[EM] Problem solved (for pure ranked ballot)
Forest W Simmons
fsimmons at pcc.edu
Thu Jan 25 17:54:36 PST 2007
Here's a slightly simpler approach for a slightly weaker result. I show
that (in the case of pure ordinal ballots) you cannot have all three of
Monotonicity, Clone Independence, Pareto, and the Strong FBC.
To be very careful we explicitly list the assumptions:
1. Strictly ranked ordinal ballots.
2. Neutrality (permuting the candidate names on the ballot rankings
permutes their winning probabilities in the same way.)
3. Anonimity (Permuting the ballots does not affect the outcome.)
4. Determinism (Chance is not employed except where required by
symmetry, i.e in the case of true ties.)
5. Monotonicity
6. Majority wins when there are only two candidates. (Actually, this
property is a consequence of properties one through six above. But we
mention it explicitly for convenience.)
7. Pareto (An alternative X that is ranked lower than Y on every ballot
must be have zero probability of winning.)
8. ICC. The probabilities assigned by the method to members of a clone
set must add up to the probability that the method would give to a
single alternative replacing the clone set on each ballot.
9. Strong FBC. To each strategy that allows a faction to achieve a
certain expected level of utility, there is another strategy that
allows that faction to rank favorite over all other candidates without
sacrificing that expected level of utility.
Suppose that we have three voters with ranked preferences of
1 A>B>>C
1 B>>C>A
1 C>>A>B
Their nearest expression in the pure ordinal ballots of our method
would be
1 A>B>C
1 B>C>A
1 C>A>B .
Properties one thru four demand that each alternative be given equal
probability P(A)=P(B)=P(C)= (1/3) .
The A>B>C voter's expectation is below their utility of their
compromise B, while the other two voters' compromises have less utility
than their expected utilities.
A sure win by B would be an improvement for the first two voters.
The first voter can accomplish this unilaterally by burying favorite:
1 B>C>A
1 B>C>A
1 C>A>B
Properties 6, 7, and 8 (applied to the clone set {C, A}) make B the
winner with 100 percent probability under our method with this ballot
set.
Now property 9 says we can do as well without moving A downward from
the top. With that constraint all we can do is switch B and C:
1 A>C>B
1 B>C>A
1 C>A>B
For future reference we will call this ballot set S.
Here the {A,C} clone set has a majority over B, so by property 6, all
of the probability goes to {A,C}. And since there is no symmetry that
would require chance, either A or C gets 100 percent of the probability
by our axiom of determinism.
We know that C is not the one with 100 percent probability because the
FBC guarantees that our result will be no worse than B.
So our method gives the win to A, given this ballot set.
Now here's the tricky part:
We change this last ballot set S into the following one S' by two
different means: (1) raising A over C on the middle ballot. OR
(2) swapping A with C on every ballot and then swapping the first and
last ballot.
1 A>C>B
1 B>A>C
1 C>A>B
Monotonicity ensures that A is still the winner, since all we did was
raise A relative to C on one ballot (at least from the first point of
view).
But from the point of view of the second way of getting from S to S',
neutrality and anonimity require that the win switch from A to C.
This contradiction shows that the stated conditions are incompatible.
Q.E.D.
Forest
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