# STV for party candidate lists?

Norman Petry npetry at sk.sympatico.ca
Tue Jul 28 08:47:06 PDT 1998

```Herman,

I've been lurking on this list for some time and thought now would be a
good time to jump in...

For what it's worth, I think your proposal sounds reasonable, although not
ideal.  At least a couple of good things can be claimed for the system you
propose:

1) The counting of ballots is relatively straightforward, and only needs to
be done once.  As you noted, the final ranking is independent of the number
of candidates elected, unlike conventional STV where the ranking is
quota-dependent.

2) No matter how many candidates are elected to parliament, the largest
possible number of party members will have at least ONE member that they
voted for in your internal elections.  In general, the most popular party
members will be elected first.

The problem with this method is that unlike STV, your method (STV with an
"infinite" quota, so all transfers occur through elimination) does not
provide intra-party proportional representation.  In other words, different
factions that might exist within your party will not necessarily be fairly
represented within your party's "fractie" (what we call a "caucus" in
(Canadian) English).

Consider the following example:

36 voters, 2 factions: A (24 voters), and B (12 voters).  Each faction
nominates 4 candidates (A1..A4, B1..B4).  Assume A1>A2>A3>A4, B1>B2>B3>B4,
in general.

Rank	Candidate	Stages (vote totals only)
1	A1		17	17	19	19	19	24	24	36
4	A2		4	5	5	5	5	0
7	A3		2	2	0
8	A4		1	0
2	B1		3	3	3	4	6	6	12	0
3	B2		3	3	3	4	6	6	0
5	B3		3	3	3	4	0
6	B4		3	3	3	0

The table below shows the final rank order, with the number of seats in
caucus each faction will receive if <rank> candidates win.

Rank	Candidate	Seats-A		Seats-B
1	A1		1		0
2	B1		1		1
3	B2		1		2
4	A2		2		2
5	B3		2		3
6	B4		2		4
7	A3		3		4
8	A4		4		4

As you can see, this ranking method can produce very unproportional results
(where 3 or 6 candidates are elected, Faction B will get twice as many reps
as Faction A, even though A has twice as many supporters!)  Faction B has
been able to consistently outperform Faction A by having its supporters
divide their votes equally among all the candidates, thereby leading to the
early elimination of as many A candidates as possible.  Interestingly, I've
heard that this strategy is actually employed by political parties in
Ireland, although the transfer of surpluses under conventional STV greatly
reduces its effectiveness (final results in Irish elections are very
party-proportional, with each party's share of seats usually within 5% of
their popular vote, based on first-place rankings).

To me, the above problem seems much more significant than the fact that STV
is not monotonic.  Ideally, your final ranking would ensure that each
faction receives a proportional share of seats no matter how many are
eventually elected to parliament.

The best and simplest way to do this would be if you could submit an
unordered (or alphabetical) list of candidates to the public prior to the
parliamentary election, and then hold a conventional STV election within
your party among those not directly elected to fill the remaining seats of
your caucus.  That way, you'd know what quota to use.  You could of course
have all the ballots submitted in advance, so it shouldn't take long.
Alternatively, if you want your party members (rather than the public) to
have the greatest possible control over the composition of its caucus, just
include all candidates (even those directly elected) in your STV election,
but with any directly elected candidates protected from elimination (since
they must ultimately be elected, of course!).  This would prevent
over-representation of factions which are publicly popular, but less
popular among party members.

Unfortunately, your election laws may require you to submit a ranked party
list.  In that case, the best method I could come up with is to:

1) Choose the top-ranked candidate by conducting an STV count for one seat.
If you want to get fancy, you might instead use a good pairwise method
(Smith//Condorcet, or whatever), since these methods can produce better
single-winner results than STV (effectively MPV/IRO in this case).  This is
probably needless complexity though, since your party's chances of winning
exactly one seat in parliament are very low, and the chance of that one MPV
winner also not being the Condorcet winner are even lower.

2) Fill the remaining ranks by holding repeated STV elections with
progressively lower quotas (Droop Quota = [votes/(currentRank+1)]+1).  For
these STV counts, you would need to employ an additional rule: no candidate
whose ranking has already been determined may be eliminated (even if they
are ranked lowest).  This ensures that all candidates ranked in the
previous stage become part of the new ranking.  The one additional
candidate which has been added is ranked last.  Repeat until all candidates
have been ranked.

This method should ensure intra-party proportional representation no matter
how many candidates are elected.  It's a computationally intensive method
though; perhaps only suitable if you're using a computer to count ballots.

Regarding the quota, I'd suggest using Droop rather than Hare; in this
voting procedure, I think the lower quota should actually help the
minorities in your party to get slightly higher rankings, while not being
unfair to larger factions.  For example, if you have a minority faction
which has 1 candidate with just over 10% first-choice support, they would
get ranked 9th using the Droop quota and 10th using Hare.  For this method,
the Hare quota is needlessly high.

When I applied the above method (very tedious by the way!) to the example
shown above (using the Droop quota, and making reasonable assumptions about
transfers) I obtained the following result:

Rank	Candidate	Seats-A		Seats-B
1	A1		1		0
2	B1		1		1
3	A2		2		1
4	A3		3		1
5	B2		3		2
6	A4		4		2
7	B3		4		3
8	B4		4		4

The method appears to work.  For the first six rankings, the result is
precisely proportional  between factions (within limits of accuracy, of
course).  This breaks down in ranks 7-8, where B rises to 50% of the total
representation, but only because faction A failed to nominate sufficient
candidates.

One final suggestion -- I strongly suggest you AVOID using Condorcet's
method (or some variant) to determine all the rankings.  This is because
creating a so-called Condorcet series (ie: a succession of Condorcet
winners obtained by successively removing the Condorcet winner, and holding
another pairwise contest among the remainder) does NOT produce a
representative group of winners.  Instead, it produces a succession of
compromises, which is something very different.  I suspect that using this
system would leave you with the same problem you have now using Borda's
rule -- a party list which in which those at the top are a collection of
bland centrists who's opinions are representative of only a small
cross-section of your party's membership.  Such leadership is unlikely to
lead to strong support from your party's members.  Also, since these
(probably uninteresting) candidates are the ones the public ultimately see
and judge your party by, your party's fortunes might continue to
deteriorate.

Best of luck with your party's electoral reform!

Norm Petry.

----------
From: 	Herman Beun[SMTP:chbeun at worldonline.nl]
Reply To: 	election-methods-list at eskimo.com
Sent: 	July 26, 1998 5:37 PM
To: 	election-methods-list at eskimo.com
Cc: 	<New Democracy
Subject: 	Re: STV for party candidate lists?

New Democracy <donald at mich.com> wrote:

>      You do not wish to use STV because the results may be
> different with a different number of seats.

Well, I _do_ wish to use STV, even very much so! The problem is that
I do not know whether my _method_ (my _ordering_ method) is correct.

>From the various replies I received so far I got the impression that
we are not all of us talking about the same thing. So either I do not
understand your (plur.) answers, or there is a misunderstanding at
least among some here about what my actual question was. Let me try
to rephrase it.

Objective
=========
Finding a way to get STV to produce a _ranking_ of candidates, in
such a way that the candidate ranked 1st would be the candidate
normally elected in a regular 1 seat STV election, the candidates
ranked 1st and 2nd would be the two candidates normally elected in a
regular 2 seat STV election, etc.

(Background: the list produced in this way will be used as the D66
party list in national elections. The national elections are
"pseudo-open" party list elections, i.e. they are open-list in name,
but can be considered closed-list for all practical purposes. I do
wish to change that election system too, but not now.)

Hypothesis
==========
I _think_ I can solve the above problem (and no other) in the
following way:

I start off from (one of) the regular procedure(s) for a 1 seat STV
election. The quota = votes/seats. The procedure:

1st round: Count first choices. Assuming no candidate has 100% of the
votes, eliminate the candidate who got the fewest votes. This
eliminated candidate comes _last_ on the outcome list. Redistribute
the eliminated candidate's ballot papers over the respective voters'
second choices.

2nd round: Count redistributed votes. Eliminate again the candidate
with the fewest votes. This candidate comes second last on the
outcome list. Redistribute this candidate's ballot papers over the
respective voters' second choices.

Etc. until all but one candidate has been eliminated and the list has
been filled.

Questions
=========
1: Does this method meet the objective, i.e. does it solve the
problem _in the way specified_?
2: Any experiences with this system or a similar one?
3: How do other democratic parties forced to operate within a
closed-list national election system (s)elect their candidates?

> That is true but I feel that the STV error is
> small and will be lost in the greater error caused by
> the difference between how the 3000 party members would
> vote and how the 770,464 voters would vote. The 770,464
> voters, if allowed, will rank the candidates far
> different than the 3000 party members. You must accept
> the greater error because it is the system you must work
> within.

Exactly! For practical purposes the possibilities that exist for the
770,464 voters to change the list in the general elections are
negligible. That _should_ be different of course, but the law leaves
us no choice and we do not have the power to change it _at present_.
BTW, I wouldn't characterise any changes the 770,464 voters make in
the list order during the general elections as an 'error' ;-).

>      I would suggest that you also accept the smaller STV
> error so you can use STV to rank the candidates. I contend that
> the error caused by STV will be less than the error caused by
> the elimination run-off order. Any changing of the STV order
> within the first 14 will not change who gets elected from
> your list.

I am not sure if I understand what you mean here. Are you suggesting
we should compose our list by holding a 14 seat STV election among
our members and rank them in an arbitrary order? I don't think we'd
want to do _that_. That number of 14 seats is not very fixed either,
especially for our party; in the last parliament we had 24 seats, in
the one preceding it 12... I'd say it _does_ matter that the D66
(s)election procedure produces an order that gives an STV composition
for the group of our members in parliament for any number of seats we
might get in the general elections.

>      It is best not to use the elimination order to rank more
> than one candidate because it does not give equal power to all
> votes. An example may explain my viewpoint: Suppose your party
> had one candidate that was head and shoulders above the rest -
> and that candidate received fifty percent(1,500) first choices.

_First_ choices, yes. Knowing how strongly the Dutch election system
makes the media focus their attention on party leaders, this is not
even a very unlikely situation.

> That candidate will receive the number one ranking on the
> list - good for him, but that is not the problem. The
> problem is that the first fifty percent of the voters will
> have only placed one candidate on the list. All the other
> positions on the list will be determined by the second fifty
> percent.

Well, no, I don't think that is the case. I think the party members
voting in the D66 (s)elections do have a clear picture in mind of
what they want the future "fractie" (1) as a whole to look like, and
definitely fill in more than only a few names. In reply to a question
from Demorep: no, I have no figures of this, but I would say the
party "culture" is like that. Also, I think voters (members) in D66
candidate elections _are_ able to form an opinion about more than 6-8
candidates because all members receive a "candidate book" with all
the names, personal details, CVs and a motivation written by each
candidate. Maybe they are less sure of the _exact_ order in which to
rank especially the candidates lower on their list, but they do have
an approximate picture in mind. I at least certainly do ;-). And
those who do fill in only a few names apparently had no opinion about
who the other candidates should be, so then it is only reasonable
that the other voters fill in those places.

(1) "fractie" is the Dutch term for the entire group of parliament
members that belong to one party (in this case D66). I wasn't able to
find an adequate, short, translation for it in my English dictionary
(it says "party", but that would be confusing since I am only talking
about the MPs).

>      If you say that you do not have a single candidate that
> will pull fifty percent of the vote then I say that every case
> you do have of a candidate with surplus votes is a case in which
> voters are not receiving the full value of their vote.

Why? Surplus votes are transfered too of course!

> Some candidates do gain sizeable amounts of surplus votes. In
> a recent election in New Zealand, a women candidate received
> more than three quotas. If you had a candidate like her in your
> elimination run-off those three quotas of voters would only be
> able to rank one candidate when they should be able to rank
> three. If we are to rank more than one candidate we must use
> a proportional method - STV is best to use.

Er, yes of course, but... I was not talking about elimination
run-off! Maybe my use of the term "elimination order" (refering to
the order in which _STV_ eliminates the candidate with the fewest
votes each time, see description above) caused the confusion?

>      Is it possible for you to give an estimate list before
> the election based on the number of seats you expect to win -
> and then adjust the list after the election based on the actual
> seats your party has won?

No, I think, because in that case the 770,464 general voters would
not know which candidates they were voting for! Although the
possibilities for those voters to change the order on the candidate
lists maybe limited, they are not zero either (one or two candidates
in the entire 150 member parliament do get elected in this way each
time). Voters should always have the possibility to overrule the
order devised by the parties.

However, I must admit that I came across a confusing piece of
information that _seems_ to contradict that parties cannot change the
order of candidates after the elections. But I am not sure, and it
doesn't seem logical either.

>      You gave another objection to STV when you wrote:
> "However, STV normally does not produce a ranking order of the
> candidates; after all, in principle each candidate is finally
> elected with a number of votes that is exactly equal to the
> election quota."
>
>      Your statement is not exactly true. There will be enough
> differentiation between candidates in order to produce a ranked
> order.

Yes, but the usual outcome of an STV count is only a number of names
to fill the seats with, no ranking order. I'll _have_ to device
something (new, it looks like) to get a ranking order.

>      One: We will have candidates with surplus votes. These
> candidates can be ranked according to highest number of surplus
> votes. If two are tied we look to the next choices of the votes
> of these two candidates.

Cannot be used since we do not know the number of seats to be filled
(N) in advance: We need N in order to calculate the quota, and we
need the quota in order to calculate the number of surplus votes. The
ranking method should be independent on N.

>      Two: We will have candidates that reached quota from
> the transfer of the surplus votes. If more than one candidate
> reaches quota, then we rank them according to the highest number
> of votes over quota - before the excess votes are transferred
> a second time.

Cannot be used for the same reason as above.

However, this transfer of votes does take place when we do know N in
advance (i.e. in what I called "a regular STV procedure"). The big
question I want to solve, is: does this transfer affect _our_
outcomes. I.e.: suppose we perform a regular STV count on the
member's ballot papers _after_ the elections, so when we know the
value of N, would that result then be different from the top N names
on the candidate list composed by my method? Or would the result of
my method be different, but nonetheless "good STV"? How do we define
(and measure!) "good STV"? Does an election for a given number of
seats perhaps have more than one "good STV" solution?

Or: Is "good STV" simply defined as the outcome of an STV election?
In that case I wouldn't have to worry too much, if the result of my
method is _usually_ close to a regular STV outcome.

Regarding "good STV": A good election outcome would IMO have the same
"political middle" as the voters themselves have, and an equally
broad distribution around it. In this way, each minority view gets a
proportional share of the representatives. From my simulations, I get
the strong impression that the method I propose for party lists has
this property.

>    Three: We will have candidates that reached quota from the
> transfer of votes from the eliminated candidates. These candidates
> can be ranked according to the order of the eliminated
> candidates. If two or more candidates reach quota on the same
> transfer then again we look at the votes over quota.

Here too: quota unknown.

>     Four: About 7 of the top 14 candidates will not reach
> quota because of exhausted ballots - provided the Hare Quota
> is used. These candidates are to be ranked according to the
> votes they each did receive.

I wanted to use the simple votes/seats quota instead. This is partly
personal preference (for a 1 seat election, I want a candidate who
has 100%, not 50%, support -- leaving aside that I'd prefer using
Condorcet instead for the 1 seat case), partly because I fear my
'elimination method' would not work otherwise: I think it is
important that no candidate reaches the quota until all others have
been eliminated, so for N=1, the quota should be 100%.

>     Five: We have the eliminated candidates. These candidates
> are to be ranked according to the number of votes they had
> when they were eliminated.

Yes, that was my proposal. This order is the same as the order in
which they were eliminated (since by definition, each time the
candidate with the fewest votes is eliminated).

>      All the above can be avoided if you can sway your country
> to use open party list - or better yet, use STV in the one
> election area. But, in the meantime you must work within the
> system.

I agree (all of the above).

>      If you use STV to rank your list you will be setting a
> good example and pointing the way for other parties and the
> government to use STV in the future - your step two.

Yep! :-)

Regards,

---------------------------------------------------------------------
Herman Beun                                                    Arnhem
http://home.worldonline.nl/~chbeun/                        Gelderland
CHBeun at worldonline.nl                                       Nederland
EU
**** Representative democracy is a contradiction in 4 year terms ****
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