[EM] A much simpler proportional Condocet Method - BTR-STV

[Antonio Oneala]

raphfrk at netscape.net wrote: 
> Now, first I'm going to introduce to you the idea of the proportional 
Condorcet winner.
> The are several of these in an election, and the majority Conorcet 
winner is always
> among them. A proportional Condorcet winner is the Condorcet winner 
of the
> minority of the electorate. For instance, in a four seat election the 
a Proportional
> Condorcet winner would be the Condorcet winner of 25% of the 
populace.

That is not a very clear definition.  Are you saying that a candidate 
is a proportional
condorcet winner if there exists a set of votes amounting to 25% (or 
more) of the
electorate in which he counts as a condorcet winner ?

If so, then in a 4 seat constituency, there is potentially more than 4 
proportional
condorcet winners.

For example (just listing first preferences)

A: 15%
B: 15%
C: 14%
D: 14%
E: 14%
F: 14%
G: 14%

Any of the candidates in the above election is a proportional condorcet 
winner.  The 25%
that each candidate would pick would be his own supporters and 11% of 
the vote from
some other supporter.  He would be the condorcet winner as 14% rank him 
first and at
most 11% rank any other candidate ahead of him.  This means that there 
could be 7
proportional condorcet winners in a 4 seat constituency (maybe more).  
At least 3
of them must be eliminated.


Let me be more specific.  A proportional Condorcet winner is the winner of one of the segments of the populace that is represented by a proportional method.  A 4 seat election represents 4 sets of 25% of the populace.  There is a proportional Condorcet winner for each 25%.  

Also, I was wrong in saying that the majority Condorcet winner will always be one fo them.  Condorcet gravitates to the center point of a group.  Eeach 25% may have a different center of opinion than all of them coming together, since they all effect each other once they come together.




> Now, this method that I'm proposing differs slightly. It uses a 
normal STV method
> for transferring the surplus votes. Then in the elimination of 
undervotes round,
> instead of eliminating the one with the least votes and then 
transferring their
> votes, you take a set of the least-vote getters equivalent to the 
amount of seats
> in the election, plus 1, and eliminate the one who is the least 
preferred on the
> ranked ballots by the the electorate, then transfer that person 
votes. Therefore,
> a proportional Condorcet winner will never be eliminated, and they 
will always rise
> to the top.

You haven't said how the least preferred is to be determined (condorcet 
loser?).
So, in a 4 seat constituency, the bottom 5 candidates are compared and 
the
condorcet loser of that election is determined based on all the votes.  
Since, I
don't know what you mean by proportional condorcet winner, I don't know
what your claim that none of them can be eliminated actually means.

Let's take this election as an example:

Election for 2 seats, 3 candidates

100: A > B > C
10: B> C > A
105:  C > B > A

The transfer of surplus votes isn't going to change the winner, so let's just ignore that part.

Now, to find which person in this election we are going to eliminate we take the bottom three with the least votes and compare them against each other.  This is all of them in this election.  Why the bottom three?  In BTR-IRV, you compare the bottom two people with the least votes because it is an election for a single winner.  The Condorcet winner will always win in this head to head matchup, so we know he will never be eliminated.  In a multi-winner election, however, this would make little sense because the majorities choice would rise to the top.  No, we take an amount of candidates equivalent to the amount of seats we are electing, and 1 more.

To compare the candidates, in this imaginary round we simply have each person "vote" for the candidate among that amount that they prefer the most.

That's

100 for A
10 for B
105 for C

A and C won.

No additional rounds are needed, although you probably would need more in a real election with a realistic amount of candidates.  Note, however, that B was the majority Condorcet winner here.  A and C, however, each had a majority in the set of the populace that was reprented by a 2 seat election.  The amount of people represented can be determined by the number of seats + 1.  It would therefore be irrational to elect B, but methods like Single Transferrable Vote by Condorcet loser elimination would elect B.

I believe CPO-STV would always elect a proportional Condorcet winner as such.

Here's a much more realistic election:

3 seats, 6 candidates

24 A1 > A2 > C1 > C2 > B1 > B2  > B3
20 B1 > B2 > B3 > C1 > C2 > A1 > A2
10 C1 > C2 >  A1 > B1 > A2 > B2 > B3
5   A2 > A1 > C1 > C2 > B1 > B2 > B3
4  C2 > C1 > A1 > B1 > A2 > B2 > B3
3  B2 > B1 > B3 > C1 > C2 > A1 > A2
2  B3 > B1 > B2 > C1 > C2 > A1 > A2

You need 15 votes be declared elected By the Droop quota, so let's redistribute the votes of people who have more than that:

A1 tranferrs 9 votes to A2.
B1 tranferrs 5 votes to B2.

A1 and B1 are immediately elected, and the new vote totals (only counting first place votes, of course) are:

B1: elected (15 votes consumed)
A1: elected (15 votes consumed)
C1: 10
A2: 14 
C2: 4
B2: 7
B3: 2

Now, let's take the bottom 4 candidates with the least votes and compare them against one another

Each voter attaches themself to the candidate out of the set they have ranked highest:

B3: 2 preferred him more than any of the other candidates in this round
B2 : 7
C2 : 4
C1 : 19

(I'm not exactly sure if consumed votes should be counted in this round...  I believe that you shouldn't, however, so I'll just think about it later)

B3 has the least people who prefer him.  He tranferrs his votes to his second choice that is not yet elected, which is C1.

The vote totals are now:

A1: Elected (15 votes consumed)
B1: Elected (15 votes consumed)
C1: 12
 A2: 14 
 C2: 4
 B2: 7
 B3: Eliminated (0 votes)

Now let's repeat the elimination round:

C1: 12
A2: 14
C2: 4
B2: 7

C2 has the least people who prefer him the most out of anybody in this round, and so his votes are transferred to C1, his next choice. 

The vote totals are now:



 A1: Elected (15 votes consumed)
 B1: Elected (15 votes consumed)
 C1: 16 votes
  A2: 14 
  C2: Eliminated (0 votes)
  B2: 7
  B3: Eliminated (0 votes)


So now we have 3 candidates with more than 15 votes.  There is no reason to transfer C1's surplus.

I believe that STV would've elected A1, B1, and A2, which illustrates the difference in the methods.  CPO-STV probably would've elected the same slate.

This deals with the problem of sequential exclusions differently the CPO-STV, however.  Instead of comparing each set of outcomes to make sure that no one is excluded that would be later elected, it compares a slate of candidates to make sure that no one who would be elected later on could be eliminated.  In the event of a majority rule cycle, however, CPO-STV would most likely give better results, as Ranked Pairs or the Schulze method give better results in those cases than BTR-IRV, Nansons method, and all the other methods that find the Condorcet winner by making sure that he is never eliminated.  

Really, though, how often is a majority rule cycle going to occur?  I believe the simplicity of explaing these kinds of methods and making the electorate understand is worth the fact that you may or may not get a less preferred candidate in the case of a majority rule cycle.





In any case, there is an interesting effect if you use the Droop quota 
for electing
candidates.  After a candidate is elected, and their votes transferred, 
you can
consider the remainder of the election as an election for one fewer 
seats.

For example, if there was a 4 seater, the rounds might go like:

Round 1:  nobody elected (20% required)
Bottom 5 candidates are compared

Round 2: candidate gets >20% so is elected
votes redistributed

Election now considered a 3 seat election.
Votes remaining: 80% of the original
Target is droop quota = 80%/(3+1) = 20% of original vote so stays 
constant

Round 3: nobody elected
Bottom 4 candidates are compared (instead of 5)

and so on.


This holds for my method also, unless I've made some kind of drastic mathmatical errors.

		
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