[EM] MMPO satisfies FBC in the general case
Kevin Venzke
stepjak at yahoo.fr
Sun May 22 18:19:43 PDT 2005
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
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