[EM] Proportional Approval System
Toby Pereira
tdp201b at yahoo.co.uk
Sat May 31 09:32:07 PDT 2014
I've come up with a system of proportional representation using approval voting that I think avoids many of the problems of other systems.
For each voter, their score is the number of elected candidates that they "possess". If v voters have voted for a particular candidate, then each voter possesses 1/v of that candidate. If there are c candidates and v voters, then as long as each elected candidate has received at least one approval, then the average possession score for a voter is c/v. Full proportionality is achieved if every voter has a score of c/v. The measure of a set of candidates is the average squared deviation from c/v for the voters' scores (lower deviation being better). For example:
4 to elect
3 voters: A, B, C, D
1 voter: E, F, G, H
The obvious proportional result is A, B, C, E. And you can see that the 3 ABCD voters would each possess 1/3 of each of A, B, C, so would have a score of 1. The single EFGH voter would fully possess E, so would also have a score of 1. c/v in this case is 1 so this is trivially a proportional result.
But this also works where there are commonly rated candidates. For example:
6 to elect
20 voters: A, B, C, D, E, F
10 voters: A, B, C, G, H, I
Reweighted Range Voting http://www.rangevoting.org/RRV.html would elect A, B, C, D, E, F on the basis that 20 voters get 3 candidates and 10 voters get 3, making it "proportional". But I would argue it's best to ignore candidates approved by all when considering who else to elect. In this case the 20 ABCDEF voters would all have a score of 3 * 1/30 + 3 * 1/20 = 1/4. The 10 ABCGHI voters would score 3 * 1/30 = 1/10.
Whereas with A, B, C, D, E, G, the 20 ABCDEF voters would score 3 * 1/30 + 2 * 1/20 = 1/5. The 10 ABCGHI voters would score 3 * 1/30 + 1 * 1/10 = 1/5. The equal scores imply a proportional result. To check, c/v = 6/30 = 1/5 so it all adds up.
This also works for other examples that I've tried. Where there isn't an exact proportional result available, the squared deviation seems to work as the best measure. For example:
4 to elect
5 voters: A, B, C, D
3 voters: E, F, G, H
1 voter: I, J, K, L
I won't do the maths here, but this would give an exact 3-way tie between ABCE, ABEF and ABEI. This is the same as the result that Sainte-Laguë would give (which I would argue is objectively more proportional than D'Hondt).
It's not obvious how to score a set of candidates where at least one candidate has no approvals, but this should never cause a problem. As long as there are at least as many approved candidates as seats, there is no need to consider unapproved candidates. If there aren't, you would always elect all the approved candidates anyway.
For a similar reason, it's not obvious what to do with score voting. For example, if a single voter has given a candidate a score of 1 out of 10, then does this voter possess 1/10 of this candidate or the whole candidate? And if 10 voters have given a score of 1/10 then do they each possess 1/100 or 1/10? If it's 1/100, then candidates won't be "fully possessed" and we end up with a no obvious way of working out the best set of candidates because the average c/v would not be the same for each set of candidates and we'd be looking at more than just the squared deviation. We'd have to consider both average and deviation, which would make it messy. But otherwise we end up with a voter fully possessing a candidate that they've given a low score to, and this would count against that voter. So at the moment, this is just a system for approval voting.
I think this is superior to the approval version of Reweighted Range Voting because it is independent of commonly approved candidates, and so it also avoids the proportionality problems that this can cause, which I have discussed elsewhere.
It is also superior to STV methods because in these methods a voter is considered to just have a single vote and if that vote is fully assigned to a candidate then it doesn't matter what else happens. For example:
4 to elect
10 voters: A, B, C, D
10 voters: A, B, C, D, E, F
The best result here seems to be A, B, C, D, but STV is indifferent between this and, say, A, B, E, F. The 10 ABCD voters can have their votes assigned to A and B (5 to each) and the 10 ABCDEF voters can have their votes assigned to E and F (5 to each). This would heavily favour the ABCDEF voters. In practice, most STV methods wouldn't do this because of the sequential way they operate. But STV methods that consider whole sets of candidates non-sequentially would be indifferent to these results. So avoiding this result would be more accidental than ideological.
However, my method is also indifferent between results that you might expect a method not to be. For example:
2 to elect
10 voters: A, B, C
10 voters: A, B, D
You might think that A, B is the obvious result. But my system would say this is equal to C, D. So in this sense it could be said to fail monotonicity. However, I do think it makes sense in the way that proportional systems work. If A and B are elected, then they are both "shared" among all 20 voters. Where if it's C and D then although each voter only has one candidate, they share the candidate with half as many people, and their possession score would be the same under either result.
If this all stands up, I think this is better than the other systems of proportional representation based on approval voting that I've seen.
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