[EM] General PR question (from Andy Jennings in 2011)
Toby Pereira
tdp201b at yahoo.co.uk
Fri Oct 3 16:11:00 PDT 2014
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 seats3 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): 2.53 voters (B): 1.5
And the seats owed to the groups now are:
5 voters (A): 0.53 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
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