Jameson Quinn wrote (5 Sep 2010):
Here's my latest Bucklin variant, which, pending the results of the
naming poll <http://betterpolls.com/do/1189>, I'm calling RMCA (because of
the catchy music). (Of course, if it's OK to appropriate the name MCA, the
editorial headline writes itself...)
Start with two-rank Bucklin ballots: Preferred, Approved, or Unapproved. The
highest majority preferred, if any, wins it. If not, find the highest number
of approved-or-preferred ("approval winner", AW), and the highest range
score ("range winner", RW), counting 2/1/0 for P/A/U. If those are the same
candidate, that candidate wins; otherwise, those two go into a runoff. (If
either of these measures gives an exact tie, then the two tied candidates go
to runoff.)
The first (to me, surprising) result is that any Condorcet winner which is
determinable from the ballots must get into the runoff. Proof: Say that the
AW is not the CW. Then the number of ballots n with CW>AW is greater than m
with AW>CW. On a ballot where X beats Y, X has a range advantage of either 1
(X>Y) or 2 (X>>Y). Sf n2 where CW>>AW is greater than (m2 where
AW>>CW)+(n-m), then the CW is the RW. And if n2 < m2, then there are more
ballots which approve the CW and not the AW than the reverse, which
contradicts the assumption that the AW is the AW. QED.
Adapting an example from Douglas Woodall:
4: A>B
6: A>C
6: B>A
2: B>C
3: C>B
The Condorcet winner is B, but Jameson's suggested "condorcet compliant"
method elects A.
Chris Benham
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