Kevin,
good work!
On Thu, Oct 10, 2013 at 9:23 AM, Kevin Venzke step...@yahoo.fr wrote:
Hi Forest,
...
Unfortunately, I realized that an SFC problem is possibly egregious:
51 AB
49 CB
B would win easily, contrary to SFC (which disallows both B and C). But
more alarmingly it's a majority favorite problem.
So it is non-majoritarian in the same sense that Approval is. In this case
the count is too close for approval voters to drop their second
preferences, so B will be the Approval winner. Of course with perfect
information, they would bullet, and A would win. Philosophically, in this
situation I sympathize with electing the candidate broader support (the
consensus candidate) over the mere majority favorite, which is why
Approval's failure of (one version of) the Majority Criterion has never
bothered me.
Jobst and I have gone to a lot of trouble to contrive methods that make B
the game theoretic winner in the face of such preferences. I'm sure you
remember his challenge to find a method that makes B the perfect
information game theoretic winner when utilities are given by (say)
60 A(100), B(70)
40 C(100), B(50)
It seems that only lottery methods can solve this challenge in a
satisfactory way. We co-authored a paper with the double entendre title of
Some Chances for Consensus on this topic for the benefit of people who
take the tyranny of the majority problem seriously.
In sum, my non-majoritarian leanings allow me to bid adieu to SFC without
too much regret.
Now here is a proof of the fact that you observed ... that not all
candidates can have greater MPO than IA:
Suppose to the contrary that (for some ballot set) every candidate has a
greater MPO than IA.
Let C(1), C(2), C(3), ... be a sequence of candidates such that C(n+1)
gives max opposition to C(n).
Since the MPO for C(n) is greater than the IA for C(n), and the IA for
C(n+1) is at least as great as the opposition of C(n+1) to C(n), we can
conclude that the IA for C(n+1) is greater than the IA for C(n).
Therefore,IA increases along the sequence C(1), C(2), C(3), ... so the
sequence cannot cycle. therefore there must be an infinite number of
candidates.
This impossibility shows that the existence of the posited ballot set.was a
false assumption.
In light of this fact I propose the following variation on our method:
1. Eliminate all candidates that have higher MPO than IA.
2. Elect the remaining candidate with the greatest difference between its
IA and its MPO.
I like differences better than ratios in this context, but I used ratios in
IA/MPO because I worried about people who couldn't easily agree that (25 -
30) (72 - 90) , for example. But now that we know eliminating all of
the negative differences is possible without eliminating all of the
candidates, let's switch to differences.
Forest
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