in larger factions would obviously have more voters approving each of their candidates, so balancing out the fact that this is supposed to be bad for voters would provide a majoritarian shift. So any transformation on the satisfaction scores when using D’Hondt divisors would mean a method even more majoritarian than D’Hondt. You could use Sainte-Laguë divisors, but the transformation would then just push it towards D’Hondt, so you might as well just use D’Hondt and say that you’re doing so for the reasons above.

In conclusion, I don’t think this would work as a magic way of transforming Thiele into a different method with its own desirable properties.