[EM] Condorcet Flavored PR Methods

[Forest Simmons]

Here's the next installment on the above topic (Condorcet Flavored PR
Methods).

Suppose we are making an head-to-head comparison of candidate subsets A
and B.  What numbers should we enter into the (A,B) and (B,A) positions
of the pairwise matrix associated with the ordinal ballot

              C1>C2>C3>C4>...>C19>C20 ?

Of course, the answer depends on how the sets A and B are distributed
among the C's.

But, as we saw in our last installment, even knowing which of the C's are
in which circles of candidates, we still have many possible ways of
answering that question.

Here's my best effort along these lines so far:

Let m1 and m2 be the respective medians of the sets (A-B) and (B-A). Let M
and m be max(m1,m2) and min(m1,m2).  In other words M is the larger of the
two medians, and m is the smaller.

Let kA and kB be the respective numbers of members of A and B ranked above
M, and let jA and jB be the respective number of members of A and B ranked
below m.

When M or m falls right on a member of A or B (rather than in the space
between two candidates), then such a member adds 1/2 to the count of kA,
kB, jA, or jB, depending on which set (A or B) and which median (M or m).

The (A,B) entry in the pairwise matrix for this ballot is
     1+1/2+...+1/(kA+jB),
and the (B,A) entry is
     1+1/2+...+ 1/(kB+jA).

Note that reversing the preference order transposes the pairwise matrix,
so that if Ranked Pairs or some other Condorcet Method satisfying Reverse
Symmetry is applied to the pairwise matrix, the Condorcet Winner and
Condorcet Loser are interchanged, i.e. "best" and "worst" candidate
subsets are interchanged.

How many other PR methods do you know that satisfy this property?

How do we know that this method gives Proportional Representation?

Heuristically, the bigger kA+jB, the more satisfaction the voter feels if
subset A wins over subset B, i.e. there are kA+jB "layers of
satisfaction."

The first layer counts fully, the next 1/2, the next 1/3, etc. just as in
PAV (Proportional Approval Voting).

In fact, in the case where every member of A is preferred to every member
of B, the (A,B) entry is identical to the PAV value for the set A.

These boundary cases are the easiest to check for proportionality.  Since
PAV satisfies PR, these boundary cases do also.

Another instructive case: suppose that we have

  C1>C2>C3>...C20,

and C3 and C10 are the only members of A-B and B-A, respectively.

Then M=C3 and m=C10, and kA=2.5, kB=2, jA=10, and jB=10.5, so the
(A,B) and (B,A) entries in the pairwise matrix for this ballot are
respectively

1+1/2+...+1/13  and  1+1/2+1/3+...+1/12 .

The margin of win for A over B is the difference 1/13.

>From this example it is easy to see that whenever the symmetric difference
has only two members, the margin magnitude is the reciprocal of one more
than the number of candidates who are not between those two members.

In other words, in this case the magnitude of the margin depends only on
how far apart in the ordering the two oddball candidates are and on how
many distinct candidates there are in the union of the two subsets.

So if the two oddballs were M=C8 and m=C15 instead of C3 and C10, the
magnitude of the margin would still be 1/13.

To be continued ...

Forest

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