[EM] More Condorcet Flavored PR examples

[Forest Simmons]

I should have known that the Borda Count is related to the number of
pairwise wins: that's why the row sums of the pairwise matrix give the
Borda counts.

So here's the latest stab at the definitive Condorcet Flavored PR Method
(CFPRM) :

To specify a CFPRM it is sufficient to tell which among two arbitrarily
given candidate subsets S1 and S2 (of the same size) an arbitrarily given
ballot B will favor, and what weight W to attach to that choice.

[Whether to use Ranked Pairs, SSD, or MinMax on the sum of the resulting
pairwise matrices is a secondary consideration.]

For simplicity let's assume for now that we are dealing with a ballot that
has the candidates fully ranked.  Later we will state the more general
version that can be applied to partially ranked ballots, grade ballots,
etc.

Here we go:

The ballot B favors subset S1 over S2 if and only if the number of
pairwise wins of members of S1 against members of S2 is greater than the
number of pairwise losses of the same type.

Equivalently, (temporarily) delete from the ballot all candidates that are
not members of the union of S1 and S2. Then S1 is favored over S2 if and
only if the Borda Count of S1 is greater than the Borda Count of S2.

Equivalently, S1 is favored over S2 if and only if the average rank R1 of
members of the difference S1-S2 is greater than the average rank R2 of the
members of the set difference S2-S1 (all relative to the ballot order
restricted to the union of S1 and S2).

Now that we know which set is to be favored (if either), how much weight W
do we assign in order to bring about proportional representation (while
taking into account the short comings of 1/n method brought to our
attention by Ollie Salami)?

To answer this question, let's assume without loss in generality that S1
is the favored subset.  Then we take W to be the number of members of S2
that are ranked strictly below the R1 rank defined above.

The greater the value of W, the greater the cause for regret if S2 were to
win rather than S1.  Furthermore this value is calibrated to bring about
proportional representation, as can be seen from the following examples:


Example 1.

There are five seats with two opposing factions:

60 C1>C2>C3>C4>C5>C6>C7
40 C7>C6>C5>C5>C3>C2>C1

Suppose that S1 is the subset {C1,C2,C3,C4,C7}, and that S2 is the (exact)
PR subset {C1,C2,C3,C7,C6}.

Then the larger faction prefers S1, and R1 is just the rank of C4, the
only member of S1-S2.

According to this faction, there are W1=2 members (C7 and C6) of S2 that
have lower rank than C4.

The smaller faction prefers S2, and R2 is the rank of C6 the only member
of S2-S1.

According to this faction, there are W2=4 members (C1, C2, c3, and C4) of
S1 that have lower rank than C6.

In summary, the larger faction has 60 ballots each with a weight of 2,
while the smaller faction has 40 ballots, each with a weight of 4.  Since
40*4 is greater than 60*2, the PR subset, S2, favored by the smaller
faction wins this head-to-head comparison.

I leave it as an exercise to convince yourself that under this method S2
beats all other subsets (of size five) head-to-head in this example.

Example 2.

This is adapted from Ollie's recent posting under the heading of D'Hondt
Without Lists. There are nine seats to be filled.

There are two factions:

4200 A>B>C>D>E>F>G>H>I>J=K=L=M=N=O
1710 A>B>C>J>K>L>M>N>O>D=E=F=G=H=I

The candidate subsets being compared are S1={ABCDEFGHI} and
S2={ABCDEFGJK}.

The larger faction prefers S1.  The rank R1 is midway between H and I.
According to this faction the candidates of S2 having lower rank are J and
K. So W1=2.

The smaller faction prefers S2.  The rank R2 is midway between J and K.
According to this faction candidates D, E, F, G, H and I are below this
rank. So W2=6.

Since W1*4200 is less than W2*1710, the S2 subset wins the comparison.

It turns out that S2 wins all pairwise comparisons.

As a result, ultimately due to the shared preference of A, B and C, both
factions get superproportional representation.

To be continued with more examples ...




On Wed, 20 Nov 2002, Forest Simmons wrote:

> 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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