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Previously, in the "Condorcet elimination PR" thread, I wrote:<br>
"Further on the subject of my proposal to modify STV-PR by , after transfering
supluses, eliminating the Condorcet loser among those votes and fractions
of votes not tied up in quotas: I am afraid I have discovered that this
method fails FBC (Favourite Betrayal Criterion). <br>
To demonstrate the problem, I have slightly modified your example by introducing
candidate X. <br>
<br>
300 votes, 3 seats, Droop quota = 75. <br>
<br>
74 A B C D X E <br>
39 B A C D X E <br>
75 C <br>
37 D X E C B A <br>
73 E D X C B A <br>
2 X E D C B A <br>
<br>
Compared to your example, all that has changed is that X has stood and taken
2 first preference votes from D and has occupied the runner-up Condorcet
loser position. The result, CBD, is unchanged. <br>
The problem is that if those 2 X supporters had reversed their top two
preferences, then E would have got a quota and the result would have been
CEB. By insincerely down-ranking their favourite, those two voters could
have caused their second prefernce to be elected instead of their third."<br>
This was maybe not so clever because this is just the mildest form of FBC
which no ranked-ballot method can avoid. For a simple single-winner example,
I lifted this from <a class="moz-txt-link-freetext" href="http://www.condorcet.org">http://www.condorcet.org</a><br>
<br>
45 ABC <br>
35 BCA <br>
25 CAB
<p>"In this example, <b>A</b> would normally win. However, the <b>BCA</b>
voters could improve the outcome (from their perspective) by voting <b>CBA</b>,
which would result in <b>C</b> winning with a majority of first place votes."<br>
With sincere voting, A is the clear beats-all CW (as well as the Plurality
and IRV winner), but by "betraying" their favourite, the B voters can elect
their second preference C.<br>
</p>
<p>I now propose a non-elimination version of ranked-ballot PR, which combines
Bucklin and Condorcet.<br>
Ranked ballots, equal preferences ok. Count the first preference votes. Equal
preferences are divided into equal fractions (which sum to 1).If any candidates
have a Droop quota they are elected, and then reduce the values of the ballots
which have elected members by an amount which sums to a Droop quota.<br>
If more than one place remains unfilled, proceed to to the second round.
Add the second preference votes to the first preferences (based on the value
of the ballots after the any reductions that were made the previous round).
If this gives any candidate a Droop quota, then elect the candidate with
the highest tally. If there is a tie, then elect the tied candidate who
had the bigger tally at the last round, if still tied then the round before
that if there was one, otherwise the Condorcet winner of the tied candidates
based on the ballots after the most recent devaluations. Reduce the value
of the ballots that elected this winner by an amount that sums to a Droop
quota. If there is still more than one place unfilled and if after the latest
devaluing of ballots any candidates have a Droop quota, then elect the one
with the highest tally (same tie-breaking proceedure) and so on.<br>
If there is more than one place to be filled, then add the third preference
votes to the tallies of first and second preferences and if that gives any
candidate a Droop quota, then as before the candidate with the highest tally
is elected and so on.<br>
If proceeding in this way leads to the situation where there is one and
only one more place to be filled, then based on the ballots after all the
devaluations elect the Condorcet Winner.<br>
<br>
This method is I think simpler to count than any version of STV, and it might
be monotonic.It performs very well in all the examples that I have seen that
showed up Condorcet Loser elimination STV. In the simple example at the top,
it elects C A E in that order.<br>
Soon I will post more on this, going through some examples.<br>
<br>
Chris Benham<br>
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