Kevin,
MMPO sounds interesting -- even though it fails the Condorcet criterion.
I note that it selects the candidate with the minimum of the maximum
pairwise votes against. Just out of curiousity, have you (or anyone
else, as far as you know) considered the converse, selecting the
candidate with the maximum of the minimum pairwise votes in favor?
--Russ
Kevin Venzke stepjak-at-yahoo.fr |EMlist| wrote:
Hello,
I said recently that I believed I could prove that MMPO (MinMax
Pairwise Opposition, aka Simpson-Kramer) satisfies FBC. This is
my attempt at this.
This is what I intend to show: Suppose that a set of like-minded
voters rank their favorite candidate A insincerely low. And
suppose that another candidate X is the MMPO winner. Then, these
voters have a way of raising A to equal-first which does not
result in the election of some candidate liked less than X.
First, I'll define MMPO, for clarity: The voters rank the
candidates; any preference order is admissible. Let v[a,b]
signify the number of voters ranking A above B. Let o[a]
signify v[x,a] where X is the candidate whose selection maximizes
this value. Then the MMPO winner is the candidate Y who minimizes
o[y].
An important feature is that increasing the value of v[a,b]
can only increase o[b], and thus can only harm B and no other
candidate. Conversely, decreasing v[a,b] can only help B, and
no other candidate.
Back to the situation described above: X wins, which means
o[x] is the minimum.
Now, suppose our like-minded voters all raise A and X to equal
first. This could decrease o[a] and o[x], but not increase it.
This is because A and X are only being raised above other
candidates, so that votes against A and X may only decrease.
For another candidate Y, raising A and X to equal-first can
increase o[y] but not decrease it. This is because Y can
only receive more votes against him, not less.
Therefore, now either o[a] or o[x] is minimum, and the new MMPO
winner is either A or X.
(I welcome any critiques of this demonstration. If it's true
that MMPO satisfies FBC, then we have a method which satisfies
FBC, LNHarm, and three-candidate Participation.)
I investigated also CDTT,RandomCandidate, but it appears to violate
the spirit of FBC. I have supposed recently that, for methods
that only use the pairwise matrix, FBC seems to imply a
criterion which says "Increasing v[a,b] must not increase the
probability that the winner comes from {a,b}."
But suppose there is a majority-strength cycle B>C>D>B, with
another candidate A with no majority-strength wins or losses.
The CDTT is {a,b,c,d}. But if we increase v[a,b] so that A
obtains a majority-strength win over B, now the CDTT is just
{a}, so that the odds that the CDTT,RC winner comes from the
set {a,b} have increased from 50% to 100%.
This leads me to believe that there are cases where you
must vote A>B to get the best outcome, even when B is your
favorite candidate.
Kevin Venzke
----
Election-methods mailing list - see http://electorama.com/em for list info
