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

Kathy Dopp kathy.dopp at gmail.com
Thu Oct 9 12:52:58 PDT 2014

```On Thu, Oct 9, 2014 at 10:09 AM, Toby Pereira <tdp201b at yahoo.co.uk> wrote:
> The method is:
>
> Voters cast approval ballots. If a particular candidate receives n votes,
> then if this candidate is elected, each voter of that candidate will get a
> representation of 1/n from that candidate. Each voter who did not vote for
> that candidate gets 0 representation from them. So the total representation
> received from that candidate will be 1.
>
> A voter's total representation is the sum of the representation they receive
> from each elected candidate. Assuming that each elected candidate has
> received at least one vote, then for c candidates, the sum of the
> representation of all voters will be c. The arithmetic mean will always be
> c/v (for v voters).
>
> Full proportionality is achieved if every voter has representation of c/v.
> The measure of a set of candidates is the total of the squared deviation
> from c/v of the voters' representation (lower being better). But also,
> because the variance of x is mean (x^2) - (mean x)^2, where in this case x
> represents voters' representation levels, (mean x)^2 will always be the same
> - it will be (c/v)^2 - so we can remove it from the equation.
>
> This means that the winning set of candidates will be the set that minimises
> the sum of the squares of the voters' representation levels. [end of
> description]
>
> Because the best result is based on the sum of the squared of the
> representation levels, then if faction A has 2 seats and faction B 0 seats
> or they have 1 seat each, the relative proportionality of each of these will
> be unaffected by a small C faction.

Yes. I see.  Do you think these three results *should* be equally

With nonoverlapping support:

# voters,  winning candidate set
150, 1
50,   1
50,   0

150,  2
50,    0
50,    0

150, 1
50,   0
50,   1

All three candidate sets for two winners for this scenario get the
same value of your measure:  0.021462585

>>> But what I mean is that if a large faction (with say 50% of all voters)
>>> is
>>> divided into two (say 25% each) because of a single controversial
>>> candidate
>>> who appears on half of that faction's ballots but not the other half,
>>> then
>>> if that faction receives half the candidates (and the one controversial
>>> candidate is not elected), then it will be measured as unproportional
>>> because each faction will have each contributed to 50% of the candidates
>>> but
>>> will only be 25% of the electorate each.
>
>>I don't see the problem. Could you possibly provide an example where
>>you believe this is situation would be a problem?
>
> Let's say we have 4 to elect and the following approval ballots:
>
> 5 voters: A, B, C, D
> 5 voters: A, B
> 10 voters: E, F
>
> The 10 E, F voters will get both these candidates elected because they are
> half the voters and they get half the candidates. But for the other two
> candidates to be elected, if we elect AB, then each AB subfaction will have
> elected half of the candidates when they "should" only have elected a
> quarter, so this is seen as disproportional. AC would be seen as more
> proportional by your metric because while one subfaction would still have
> elected half of the elected candidates, the other would have elected a
> quarter, so it would be seen as more proportional overall. But this is an
> unbalanced allocation.

OK. I see what you mean. Yes. That does not seem like the best winning set.

>
>>How do you apply your method sequentially?  Many sequential methods
>>I've seen are fundamentally unfair (IRV, for example that treats
>>voters' votes unequally) and can tend to produce undesirable results.
>
> Instead of finding the slate that minimizes squared deviation, you could
> elect them one at a time. So initially the candidate that has the most
> approvals. Then the most proportional two-candidate set that includes the
> first winner. Then the most proportional three-candidate set that includes
> these two and so on. It would save computing time even if it doesn't find
> the overall most proportional candidate set.

At first glance, that seems like a good  algorithm.  (I'm busy with
another project now, but it seems OK to me.)

--

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
```