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

[Toby Pereira]

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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