[EM] Condorcet Violates Strong FBC

[Alex Small]

Suppose we have an election method M that elects the CW when one exists
and uses some unspecified auxillary procedure when no CW exists.

I will argue that this method cannot satisfy the requirement that no voter
ever has an incentive to rank another candidate above or equal to his
favorite.  By assuming strict preference orders with no indifference
between candidates we eliminate weak FBC from consideration (no incentive
to rank another candidate above your favorite).  I know that earlier I
argued that no monotonic method which always elect the favorite of a
majority can satisfy strong FBC, but not all Condorcet methods are
montonic (e.g. IRV-completed Condorcet).

Begin with 2m+1 voters:

1 A>B>C
m B>C>A
m C>A>B

Here we have a cycle.  If M elects C then the A>B>C voter has an incentive
to switch his vote to B>C>A, in which case B is the CW.  The only way to
satisfy strong FBC while leaving that incentive intact is to say that the
A>B>C voter could cause A or B to win by specifying the preference A>C>B. 
In that case the voter has no reason to betray his favorite because
another strategy can give him an equally good or better outcome.  However,
if the first voter in the example reports A>C>B then C is the Condorcet
winner.  Hence, the method M cannot elect C in this case.

Now suppose that we have the situation

2   A>B>C
m-1 B>C>A
m   C>A>B

Once again, M can't elect C.  One of the voters with the preference A>B>C
could always switch his reported preference to B>C>A.  We then have the
previous situation, which I just argued must elect A or B to satisfy
strong FBC.  Also, if one of those voters switches his preference to
A>C>B, C now beats B (the B vs. C pairwise contest has a margin of 1) and
is the CW.

By induction, if we say that A or B wins in the situation

n      A>B>C
m-n+1  B>C>A
m      C>A>B

then A or B must win in the situation

n+1    A>B>C
m-n    B>C>A
m      C>A>B

Otherwise one of the A>B>C voters would have an incentive to betray his
favorite by switching to B>C>A, creating a situation which elects A or B. 
Also, if the A>B>C voter tries to use an alternate non-favorite-betrayal
strategy, switching to A>C>B, C is then the Condorcet winner, since in
both cases B beats C by a single vote.

So, A or B must win in the situation

m A>B>C
1 B>C>A
m C>A>B

But if A wins then the B>C>A voter has an incentive to betray his favorite
by switching to C>A>B, causing C to become the CW and win.  Also, if that
voter switches to B>A>C then A will beat C and become the CW, since the A
vs. B contest had a margin of one vote.  So B must win.

Now, use the same process, moving voters from C>A>B to B>C>A.  In each
case, B must still win by the logic of the first example.  Otherwise one
of the elect C by switching to C>A>B.  And in each case, the A vs. C
contest has a margin of one vote, so switching to B>A>C will elect A, the
voter's least favorite.

Therefore, B must win in the case

m A>B>C
m B>C>A
1 C>A>B

But in this case, the C>A>B voter can switch his vote to A>B>C and elect
A, so he has an incentive for favorite betrayal.  If he tries to
accomplish the same result without betraying his favorite, by switching to
C>B>A, then B (his least favorite) will be the CW since the A vs. B
contest has a margin of one vote.  So while our arguments above based on
strong FBC require that B win in this case, strong FBC prohibits B from
winning in this case.  We have a contradiction, and hence we cannot
simultaneously elect the CW whenever he exists and still satisfy strong
FBC.

Weaknesses of this argument:

1)  It ignores truncation, an important strategy to consider when
analyzing various Condorcet methods.

2)  It is specific to 3 candidates.  Strong FBC is satisfied if a voter
can obtain the same (superior) result by either ranking somebody else
above his true favorite OR by changing the relative rankings of candidates
lower on his list.  With 3 candidates I only have to consider one possible
insincere ranking of the lower choices.  With 4 candidates I'd have to
consider 5 such rankings.  It isn't clear to me how to invoke the
Condorcet criterion when there is a cycle A>B>C>A and I'm considering a
swtich from A>B>C>D to A>B>D>C (assume all candidates defeat D).

Anyway, that's all for now.



Alex


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