EntropyStatisticWeighted Approval Voting

@cfrank Yes, I think you're right that entropy is a better choice of information measure. Using the variance wouldn't work for approval ballots.
I'm thinking more about similarities in the end resultwe're trying to assign more weight to more informative ballots, which I think is a good idea in principle, but in practice we're left open to very easy manipulation by strategic nomination/voting for hopeless candidates.

Perhaps we could do better by weighting each candidate's entropy contribution by their number of approvals? So that way, voters who cast ballots that assigns only 1 approval and 9 disapprovals in the 10 highest candidates, then approve/disapprove of 50% of the bottomranked 200 candidates, are treated as lowentropy/information (rather than highentropy/information) votes.

@lime yes strategic approval for hopeless candidates is an issue. However, we could also do something like, for example, a raw approval rate threshold, and then restrict our approval election only to candidates above that threshold. This would tamp down on the effectiveness of that kind of tactical nomination. There could even be a dynamic elbowpoint cutoff for approvalranked candidates to remove candidates with low raw approval rates prior to the entropy computation.
For example, suppose that there is a ballot such as:
10: A  B C D E F G
15: A B  C D E F G
12: B C D  A E F G
7: C D F  A B E G
1: E F  A B C D GThen the raw approval counts ranked in descending order are
B[27] A[25] C[19] D[19] F[8] E[1] G[0]
Or as fractions, these are approximately
B[0.6] A[0.556] C[0.4222] D[0.4222] F[0.1778] E[0.0222] G[0.000]
A reasonable tailend elbow detection method would detect E as the elbow point, and we could remove E and G from the election. Or, for example, a 1/7=0.14285... approval rate threshold (7 being the number of candidates) would accomplish the same.
This would leave candidates
B, A, C, D, F
Now we have ballots, with binary entropies H2 as
10: A  B C D F > H2 = 0.721928...
15: A B  C D F > H2 = 0.97095...
12: B C D  A F > H2 = 0.97095...
7: C D F  A B > H2 = 0.97095
1: F  A B C D > H2 = 0.721928...From these, we compute the final scores as
A[21.7835...] B[26.215666...] C[18.448...] D[18.448...] F[0.7219...]
and B (who was actually the original approval winner) still wins, and by a wider margin.
I do think there are problems with this method from the standpoint of majoritarianism. It seems plausible that a minority could obtain disproportionate voting power over a bullet majority.
If the ballots were, for example,
51: A  B C D E F G
25: G F  A B C D E
12: E D C  A B F G
12: B  A C D E F Gthen I think we have a problem.

@cfrank said in EntropyStatisticWeighted Approval Voting:
Let each voter submit an approval ballot.
Weight his approvals by
P*log(1/P)+(1P)*log(1/(1P))
where P is the fraction of candidates he approved.
Consider it and let me know what you think.What is the mathematics/motivation behind this particular formula? I don't think we've been given much to go on.
Edit  But anyway, it seems that basically you get more weight per approval if you approve more candidates, though I'm not sure where this formula comes from other than it's possibly something to do with entropy.
But obviously it's a bad idea. There's no reason to punish people who approve fewer candidates, and it encourages cloning.

It rewards those who approve close to half the candidates and punishes those who disapprove just one or who disapprove all but one.

@jackwaugh Right, I see. Though I think it's no better or worse than what I thought it was. Just arbitrarily favouring voters who approve a particular number of candidates. And it encourages putting up clones or nonentities accordingly.

I don't approve of it.

@tobypereira yes you’re right, it was just a thought that occurred to me when I was thinking about how to discourage bullet approvals, but it has irreconcilable flaws that are now apparent.

@cfrank While I don't think it would be a good method in practice, there may still be some theoretical interest in it. Perhaps another way of implementing it would be to have the peak at the mean number of approved candidates, rather than half.
Instead of looking at the number of states for each voter/candidate being two (approved or not approved), if e.g. 1/3 of voters approve a candidate, we could see it as one state for approved and two for not approved. In such a case I think the highest entropy state would be for a voter to approve the mean number of candidates.

@tobypereira said in EntropyStatisticWeighted Approval Voting:
While I don't think it would be a good method in practice
The 2 most popular voting systems in practice are IRV and plurality. Anything is a good method in practice