[EM] Cycles in sincere individual preferences and application to vote-collection

[Paul Kislanko]

Evidently examples matter more than logic, so here's one last try in support
of Jobst's argument that pairwise counting methods should have a pairwise
collection method.

 

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.

 

If you ask me to compare B, C or D to E I'd rank E>any. 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.

 

To suggest that you can infer my sincere pairwise preference between any two
alternatives who are not my first choice among many is unwarranted. 

 

On the other hand, if you give me the option to explicitly express my
sincere pairwise preferences, you can directly fill in the pairwise matrix
with my sincere preferences. You can even use my answers to fill out a
"ranked ballot" so long as you allow ties and truncation ( { X, Y, either,
neither} being my choices for each pair).

 

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). 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).

  

In theory, there is no difference between theory and practice; In practice,
there is. - Chuck Reid 

---------------------- 

Paul Kislanko

 

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