[EM] how come i hadn't heard about "Meek STV" before on this list?

[robert bristow-johnson]

---------------------------- Original Message ----------------------------

Subject: Re: [EM] how come i hadn't heard about "Meek STV" before on this list?

From: "Kristofer Munsterhjelm" <km_elmet at t-online.de>

Date: Tue, March 15, 2016 11:21 am

To: rbj at audioimagination.com

election-methods at electorama.com

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> On 03/15/2016 02:39 AM, robert bristow-johnson wrote:
>...

>> also, what to others on this list think of Meek STV? i would love to

>> hear pros and cons?

>

> Consider a complexity line for STV. On the very left is the simplest

> possible STV method, which goes something like this:

>

> 1. Create n piles, one for each candidate.

> 2. Place each ballot in the pile corresponding to the candidate it ranks

> first.

> 3. As long as there's any pile has more ballots in it than the Droop

> quota Q:

> 3.1. Elect the candidate that pile belongs to.

> 3.2. Eliminate that candidate from the count.

> 3.3. Let the number of ballots in the pile be n. Draw n-Q ballots at

> random from this pile and place the ballots in piles according to their

> first uneliminated preference.

> 4. Eliminate the unelected uneliminated candidate with the least votes

> (thinnest pile) and redistribute the ballots in that pile according to

> their first uneliminated preference.

> 5. Repeat from 3 until all candidates are either eliminated or elected.

>

> This is only approximately fair because of the random component, and is

> also vulnerable to Woodall free riding (where you vote for candidates

> who have no chance so that you don't get stuck early in an elected pile).

>

> You can then go further to the right by transferring every ballot at a

> fractional weight rather than a random subset at full weight in step

> 3.3. That's not very practical in the real world, but simple enough for

> an algorithm to do.

>

> So as you go to the right, the method gets more complex but also more

> fair. Meek, in particular, deals with two things: first, it acts like a

> method that restarts the count whenever a candidate is eliminated (i.e.

> no candidates are considered elected and they're back in the running).

> This makes Meek impervious to Woodall vote management. Second, it tries

> to be more fair with votes that transfer to someone who is already

> elected. E.g. suppose we have a ballot

> X>Y>Z

> where Y was elected earlier on and we're transferring away from X

> (either because X is eliminated or because he's elected). Simpler rules

> would give that ballot to Z. But that means that this ballot isn't

> subjected to the redistribution that every ballot that got Y elected is.

>

> Meek fixes that problem by giving each candidate a weight between 0 and

> 1 (or 0 and 100%). If candidate X has a weight of 80%, then a ballot

> voting for X first will give 80% of a vote to X, and 20% to whichever

> candidates come after X on that ballot (recursively, e.g. if the second

> ranked candidate is Y with weight 70%, the ballot would give 20% * 70% =

> 10% to Y and so on).

> If a candidate X ends up with more than a Droop quota, his weight is

> adjusted downwards. In turn, that may cause other candidates to get more

> than a Droop quota as more ballot weight flows past X to others, so the

> process is repeated with them. The method eventually converges, but it

> can take a long time.

>

> So in short, Meek is a more complex but more fair type of STV. It isn't

> fundamentally different from other types of STV, since its single-winner

> version is still IRV. But it treats ballots more equally and so is

> better than the manual methods if you can handle the complexity.

>
it's unbelievably complex in my judgement.  i can't believe it gets as much traction as it does (as i found out at Stack Exchange).
 


>> and can we discuss Condorcet methods for multi-winner elections?

>

>...

> A problem with multiwinner Condorcet methods is that they're generally

> very complex. It's hard to come up with a multiwinner analog to the

> Condorcet criterion,
why?
1. start with NumRemainingCandidates = NumCandidates and NumRemainingSeats = NumSeats .
2. compute the "Defeat Matrix" like you would for single winner Condorcet, which i still think should be triangular in shape and not rectangular.
 everything else is determined solely from that Defeat Matrix.
3. determine whom the top-preferred candidate is (using whatever variant of Condorcet you like best, Schulze, RP, minmax, margins, winning votes, whatever to resolve any cycle).
4. elect that person and eliminate that
person from the set of candidates.  decrement both NumRemaningCandiates and NumRemainingSeats by 1.
5. with the set of remaining candidates, if NumRemainingSeats > 0 go to step 3.  else end and pack up the ballots into the ballot bag and seal it.
how hard is
that??
 
and here is what we tell voters how the system works.  first this is what we tell them for single winner:
"If more voters prefer Candidate A over Candidate B than vise versa, then do not elect Candidate B."
with  multiwinner races, simply add
four words at the end getting
"If more voters prefer Candidate A over Candidate B than vise versa, then do not elect Candidate B before electing Candidate A."
who can argue with that?
even with the need to resolve cycles, it's a *far* simpler, more meaningful, ostensibly
fair and democratic (majoritarian) rule or principle than any way you can explain **any** STV which, even at it's simplest, is a kabuki dance of transferred votes that will make a lot of people suspicious that you're trying to pull something over on them.  Meek seems like it's trying to be
fair and proportional, but it's complicated and, like other STV, has contrivances (totally opaque to second ranked votes until the first choice is eliminated) and i am sure goofy pathologies like single-winner IRV has.
	 

> so usually what the methods do (like Schulze STV)

> is compare *assemblies* as if they were candidates in a Condorcet method

> (e.g. "elect A, B, and C" beats "elect B, C, and E" pairwise). But since

> there are up to (numcands choose numseats) of these assemblies, it can

> get unwieldy fast.
i don't see why it would have to.
sorry to sound like a Condorcet partisan, but i am that.  especially after our experience in Burlington Vermont in 2009 and 2010.
 
thanks for replying, Kristofer.
BTW, if i may share a fun little fact:
i am an elected Inspector of Election for Ward 7 in Burlington Vermont ( https://www.burlingtonvt.gov/CT/Ward-Clerks-and-Inspectors-of-Election ).  a certain very cool U.S. presidential candidate lives in Ward 7 and i got to greet him and his wife when they voted last March 1 (Super
Tuesday).  my sister told me on facebook that she saw me on the Today Show that morning.



--
r b-j                  rbj at audioimagination.com
"Imagination is more important than knowledge."


 
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