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Paul Kislanko wrote:<br><br>
<blockquote type=cite class=cite cite><font face="arial" size=2>Suppose I
were a staunch pro-life believer, so anti-abortion is my most important
criterion. There are 5 candidates in the race, and A & E are both
anti-abortion, but have opposite views on gun control (A for, E against)
and capital punishment (A against, E for). B, C, and D are all
pro-choice, and either pro gun control or anti-capital punishment or
both. When asked to rank all 5 I give A>B>C>D>E.<br>
</font><br>
<font face="arial" size=2>If you ask me to compare B, C or D to E I d
rank E>any.</font></blockquote><br>
Then... why on earth would you rank E behind them all? That runs
contrary to all three pairwise preferences you purport to have.<br><br>
<blockquote type=cite class=cite cite><font face="arial" size=2> If
you ask me to compare B, C or D pairwise to each other, the abortion
issue isn t a factor, and my sincere preference might be D>either B or
C because of fiscal policy and a virtual tie on the other pro-life
issues.</font></blockquote><br>
Then... why do you rank D behind B and C? Your preferences appear
to be completely transitive as A>E>D>B?C.<br><br>
<blockquote type=cite class=cite cite><font face="arial" size=2>To
suggest that you can infer my sincere pairwise preference between any two
alternatives who are not my first choice among many is unwarranted.
</font></blockquote><br>
Only because you appear to have picked your ranked order arbitrarily,
aside from the first choice.<br><br>
<blockquote type=cite class=cite cite><font face="arial" size=2>To prove
that construction of a pairwise matrix from ranked ballots is always
possible, I think you d need to show inductively that all orderings by
any voter of N candidates will always be the same for those as those
obtained by asking each voter to order N+1 candidates (with respect to
the N original candidates). </font></blockquote><br>
I agree with all of that, more or less.<br><br>
<blockquote type=cite class=cite cite><font face="arial" size=2>I believe
a logical consequence of such a proof would be contrary to Arrow s
theorem, and therefore is impossible. (Just substitute issue for
individual and ranked ballot for group and the same logic
applies).</font></blockquote><br>
(I think you mean, substitute issue AND individual WITH ranked ballot AND
group?)<br><br>
And therein lies my objection. I don't think you can simply
substitute individual for group. A group can have cyclic
preferences, and on that fact rests Condorcet's paradox and Arrow's
theorem.<br><br>
But I do NOT believe that an individual can have such preferences.
Or, more accurately, an individual may have such preferences, but I do
not consider them logical, and I have absolutely no interest in factoring
such preferences into a social choice algorithm.<br><br>
I guess this makes me a "transitive preference elitist" of
sorts. I'm comfortable with that.<br><br>
-Adam</html>