[EM] Condorcet loser elimination PR

[Chris Benham]

  Marcus,
In response to my proposed  version of  STV- PR:

Ranked ballots, equal preferences ok, truncation ok,truncation treated
 just like equal prefernces. Elect those candidates with a Droop quota 
(1/number of seats to be filled + 1) of first preferences (equal first 
preferences count as fractions). Distribute the surplus preferences 
("the overflow") of those candidates all-at-once  as fractions, elect 
any who now have a quota and in the same way distribute the new 
surpluses and so on. If at the end of this process there are still seats 
unfilled, NOT counting those votes and fractions of votes that make up 
the quotas of  those already elected candidates, determine among the 
remaining candidates the Condorcet Loser and eliminate that candidate. 
Distribute that candidate's preferences, elect any who have a Droop 
quota, distribute any surplus and so on until all the seats/positions 
have been filled.
You wrote:
"I haven't yet understood your proposal. Please
explain your proposal with the following two examples.

Example 1:

   100 voters; 2 seats; 5 candidates

   26 BCAED.
   24 DCEBA.
   10 EADBC.
   08 ABCED.
   07 EABDC.
   07 EDBCA.
   06 CDEBA.
   06 DEBCA.
   03 DCEAB.
   02 EBADC.
   01 DCBEA.

Example 2:
   In example 1, change a single DEBCA ballot to BDECA."

This method may not be the best possible ranked ballot PR method, and
so may not be my last word on the subject. James Green-Armytage has shown
that it tends to tidy up towards the centre.
I have worked through your examples. For "Condorcet loser", I have used the 
Schulze loser. To clean up some fractions, I multiplied the votes tallies by
24. For the first example this gives:
   
   2400 voters; 2 seats; 5 candidates
    
    624 BCAED
    576 DCEBA
    240 EADBC
    192 ABCED
    168 EABDC
    168 EDBCA
    144 CDEBA
    144 DEBCA
     72 DCEAB
     48 EBADC
     24 DCBEA
 
Droop quota is 2400/3 = 800.
No candidate has 800 first preference votes, so we eliminate the Schulze loser,
which is A.(I used an on-line ranked ballot calculator.) Now the 192 ABCED ballots
merge with the 624 BCAED ballots, to become 816 BCED ballots. B now has a quota and
is elected.  800 of those 816 ballots take no further part in the election.
The remaining active ballots are:
   
    16 CED
   672 DCE
   624 EDC
   144 CDE
   144 DEC

No candidate has a quota, so based on these ballots, we eliminate the Schulze loser,
which is E. So now the ballots reduce to:
   
   1440 DC
    160 CD

So D has a quota and is elected. The final result is BD.

For your second example we change 24 ballots from DEBCA to BDECA.

   624 BCAED  
   576 DCEBA
   240 EADBC
   192 ABCED
   168 EABDC
   168 EDBCA
   144 CDEBA
   120 DEBCA
    24 BDECA
    72 DCEAB
    48 EBADC
    24 DCBEA

No candidate has a quota, so eliminate Schulze loser which is A.
This results in B's total rising to 840. The 40 surplus votes are divided thus:
 40 x 816/840 : CED
 40 x  24/840 : DEC (these numbers are close enough to 39 and 1).  
The active  ballots now reduce to:

   672 DCE
   624 EDC
   144 CDE
   121 DEC
    39 CED

No candidate has a quota, so we elminate the Schulze loser,E.
As before D easily wins the last seat, and so the final result is again
BD.

Chris Benham







































 







 































































    







   










 In example 1, a single DEBCA ballot is changed to BDECA. 

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