[EM] Proportional Approval System

[Toby Pereira]

I sort of skimmed over this in the initial post, but where methods fail "independence of commonly rated candidates", it seems that they can also fail proportionality in a more obvious way. In my example in the message quoted below, you might not consider it to be a failure of proportionality. However, partial agreement between factions does lead to a violation of proportionality when there is a third unconnected faction. For example:

20 voters: A1, A2, A3
10 voters: A1, A2, A4
30 voters: B1, B2, B3, B4

My method and I think any STV method would elect A1, A2, A3, B1, B2, B3. But Reweighted Range Voting http://www.rangevoting.org/RRV.html, Forest Simmons's proportional approval voting and Thorvald N Thiele's version http://wiki.electorama.com/wiki/Proportional_approval_voting (which all have very similar implementation) all elect A1, A2, B1, B2, B3, B4 giving the B faction 4 candidates instead of the more proportional 3. I would argue that because of this, they are not fully proportional systems.

Had the two A factions been in complete agreement or complete disagreement, then they would have had three candidates elected between them. It's the partial agreement that counts against them.


 From: Toby Pereira <tdp201b at yahoo.co.uk>
>To: "election-methods at lists.electorama.com" <election-methods at lists.electorama.com> 
>Sent: Saturday, 31 May 2014, 17:32
>Subject: [EM] Proportional Approval System
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>But this also works where there are commonly rated candidates. For example:
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>6 to elect
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>20 voters: A, B, C, D, E, F
>10 voters: A, B, C, G, H, I
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>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.
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