__Requiring
Transitivity for Social Choice is Irrational!__

Arrow’s requirement of transitivity effectively precludes tie solutions. Why? Because a tie among profiles is intransitive. However, a tie is a perfectly rational solution for a voting outcome. To see this let’s consider 3 alternatives: a, b and c. There are 6 possible voting profiles: abc, acb, bac, bca, cab, cba. Let each profile be identified with a separate candidate: v1, v2,…,v6. Therefore we have

v1=abc

v2=acb

v3=bac

v4=bca

v5=cab

v6=cba

When a voter chooses a profile, the voter is really voting for one of the alternatives v1…v6. It’s easy to come up with a rational social choice function. Let the candidate (v1…v6) which receives the most votes be the winner as in a plurality election. If two or more of the candidates v1…v6 receive the same number of votes and this number is greater than the number of votes for any of the other candidates, there will be a tie for the winner. This is entirely possible in a plurality election and, therefore, rational since plurality elections are very common , and nobody has ever seemed to question their rationality. However, this solution, while rational, is not transitive. The solution might be, for example, a tie between the profiles abc, bca and cab. Let us denote a tie among these 3 profiles like this: {abc, bca, cab}. a is preferred to b and b is preferred to a. Similarly, b is preferred to c and c is preferred. The same is true for a and c. For a transitive solution, if a were preferred to b and b were preferred to c, then a must be preferred to c. Since this is not the case for a tie solution and a tie solution is entirely rational, Arrow’s requirement of transitivity is irrational.

If we don’t require transitivity,
we can find a solution based on pair wise comparisons which Arrow requires in
his Condition 3: Independence of Irrelevant Alternatives. For instance, if we
have aPb, bPc, and cPa for the solutions when the alternatives
are considered 2 at a time, this implies a tie among abc, bca and cab. Why? Because,
if you eliminate one of the candidates, the binary results follow. Since
effectively a third of the population voted aP_{i}bP_{i}c,
a third voted bP_{i}cP_{i}a and a
third voted cP_{i}aP_{i}b, it follows
that two thirds voted for a over b, two thirds voted for b over c and two
thirds voted for c over a which are the binary results. Therefore, a tie would
be a viable and rational social choice if transitivity is not required. The
subscript, i, denotes an individual vote; P without
the subscript denotes a social choice.

Actually, to produce the result aPb, only one more than half the population would have to
vote aP_{i}b, but it is the same as if the
population unanimously voted aP_{i}b.
Therefore, aP_{i}b for each voter represents
an idealized element of the domain which represents all the domain elements
that would result in the social choice aPb. Each
domain element represents a possible combination of all individual votes. In
the same way, the social choice {abc, bca, cab} is a social choice that
represents the ideal domain element for which a third of the population voted
for abc, bca and cab, respectively. Let the number of elements in the tie set be
defined as the order of the tie. It can be easily seen that for any number of
alternatives a, b, c,…, if there is a tie at the nth stage, that tie represents
the idealized domain element for which 1/n of the population voted for the
first element of the tie, 1/n of the population voted for the second element of
the tie etc. where the nth stage corresponds to some possible combination of n
out of the total number of alternatives.

Therefore, if one of the alternatives is eliminated from the tie solution, it is the same as if that element were to be eliminated from all the individual votes of the idealized domain element. Now we can sum up the stage n-1 elements that are similar if there are any. If there is one element that represents a larger share of the vote, then the tie collapses to that one element at stage n-1. If there are no similar elements at stage n-1, then the order of the tie remains the same. If there are 2 or more elements with the same number of votes at stage n-1 and that number is greater then the number for any of the other elements in the set, then the tie collapses to another tie of lesser order.

For example, if the stage 4 solution is a tie: {abcd, abdc, acbd, dabc, cdab} and we eliminate a d, we get: {abc, abc, acb, abc, cab}. We can then combine terms to get {3abc, acb, cab}. Since 3/5 of the (idealized) population voted for abc while only 1/5 voted for acb and 1/5 voted for cab, the tie collapses from order 5 to order 1 and the winner at this stage is abc. This is exactly analogous to the tie solution {abc, bca, cab} collapsing to aPb if c is eliminated.

The line of reasoning is that at stage 2 we abstract from the data to come up with binary pairwise solutions for all combinations of alternatives. We then can project the stage 3 solutions by demanding that they reduce to the corresponding stage 2 solutions. We can project the stage 4 solutions by demanding that they reduce to the corresponding stage 3 solutions etc. If, at any stage, the order of the solution is 1, then that domain element in which the entire population votes unanimously for the winner is the idealized domain element. If the order of the solution is n (there are n elements in the tie), each element represents the idealized vote of 1/n of the population. An idealized domain element always exists. No matter which alternative is eliminated from the tie solution representing the social choice at any stage, the remaining stage n-1 elements represent the individual votes of the population with that alternative crossed out, and so votes can be combined and the order of the tie solution at stage n-1 will either collapse to a lower order or remain the same at stage n. Consistency requires that all binary solutions at stage 2 be the same as the original binary comparisons among all voters and all alternatives.

The preceding argument is basically that a rational solution must include ties and, since the requirement of transitivity precludes ties, transitivity for social choice solutions should be abandoned. Transitivity for individuals, however, is completely rational and must be retained.