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

[Jameson Quinn]

Yes, my "proof" was flawed. The CW Cathy must beat the Approval winner
Andy's approval (1,1,0) score with her range(2,1,0) score; but she need not
beat his range score with her range score. This fact is much less useful,
because there could be a number of candidates whose range scores surpass
Andy's approval score; so you may not actually eliminate anyone for the
runoff, unless you add additional filtering criteria. Such extra critera
could not, as far as I can see, be both O(N) summable, Condorcet-compliant,
and guaranteed to cut down to two candidates.

(There is, however, one more summable, Condorcet compliant criterion based
on 3-rank Range: any candidate whose 1,0,0 "preferral"/"range minus
approval" score beats another candidate's 1,1,0 "approval" score, must beat
them pairwise. This could be used in addition to the "absolute majority"
requirement to give some extra first-round wins in cases of a clear winner
despite a fragmented electorate).

Warren Smith came up with a simpler counterexample, which I simplified
further:

#voters vote
7 CAB
5 ACB
4 ABC
3 BCA

C is the CW, A is the AW and the RW. Apparently, Condorcet himself was aware
of this possibility (though of course he was considering Borda, not Range).

JQ

2010/9/5 C.Benham <cbenhamau at yahoo.com.au>

>
> 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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