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

Kathy Dopp kathy.dopp at gmail.com
Fri Oct 3 13:03:35 PDT 2014

```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

On Fri, Oct 3, 2014 at 3:28 PM, Toby Pereira <tdp201b at yahoo.co.uk> wrote:
> I had a look at the top example, and I didn't get the same numbers as you.
>
> 51: ABCD
> 49: EFGH
>
> ABEF - 51/100 * |51/100-2/4| + 49/100 * |49/100-2/4| = 0.01
> ABCD - 51/100 * |51/100-4/4| + 49/100 * |49/100-0/4| = 0.49
>
> But either way, it rates the proportional result higher.
>
> But it doesn't work in all cases.
>
> 4 to elect
>
> 5: A1, A2, A3, A4
> 3: B1, B2, B3, B4
> 1: C1, C2, C3, C4
>
> Sainte-Laguë and my method would give an exact three-way tie between A1, A2,
> A3, B1 and A1, A2, B1, B2 and A1, A2, B1, C1. Your method gives different
> ratings to each, whereas it seems to work in the equivalent situations with
> just two of the three factions. For example:
>
> 4 to elect
>
> 5: A1, A2, A3, A4
> 3: B1, B2, B3, B4
>
> This would be a tie between A1, A2, A3, B1 and A1, A2, B1, B2. Under either
> result, every voter is 0.125 away from their "ideal" so it's clear to see
> your method would rate them as equally proportional. But these two results
> have different proportionality when the "irrelevant" C faction is added. So
> it seems that it breaks down here.
>
> Also, because it considers a faction to be voters that have all their
> choices the same, it fails to take into account overlapping factions.
>
> 2 to elect
>
> 10: A, B, C
> 10: A, B, D
>
> One might want to elect AB (actually my system would award a tie between AB
> and CD), but your system would insist on electing CD. It tries to make each
> faction have the number of seats in proportion to the number of voters it
> has, and in this case it's half the seats each. Anything more, and your
> system rates it as less proportional.
>
> In terms of outright proportionality, I honestly don't think you'll get a
> better system than my measure. It's just a case of whether its indifference
> to outright positive support is a problem.
>
> 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, 17:28
>
> Subject: Re: [EM] General PR question (from Andy Jennings in 2011)
>
> Hi.  Yes Toby.  I agree, which is why I sent a followup email to
> suggest weighting each term by the proportion of voters it contributes
> to the sum to be minimized.  I tested it against both of our examples
> and it seems to work. I believe it will always work because the
> formula is derived to measure the overall proportional fairness of the
> result. Here it is again:
>
> Sum over i of (v_i/v*Absolute(v_i/v - s_i/s))
>
> where v_i and s_i are the number of voters and  number of winning
> candidates for each group i of voters who vote for the same
> combination of candidates and where v is the total number of voters
> and s is the number of winning seats for that election.  Obviously,
> the sum could be multiplied by any number, such as v, and the minimum
> would still be the same.
>
>
> Minimizing this sum for the two examples we discussed in this thread:
> Example 1:
>
> 51: ABCD
> 49: EFGH
>
> ABEF gives 0.0051
> ABCD gives 0.4999
>
> so ABEF is more proportionally fair
>
> Example 2:
>
> 10 A
> 10 B
> 10  C
> 9  D
> 1 D,E
>
> ABCD  gives 0.00625
> ABCE  gives 0.05625
>
> so ABCD is more proportionately fair.
> -------------------------------
>
> I believe this formula is the simplest available for determining the
> most proportional result for any of approval voting election results,
> and would highly support such a process for counting approval votes if
> the public could be convinced to use it.
>
> I feel confident that minimizing this sum identifies the most
> proportionately fair result in the case of any number of seats (1+)
> and any number of candidates to be chosen for approval election
> results.
>
>
> On Fri, Oct 3, 2014 at 7:04 AM, Toby Pereira <tdp201b at yahoo.co.uk> wrote:
>> Kathy, would you be able to give some examples of this in action to flesh
>> it
>> out a bit? I'm a bit unsure about how it would work in practice. For
>> example, if 1000 people vote for the candidates A, B, C, D, E and one
>> person
>> votes for A, B, C, D, F, then that one person is a very small faction. But
>> If A, B, C, D are all elected then this very small faction (one person)
>> has
>> contributed to the election of four candidates, so I think it would cause
>> your measure to rate it as quite unproportional. But in reality the one
>> person is very closely related to the 1000 so shouldn't be seen as an
>> entirely separate faction. We could also imagine a hypothetical case where
>> although, broadly speaking, the voters are in factions, no two voters have
>> identical ballots.
>>
>> Toby
>>
>>
>>
>> ________________________________
>> From: Kathy Dopp <kathy.dopp at gmail.com>
>> To: Kristofer Munsterhjelm <km_elmet at t-online.de>
>> Cc: Toby Pereira <tdp201b at yahoo.co.uk>; EM
>> <election-methods at lists.electorama.com>
>> Sent: Thursday, 2 October 2014, 18:31
>> Subject: Re: [EM] General PR question (from Andy Jennings in 2011)
>>
>> OK. Here's the formula that will *always* work to evaluate how
>> proportional fair any election result is, given any set of voter
>> groups and the combination of approval votes each group casts:
>>
>> Sum(Absolute(v_i/v - s_i/s))
>>
>>
>>
>> or
>>
>>
>> Sum(|v_i/v - s_i/s|)
>>
>> Where v_i and s_i are, respectively the number of voters in group i
>> and the winning candidates group i voted for (for any group voting for
>> the same combination of voters)
>>
>> and where v is the total number of voters and s is the total number of
>> seats.
>>
>> Thus, for an approval vote election, one fairly simple way to find the
>> most proportionately fair set of winning candidates would be to find
>> the set of candidates who minimize this sum of absolute values of the
>> differences between the proportion of the voters in each voting block
>> that votes for the same combination of candidates out of all voters,
>> and the proportion of seats that this group contributes to electing.
>>
>>
>> I am convinced this method of counting approval ballots will never
>> fail to assign the most proportional outcomes to select the winning
>> set of candidates.  If there are more than one set of candidates with
>> the same minimum sum, perhaps toss a coin.
>>
>> I actually like this proportional voting method very much because it
>> strictly adheres to finding the most proportionately fair set of
>> winning candidates, the vote tallies are easily precinct summable and
>> auditable, the vote casting method is easy and gives voters more
>> choice and flexibility to express themselves, and the method is fairly
>> (equally) counted for all voter groups. However, perhaps the summing
>> method with its proportions and differences in tallying the votes of
>> all voter groups voting for the same candidates is a little too
>> complex for some voters to comprehend.
>
>
>
>>
>>
>>
>
>
>
> --
>
> 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
>
>
>

--

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
```