[EM] The Strong FBC (fwd)

[Forest Simmons]

In what follows you will find (most of) an argument showing that the
Favorite Betrayal Criterion (FBC) is incompatible with neutrality and
other weak assumptions when fully ranked ballots are employed.

By neutrality the ballot set XYZ+ZYX cannot yield either X or Z as winner,
so there are two main cases:

Case I.  The ballot set XYZ+ZYX yields a two or three way tie.
Case II.  The ballot set XYZ+ZYX yields Y as the winner.

In the first case there are three subcases depending on whether the ballot
set UVW+VWU yields U or V as the winner or a two way tie between them.
[The Pareto condition rules out the possibility of W winning, since V is
unanimously preferred over W.]

Let's assume (by way of contradiction) that the FBC is satisfied and that
we are dealing with the first subcase of the first case.

Suppose that your preference is BCA and the only other ballot is ABC.

If you vote sincerely, you get the worst possible outcome A, according to
this subcase.

If you keep your favorite at the top, but switch the other two, then the
voted ballots are ABC+BAC, which by neutrality and Pareto, yields a tie
between A and B.  So the best you can do without betraying your favorite
is a tie between worst and favorite (A and B, respectively).

If you betray favorite and vote CAB, then the ballot set is CAB+ABC, which
according to this subcase yields C as winner.

If your utility for your compromise C is greater than the average of your
utilities for your favorite and worst candidates (B and A), then you pay a
penalty for loyalty to your favorite.  In other words, the FBC is not
satisfied in this case.

Now let us consider the second subcase of case I.

This time suppose that the other ballot is ACB while your sincere
preference order is still BCA.

Since we're still in case I, voting a sincere ballot would yield you
either a tie between A and B or a three way tie between A,B, and C.

If you remain loyal to favorite but switch the other two, the ballot set
becomes ACB+BAC, which according to this second subcase yields your last
choice A as winner.

If you betray favorite by voting CBA, then the ballot set is ACB+CBA,
which according to this subcase yields your compromise C as winner.

Clearly it is to your advantage to betray favorite, so the FBC is not
satisfied in this subcase either.

In the third subcase let's suppose that (as in the first subcase) your
preference order is BCA and the other ballot is of type ABC.

In this third subcase sincere voting (BCA+ABC) would yield a tie between
your first and last choices B and A.

Remaining loyal to favorite you could also vote BAC which together with
the ABC ballot would yield the same tie.

But betraying favorite and voting CBA would yield (since we are still in
case I) either a three way tie or a tie between favorite and compromise.

If you happen to prefer compromise to the average utility of first and
last choice, then your best strategy is to betray favorite, so the FBC is
violated in all three subcases of case I.

In case II we still have the same three subcases.

In the first subcase suppose that your sincere preference order is BCA and
that the only other ballot is of type ABC.  Then sincere voting gets you
A, your last choice.  Staying loyal to favorite but reversing the order of
compromise and last gives ballot set BAC+ABC which yields a tie between
first and last choices, B and A.  Betraying favorite and voting CBA yields
a win for favorite B in this case II setting.

[The third subcase is the same as the first subcase except sincere voting
also yields a tie between first and last choices.]

So the FBC is violated in these two subcases.

We will now consider the remaining subcase, where UVW+VWU yields V as
winner of case II, where XYZ+ZYX yields Y as winner.

We need to know that QPR+RPQ+PQR yields P as winner in this context.

We will call this fact, "Fact 1."

To see the truth of this fact, consider the guy whose preference order is
PQR and the only other ballots are QPR and RPQ. If this guy abstains from
voting at all, then his favorite P wins the election (since we are in case
II).

The FBC implies that he can do just as well by casting a ballot with
his favorite in first place, which would be either PQR or PRQ.  But if
either of these gives the win to P, then so does the other by neutrality,
since QPR+RPQ+PRQ can be obtained from RPQ+QPR+PQR by swapping Q and R.

We also need "Fact 2," that in this context either RPQ+RPQ+PQR or
RQP+RPQ+PQR yields R as winner.

[We will prove this later.]

Now to show that the FBC is violated in this subcase of case II:

Suppose that your sincere preference order is BCA and that the only other
ballots cast are CAB and ABC.

By neutrality voting your sincere order would yield a three way tie.

Maintaining loyalty to favorite would result in the ballot set
(BAC+CAB+ABC). According to the Fact 1, above (with P,Q, and R replaced
with A, B, and C, respectively), your worst candidate A wins.

On the other hand, you can achieve C as an outcome by abandoning favorite
and voting whichever of CAB or CBA works (according to Fact 2).

In summary, we have shown in each case (from an exhaustive set of cases)
that there is an example where loyalty to favorite exacted a penalty.

The basic assumptions were neutrality and Pareto, as well as three
candidates, all fully ranked on each ballot.

Forest




----
For more information about this list (subscribe, unsubscribe, FAQ, etc), 
please see http://www.eskimo.com/~robla/em