[EM] Condorcet loser elimination PR
Chris Benham
chrisbenham at bigpond.com
Tue Jul 29 11:35:59 PDT 2003
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.
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20030729/b80d83fb/attachment-0003.htm>
More information about the Election-Methods
mailing list