[EM] General PR question (from Andy Jennings in 2011)

Toby Pereira tdp201b at yahoo.co.uk
Wed Oct 1 13:33:15 PDT 2014

It's looks to me as well as though Kathy's proposal would actually be equivalent to normal approval voting but with multiple winners, so not a proportional method.

With Proportional Approval Voting (PAV) - far too generic a name really to be given to a specific method - it fails what I call independence of commonly rated candidates. For example with three to elect here:

20 voters: A, B, C, D, E, F
10 voters: A, B, C, G

PAV would elect ABCDEF. It essentially views it as proportional because the 20 voters are getting six candidates and the 10 voters are getting 10. But because they're not entirely separate factions, I would dispute this result - and in fact there are other examples where this failure makes it look even less proportional. With my method, because ABC are approved by all, they are effectively ignored when electing the other candidates, giving the result ABCDEG. As far as I understand, any system that works by giving satisfaction scores like PAV fails this.

> From: Kristofer Munsterhjelm <km_elmet at t-online.de>
>To: kathy.dopp at gmail.com; Toby Pereira <tdp201b at yahoo.co.uk> 
>Cc: EM <election-methods at lists.electorama.com> 
>Sent: Wednesday, 1 October 2014, 19:33
>Subject: Re: [EM] General PR question (from Andy Jennings in 2011)
>On 10/01/2014 04:16 PM, Kathy Dopp wrote:
>> Tony,
>> IMO, a better measure taking your approach, would be simply the sum of
>> the measures of individual voters' representation and electing the set
>> of candidates that maximizes this sum of representation measure.  In
>> fact, using such a summation measure would be, IMO, a good way to
>> compare the satisfaction of voters with various voting methods.
>> The problem with STV is that it would, in circumstances where there
>> are three strong candidates, often result in much less satisfaction of
>> voters according to the sum of individual voters' representation
>> measure than approval voting and many other voting methods, due to the
>> unfair way it treats different voters' ballots unequally, counting the
>> 2nd and 3rd choices of some voters but not others due to its
>> sequential consideration and elimination method.
>What do you think of methods like Schulze STV that use a Condorcet-like 
>setup to consider all candidate combinations and thus avoid the path 
>dependence of the eliminations?
>> if there were 4 seats to be elected, each voter's representation for
>> set ABCE would be:
>> 10 A    1/4
>> 10 B    1/4
>> 10  C   1/4
>> 9   D    0
>> 1 D,E   1/4
>> Summing these measures for each voter would give 31/4
>> If set ABCD were elected instead, the sum would  be 39/4 because more
>> voters are have contributed towards electing a representative.  Thus,
>> the set ABCD is preferable.
>> Thus, I would gauge the representation for each voter using the
>> formula n_i/s where n_i is the number of winning candidates each voter
>> voted for.  This seems like a more straight forward, simpler, still
>> effective measure.
>How about this?
>51: ABCD
>49: EFGH
>Four to elect.
>Consider first a proportional outcome, ABEF. This would give:
>51: ABCD 2/4
>49: EFGH 2/4
>= 50.
>Then consider the extremely majoritarian outcome ABCD:
>51: ABCD 4/4
>49: EFGH 0
>= 51, so this has a greater score. But that's not proportional!
>Perhaps I got your method wrong, though, but it seems you'd need some 
>sort of deweighting so that getting two candidates you approve of is 
>worth less than twice getting one candidate you approve of. Otherwise a 
>coordinated majority can fill the council completely.
>PAV (proportional approval voting) is an example of such a rule: it 
>gives a voter 1 point for the first candidate both on the council and 
>the voter's approval ballot, 1/2 additional points for the second, 1/3 
>for the third and so on (reducing to D'Hondt in a party list situation). 
>Then it chooses winners so that the sum of points is maximized. There's 
>also birational voting, which is similar but with a different penalty 
>function; if I recall correctly, birational did better than PAV 
>according to my tests ( 
>It would be an interesting mathematical puzzle to determine such a 
>penalty function so that various desirable properties are satisfied.
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