Intuitive argument Condorcet's example is already enough to see the impossibility of a fair
ranked voting system, given stronger conditions for fairness than Arrow's theorem assumes. Suppose we have three candidates (A, B, and C) and three voters whose preferences are as follows: If C is chosen as the winner, it can be argued any fair voting system would say B should win instead, since two voters (1 and 2) prefer B to C and only one voter (3) prefers C to B. However, by the same argument A is preferred to B, and C is preferred to A, by a margin of two to one on each occasion. Thus, even though each individual voter has consistent preferences, the preferences of society are contradictory: A is preferred over B which is preferred over C which is preferred over A. Because of this example, some authors credit
Condorcet with having given an intuitive argument that presents the core of Arrow's theorem. ;
Pareto efficiency : If alternative \mathbf{a} is preferred to \mathbf{b} for all orderings R_1, \ldots, R_N, then \mathbf{a} is preferred to \mathbf{b} by F(R_1, R_2, \ldots, R_N). and
Ariel Rubinstein. The simplified proof uses an additional concept: • A coalition is
weakly decisive over (x, y) if and only if when every voter i in the coalition ranks x \succ_i y,
and every voter j outside the coalition ranks y \succ_j x, then x \succ y. Thenceforth assume that the social choice system satisfies unrestricted domain, Pareto efficiency, and IIA. Also assume that there are at least 3 distinct outcomes. {{Math proof|proof=Let z be an outcome distinct from x, y. Claim: G is decisive over (x, z). Let everyone in G vote x over z. By IIA, changing the votes on y does not matter for x, z. So change the votes such that x \succ_i y \succ_i z in G and y \succ_i x and y \succ_i z outside of G. By Pareto, y \succ z. By coalition weak-decisiveness over (x, y), x \succ y. Thus x \succ z. \square Similarly, G is decisive over (z, y). By iterating the above two claims (note that decisiveness implies weak-decisiveness), we find that G is decisive over all ordered pairs in \{x, y, z\}. Then iterating that, we find that G is decisive over all ordered pairs in X.}} {{Math proof|proof=Let G be a coalition with size \geq 2. Partition the coalition into nonempty subsets G_1, G_2. Fix distinct x, y, z. Design the following voting pattern (notice that it is the cyclic voting pattern which causes the Condorcet paradox): \begin{align} \text{voters in } G_1&: x \succ_i y \succ_i z \\ \text{voters in } G_2&: z \succ_i x \succ_i y \\ \text{voters outside } G&: y \succ_i z \succ_i x \end{align} (Items other than x, y, z are not relevant.) Since G is decisive, we have x \succ y. So at least one is true: x \succ z or z \succ y. If x \succ z, then G_1 is weakly decisive over (x, z). If z \succ y, then G_2 is weakly decisive over (z, y). Now apply the field expansion lemma.}} By Pareto, the entire set of voters is decisive. Thus by the group contraction lemma, there is a size-one decisive coalition—a dictator. Proofs using the concept of the
pivotal voter originated from Salvador Barberá in 1980. The proof given here is a simplified version based on two proofs published in
Economic Theory.
Setup Assume there are
n voters. We assign all of these voters an arbitrary ID number, ranging from
1 through
n, which we can use to keep track of each voter's identity as we consider what happens when they change their votes.
Without loss of generality, we can say there are three candidates who we call
A,
B, and
C. (Because of IIA, including more than 3 candidates does not affect the proof.) We will prove that any social choice rule respecting unanimity and independence of irrelevant alternatives (IIA) is a dictatorship. The proof is in three parts: • We identify a
pivotal voter for each individual contest (
A vs.
B,
B vs.
C, and
A vs.
C). Their ballot swings the societal outcome. • We prove this voter is a
partial dictator. In other words, they get to decide whether A or B is ranked higher in the outcome. • We prove this voter is the same person, hence this voter is a
dictator.
Part one: There is a pivotal voter for A vs. B Consider the situation where everyone prefers
A to
B, and everyone also prefers
C to
B. By unanimity, society must also prefer both
A and
C to
B. Call this situation
profile[0, x]. On the other hand, if everyone preferred
B to everything else, then society would have to prefer
B to everything else by unanimity. Now arrange all the voters in some arbitrary but fixed order, and for each
i let
profile i be the same as
profile 0, but move
B to the top of the ballots for voters 1 through
i. So
profile 1 has
B at the top of the ballot for voter 1, but not for any of the others.
Profile 2 has
B at the top for voters 1 and 2, but no others, and so on. Since
B eventually moves to the top of the societal preference as the profile number increases, there must be some profile, number
k, for which
B first moves
above A in the societal rank. We call the voter
k whose ballot change causes this to happen the
pivotal voter for B over A. Note that the pivotal voter for
B over
A is not,
a priori, the same as the pivotal voter for
A over
B. In part three of the proof we will show that these do turn out to be the same. Also note that by IIA the same argument applies if
profile 0 is any profile in which
A is ranked above
B by every voter, and the pivotal voter for
B over
A will still be voter
k. We will use this observation below.
Part two: The pivotal voter for B over A is a dictator for B over C In this part of the argument we refer to voter
k, the pivotal voter for
B over
A, as the
pivotal voter for simplicity. We will show that the pivotal voter dictates society's decision for
B over
C. That is, we show that no matter how the rest of society votes, if
pivotal voter ranks
B over
C, then that is the societal outcome. Note again that the dictator for
B over
C is not a priori the same as that for
C over
B. In part three of the proof we will see that these turn out to be the same too. In the following, we call voters 1 through
k − 1,
segment one, and voters
k + 1 through
N,
segment two. To begin, suppose that the ballots are as follows: • Every voter in segment one ranks
B above
C and
C above
A. • Pivotal voter ranks
A above
B and
B above
C. • Every voter in segment two ranks
A above
B and
B above
C. Then by the argument in part one (and the last observation in that part), the societal outcome must rank
A above
B. This is because, except for a repositioning of
C, this profile is the same as
profile k − 1 from part one. Furthermore, by unanimity the societal outcome must rank
B above
C. Therefore, we know the outcome in this case completely. Now suppose that pivotal voter moves
B above
A, but keeps
C in the same position and imagine that any number (even all!) of the other voters change their ballots to move
B below
C, without changing the position of
A. Then aside from a repositioning of
C this is the same as
profile k from part one and hence the societal outcome ranks
B above
A. Furthermore, by IIA the societal outcome must rank
A above
C, as in the previous case. In particular, the societal outcome ranks
B above
C, even though Pivotal Voter may have been the
only voter to rank
B above
C.
By IIA, this conclusion holds independently of how
A is positioned on the ballots, so pivotal voter is a dictator for
B over
C.
Part three: There exists a dictator In this part of the argument we refer back to the original ordering of voters, and compare the positions of the different pivotal voters (identified by applying parts one and two to the other pairs of candidates). First, the pivotal voter for
B over
C must appear earlier (or at the same position) in the line than the dictator for
B over
C: As we consider the argument of part one applied to
B and
C, successively moving
B to the top of voters' ballots, the pivot point where society ranks
B above
C must come at or before we reach the dictator for
B over
C. Likewise, reversing the roles of
B and
C, the pivotal voter for
C over
B must be at or later in line than the dictator for
B over
C. In short, if
kX/Y denotes the position of the pivotal voter for
X over
Y (for any two candidates
X and
Y), then we have shown :
kB/C ≤ kB/A ≤
kC/B. Now repeating the entire argument above with
B and
C switched, we also have :
kC/B ≤
kB/C. Therefore, we have :
kB/C = kB/A =
kC/B and the same argument for other pairs shows that all the pivotal voters (and hence all the dictators) occur at the same position in the list of voters. This voter is the dictator for the whole election.
Stronger versions Arrow's impossibility theorem still holds if Pareto efficiency is weakened to the following condition: ; Non-imposition : For any two alternatives
a and
b, there exists some preference profile such that is preferred to by . == Interpretation and practical solutions ==