[EM] General PR question (from Andy Jennings in 2011)
Kathy Dopp
kathy.dopp at gmail.com
Fri Oct 3 17:11:11 PDT 2014
Yes. I agree that, in the case of only two factions divided in the way
you suppose, giving the forth seat to either one of the factions is
equally proportionately fair, which is exactly what my method (and the
Sainte-Laguë method) suggests. Of course, no faction of voters should
be ignored in this calculation. If there is a third faction, it should
be considered. If there is none, then not. Thus, eliminating the
third faction, that forth representative that would otherwise be
allotted to the 3rd faction, could go to either of the first two
factions in this scenario.
The formula I've derived exactly measures proportional fairness of
approval voting for any factions of voters and for any number of
candidates. If your method disagrees with its results, then your
method is evaluating something different than proportionate
equitableness.
On Fri, Oct 3, 2014 at 7:11 PM, Toby Pereira <tdp201b at yahoo.co.uk> wrote:
> Kathy
>
> The problem I have with it is that the result changes when irrelevant
> alternatives are removed. Assign the obvious seats as before:
>
> 5 voters (A): 2 seats
> 3 voters (B): 1 seat
>
> And according to your method C is first in line for the next seat followed
> by B and then A. But what if there was no C? It should still be B ahead of
> A, right? But the ideal share of seats becomes:
>
> 5 voters (A): 2.5
> 3 voters (B): 1.5
>
> And the seats owed to the groups now are:
>
> 5 voters (A): 0.5
> 3 voters (B): 0.5
>
> This happens with any pair. A v B, A v C or B v C all produce ties for the
> final seat when the third faction is ignored. Sainte-Laguë (which I would
> argue works proportionally for party voting) also gives a three-way tie. I
> specifically came up with this scenario a while ago as a test for methods
> that seem to work for two factions, and it was important that my own method
> passed it before I deemed it acceptable.
>
> Toby
>
>
>
> ________________________________
> From: Kathy Dopp <kathy.dopp at gmail.com>
> To: Toby Pereira <tdp201b at yahoo.co.uk>
> Cc: Kristofer Munsterhjelm <km_elmet at t-online.de>; EM
> <election-methods at lists.electorama.com>
> Sent: Friday, 3 October 2014, 21:03
> Subject: Re: [EM] General PR question (from Andy Jennings in 2011)
>
> Toby,
>
> I argue that the three possible sets of winning candidates are not
> equally proportionately fair; and, further, my method finds the
> closest to proportional of the three possible sets of winning
> candidates you gave -- and here is why:
>
> First, to reiterate the example you gave:
>
> 5: A1, A2, A3, A4
> 3: B1, B2, B3, B4
> 1: C1, C2, C3, C4
>
> Simply calculate what proportions out of the total voters, each group
> is times the number of total seats:
>
> i.e. the share of the voters should, ideally, be as close as possible
> to the share of seats they receive:
>
> 5 voters: 2.222222222... seats
> 3 voters: 1.333333333... seats
> 1 voter: 0.44444444... seats
>
> First assign the obvious seats:
>
> 5 voters -- 2 seats
> 3 voters -- 1 seat
>
> Now there is one seat left to assign and the following remainders of
> seats owed proportionately to each group:
>
> 5 voters: 0.222222222....
> 3 voters: 0.333333333...
> 1 voter: 0.444444444....
>
> The winner who is still owed the largest proportion of seats is the 1
> voter. Thus, the MOST proportionately fair set of winning candidates,
> as minimizing my sum says, is:
>
> 5 voters: 2 seats
> 3 voters: 1 seat
> 1 voter: 1 seat
>
> That is the best that could be done, proportionately for 4 seats for
> those voters.
>
> The numerical results of my Proportionality measurement sum for the
> three sets of winners you suggested is:
>
> A1, A2, A3, B1: 0.148148
> A1, A2, B1, B2: 0.098765
> A1, A2, B1, C1: 0.074074
>
> Thus, the method I suggest identifies a set of winning candidates that
> is arithmetically closest to fairly proportional. Of course for those
> hypothetical voter groups any set of winners having two candidates
> from the group with 5 voters and any one each from the groups with 3
> voters and 1 voter will have equal proportionality measurement with
> the third winning set you suggested above.
>
> I believe there is no other measure that can measure closeness to
> proportional fairness of a winning set better for approval ballots,
> although the measure could obviously be changed by multiplying all
> terms by a constant since that would not change the ordering of any
> winning set relative to another.
>
> Kathy
>
>
--
Kathy Dopp
Town of Colonie, NY 12304
"A little patience, and we shall see ... the people, recovering their
true sight, restore their government to its true principles." Thomas
Jefferson
Fundamentals of Verifiable Elections
http://kathydopp.com/wordpress/?p=174
View my working papers on my SSRN:
http://ssrn.com/author=1451051
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