[EM] surprising consequence of strong FBC

Alex Small asmall at physics.ucsb.edu
Mon Jan 13 12:45:29 PST 2003

I'm still writing the proof.  I'll have it done in the next week. 
However, I realized a consequence of the result that exactly one method
satisfies strong FBC for 3 candidates:

Recall the idea of a voting machine that takes each voter's preference
order and assigns him an optimal strategy given everybody else's assigned
strategies and some election method.

Suppose that the machine assigns everybody his optimal strategy in
approval voting.  Since there's never a disincentive to approve your
favorite, the only question that the machine needs to answer for each
person is "Should this voter also approve his second favorite?"  One would
think that such a machine would never flunk strong FBC, because a voter
always wants #1 to get a vote.  There may be an incentive to give the
machine an insincere preference order (as Gibbard and Satterthwaite
proved) but it would seem that the incentive can only be to swap #2 and
#3, to prevent a strong #2 from beating #1 (in cases where #3 is weak).

However, any method satisfying strong FBC for 3 candidates must be
equivalent to "Top 2 Voting" (give one vote each to your favorite and
runner-up, zero to your least favorite).  Such a machine will not, in
general, give the same result as top 2 voting.  The reason is the
ambiguity in deciding what strategies to assign to the voters.  It's easy
to say what my optimal strategy is if I know everybody else's.  But what
if the machine must also specify everybody else's strategies?  It has a
conflict of interest.

One other interesting observation:  The method of "top 2 voting" is
equivalent to approval if we allow truncation.  Interesting that a method
that (sort of) satisfies strong FBC would be equivalent to Approval.


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