[EM] More Condorcet Flavored PR examples

[Forest Simmons]

On Wed, 20 Nov 2002 matt at tidalwave.net wrote:

> On 20 Nov 2002 at 14:11, Forest Simmons wrote:
> > Suppose that a voter ranks six candidates as follows in a three seat multiwinner race:
> >                 A1>B1>B2>A2>A3>B3 .
> > Which of the following two outcomes would this voter be most likely to prefer?
> > (1) The A team {A1,A2,A3} wins. (2) The B team {B1,B2,B3} wins.
> > ....
> > ...the Condorcet spirit seems to dictate the choice of the higher median subset, since
> > the one  with the higher median has more pairwise wins (relative to the ballot).
>
> Using the PERL scripts found at http://condorcet-dd.sourceforge.net/ I ran
> ballot2tally.pl with the voted ballot list 20:A1>B1>B2>A2>A3>B3 as the input. Then I
> ran option2outcome.pl --size 3 followed by Condorcet_DD.pl with all tie-breakers
> turned off.  The results were {A1,A2,A3} tied for sixth place with {A3,B1,B2} while
> {B1,B2,B3} tied for seventh place with {A2,A3,B1} and {A1,A2,B3}.  So your
> assumption that Condorcet dictates choice of the higher median subset was not true
> here.  First place was {A1,B1,B2} (of course), second place was {A1,A2,B1}, tied for
> third was {A1,A3,B1} and {A1,A2,B2}, tied for fourth was {A1,A3,B2} and {A1,B1,B3}
> and tied for fifth was {A2,B1,B2} and {A1,B2,B3}.
>

Let's count:

A1 beats all three B's head-to-head.
A2 and A3 each beat only B3 head-to-head.

So over all, there are 5 pairwise wins of A's over B's.

B1 and B2 each beat both A2 and A3,
with no other pairwise wins of B's over A's.

By golly, you're right!

Forest

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