Methods that pass Vote Unitarity seem to return more representative outcomes than variants of Thiele such as RRV and SPAV. Vote Unitarity methods tend to pass ULC, for example, which Thiele fails. They also don’t use the same candidate to satisfy multiple voting blocs, as PAV does in the following case I described in a different post:
In this case, PAV uses the candidates A, B, and C to satisfy the top 4 Hare quotas of voters. In contrast, Vote Unitarity methods would make the election of A, B, and C cost the top 4 Hare quotas of voters 3 Hare quotas in total, most likely split between the two. Electing F and G would cost the bottom two rows of voters 2 Hare quotas, leaving them with nothing. So the final seat would go to either D or E.
However, as a candidate’s election can be forced with less than a Hare quota, and sometimes much less, Vote Unitarity methods are also more vulnerable to Hylland free riding and vote management than Thiele variants. Here is a case where they can be used to reverse a majority:
Electing two members of party B costs the bottom row of voters 40 votes, leaving them with 11. So all 3 A candidates get elected with less than a full quota.
With Thiele variants such as RRV and SPAV, there is no incentive for a party with no support from non-party voters to divide themselves in such a manner. (I have termed this the “Limited Resistance to Hylland Free Riding” criterion.) This allows partisan voters more freedom to distinguish between candidates in their party, as they don’t just have to follow instructions from the party’s leadership, something that vote management entails. This is largely due to these methods having D’Hondt party list cases, although that isn’t sufficient to guarantee this.
I will now describe a method that attempts to compromise between the two methods. I will only describe the Approval case for now, as I have not yet decided how to extend it to Score.
All ballots start with a weight of 1. Select the largest quota that makes it possible to elect a candidate without going below the quota. The candidate that can be elected with this quota is elected. Call the quota q1. To elect the candidate, each ballot must pay weight*[q/(weighted total votes for the elected candidate)]. Proceed to the next round. The next candidate who is elected is again the one that can be elected with the largest quota size (call it q2) if the voters for the candidate elected in the previous round only had to pay q2 to elect that candidate. In later rounds, again, the candidate who is elected is the one who can be elected with the largest quota size, qn, if voters for each candidate elected in the previous round only had to pay qn to elect their candidate.
As an example, consider a 4 winner election with the following ballots:
q1=30, and A wins, because A can be elected if the quota is 30, and no other candidates can.
B, C, or D can be elected second with a quota of 12, since if A were elected with a quota of 12, then the weight of the ballots would be:
E can only be elected with a quota of 7.5, since if A were elected with a quota of 7.5, then the weight of each ballot would be:
So B wins, and q2=12.
C or D can be elected third with a quota of 7.5, since after A was elected with a quota of 7.5, the ballot weights would be as listed above, and after B was elected with a quota of 7.5, the ballot weights would be:
E can also be elected with a quota of 7.5. It seems clear that if E is elected, C will be next, and if C is elected, E will be next, so the winners are A,B,C,E.
A potential modification to this method would be to set a maximum for the quota in the early rounds. If it is possible to exceed this maximum quota, just elect the score leader for that round if the chosen maximum is used as quota. The final quota is necessarily smaller than a Hare quota, so a Hare quota would probably be a good maximum to choose. If this is done, this method behaves like Sequentially Subtracted Score until it is no longer possible to elect candidates with full Hare quotas. I will probably extend this method to Score in a way that maintains this when such a quota cap is used.