[EM] Re: A Condorcet-like method that satisfies FBC (I believe)

Araucaria Araucana araucaria.araucana at gmail.com
Thu May 19 12:27:09 PDT 2005

On 18 May 2005 at 15:14 UTC-0700, Kevin Venzke wrote:

<long explanation of method>

> For clarity, here is a brief definition of the method I'm suggesting:
> The voter places each candidate into one of three slots.
> v[x,y] is the number of voters voting X over Y.
> t[xy] is the number of voters ranking X and Y together in the top slot.
> Define a set S containing every candidate Z for whom there is no other 
> candidate W such that v[z,w]+t[zw]<v[w,z].
> If S is empty, then S contains all the candidates.
> Elect the member of S who is in the first or second slot on the most 
> ballots.
> (I admit that it's weird to have everyone compress in the second 
> stage in the same way, regardless of who is in S. In ordinary 
> Condorcet//Approval, the second stage never occurs unless S contains
> all the candidates. But I don't see any other neat way of picking a 
> winner from S, especially if S contains more than two candidates.)

Hi Kevin,

This is a very interesting idea.  But I don't see any reason why it
couldn't be applied to a 'standard' approval cutoff ranked ballot.

     v[x,y] is your notation for the standard pairwise array.

     t[xy] is your notation for another pairwise array (symmetric).

Let's consider two other summable arrays that could be tabulated from
an approval cutoff ballot:

     ab[x,y] = number of voters Approving Both x and y.  This array is
     summable, and a[y,x] = a[x,y], so it is symmetric.

     sp[x,y] = number of voters approving x but not y.  This is James
     Green-Armytage's Strong Preference array, also summable.

The kernel of your method is that you don't want to penalize a
candidate for a weak defeat; i.e., one that is has more weak
preferences (both in the top-slot in your proposal, but I'm suggesting
that both approved could be used instead -- let's discuss a class of
methods and not get specific for now) than the winning margin.  A
strong defeat is when the winning margin is greater than the number of
weak preferences.

So your first round is to check whether any candidates have no strong
defeats.  If any exist, eliminate any strongly defeated candidates.
That's all well and good -- we've eliminated candidates that a
majority has agreed should not be elected.

Your question is what do to on subsequent rounds, and you choose to
pick the approval winner among the remaining candidates.  I think this
might actually fail FBC because a lower-ranked candidate could have
higher approval.

Here's another idea:  combine this with something sort of like Bucklin:

       f[x,y]   = # of voters putting x and y in first place

       fs[x,y]  = # of voters putting x and y in either first or second
                  place and approving both

       fst[x,y] = # of voters putting x and y in 1st/2nd/3rd place and
                  and approving both

For rounds 1-4, if some candidates would remain, eliminate strongly
defeated candidates (within the set of remaining candidates) according
to the round's measure of strong defeat:

Round 1:  Strong defeat X>Y means v[x,y] - ab[x,y]  > v[y,x]

Round 2:  Strong defeat X>Y means v[x,y] - fst[x,y] > v[y,x]

Round 3:  Strong defeat X>Y means v[x,y] - fs[x,y]  > v[y,x]

Round 4:  Strong defeat X>Y means v[x,y] - f[x,y]   > v[y,x]

Round 5:  Elect DMC winner among remaining candidates.

No candidate is eliminated until it is strongly defeated by another
candidate, according to the strong defeat measure in the round.  So if
X and Y are not strongly defeated by Z, but one of them strongly
defeats Z, Z can be eliminated and then the preference between X and Y
is considered.

This is sort of off the top of my head, so play with it as you will.

araucaria dot araucana at gmail dot com
Q = Qoph = "monkey/knot" -- see http://www.ship.edu/~cgboeree/alphabet.html

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