[EM] Holy grail: a "condorcet compliant" cardinal method (MCA/Bucklin variant)

C.Benham cbenhamau at yahoo.com.au
Sun Sep 5 14:23:39 PDT 2010

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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