Independence of irrelevant alternatives

A criterion weaker than IIA proposed by H. Peyton Young and A. Levenglick is called local independence from irrelevant alternatives (LIIA).[4] LIIA requires that both of the following conditions always hold:

• If the option that finished in last place is deleted from all the votes, then the order of finish of the remaining options must not change. (The winner must not change.)
• If the winning option is deleted from all the votes, the order of finish of the remaining options must not change. (The option that finished in second place must become the winner.)

An equivalent way to express LIIA is that if a subset of the options are in consecutive positions in the order of finish, then their relative order of finish must not change if all other options are deleted from the votes. For example, if all options except those in 3rd, 4th and 5th place are deleted, the option that finished 3rd must win, the 4th must finish second, and 5th must finish 3rd.

Another equivalent way to express LIIA is that if two options are consecutive in the order of finish, the one that finished higher must win if all options except those two are deleted from the votes.

LIIA is weaker than IIA because satisfaction of IIA implies satisfaction of LIIA, but not vice versa.

Despite being a weaker criterion (i.e. easier to satisfy) than IIA, LIIA is satisfied by very few voting methods. These include Kemeny-Young and ranked pairs, but not Schulze. Just as with IIA, LIIA compliance for rating methods such as approval voting, range voting, and majority judgment require the assumption that voters rate each alternative individually and independently of knowing any other alternatives, on an absolute scale (calibrated prior to the election), even when this assumption implies that voters having meaningful preferences in a two candidate election will necessarily abstain.

IIA is largely incompatible with the majority criterion unless there are only two alternatives.

Consider a scenario in which there are three candidates A, B, & C, and the voters' preferences are as follows:

25% of the voters prefer A over B, and B over C. (A>B>C)

40% of the voters prefer B over C, and C over A. (B>C>A)

35% of the voters prefer C over A, and A over B. (C>A>B)

(These are preferences, not votes, and thus are independent of the voting method.)

75% prefer C over A, 65% prefer B over C, and 60% prefer A over B. The presence of this societal intransitivity is the voting paradox. Regardless of the voting method and the actual votes, there are only three cases to consider:

• Case 1: A is elected. IIA is violated because the 75% who prefer C over A would elect C if B were not a candidate.
• Case 2: B is elected. IIA is violated because the 60% who prefer A over B would elect A if C were not a candidate.
• Case 3: C is elected. IIA is violated because the 65% who prefer B over C would elect B if A were not a candidate.

To show failure, it is only assumed at least possible that enough voters in the majority might cast a minimally positive vote for their preferred candidate when there are only two candidates, rather than abstain. Most ranked ballot methods and Plurality voting satisfy the Majority Criterion, and therefore fail IIA automatically by the example above. Meanwhile, passage of IIA by Approval and Range voting requires in certain cases that voters in the majority are necessarily excluded from voting (they are assumed to necessarily abstain in a two candidate race, despite having a meaningful preference between the alternatives).

So even if IIA is desirable, requiring its satisfaction seems to allow only voting methods that are undesirable in some other way, such as treating one of the voters as a dictator. Thus the goal must be to find which voting methods are best, rather than which are perfect.

An argument can be made that IIA is itself undesirable. IIA assumes that when deciding whether A is likely to be better than B, information about voters' preferences regarding C is irrelevant and should not make a difference. However, the heuristic that leads to majority rule when there are only two options is that the larger the number of people who think one option is better than the other, the greater the likelihood that it is better, all else being equal (see Condorcet's Jury Theorem). A majority is more likely than the opposing minority to be right about which of the two candidates is better, all else being equal, hence the use of majority rule.

The same heuristic implies that the larger the majority, the more likely it is that they are right. It would seem to also imply that when there is more than one majority, larger majorities are more likely to be right than smaller majorities. Assuming this is so, the 75% who prefer C over A and the 65% who prefer B over C are more likely to be right than the 60% who prefer A over B, and since it is not possible for all three majorities to be right, the smaller majority (who prefer A over B) are more likely to be wrong, and less likely than their opposing minority to be right. Rather than being irrelevant to whether A is better than B, the additional information about the voters' preferences regarding C provides a strong hint that this is a situation where all else is not equal.

From Kenneth Arrow,[5] each "voter" i in the society has an ordering R_i that ranks the (conceivable) objects of social choice—x, y, and z in simplest case—from high to low. An aggregation rule (voting rule) in turn maps each profile or tuple (R₁, ...,R_n) of voter preferences (orderings) to a social ordering R that determines the social preference (ranking) of x, y, and z.

Arrow's IIA requires that whenever a pair of alternatives is ranked the same way in two preference profiles (over the same choice set), then the aggregation rule must order these alternatives identically across the two profiles.[6] For example, suppose an aggregation rule ranks a above b at the profile given by

• (acbd, dbac),

(i.e., the first individual prefers a first, c second, b third, d last; the second individual prefers d first, ..., and c last). Then, if it satisfies IIA, it must rank a above b at the following three profiles:

• (abcd, bdca)
• (abcd, bacd)
• (acdb, bcda).

The last two forms of profiles (placing the two at the top; and placing the two at the top and bottom) are especially useful in the proofs of theorems involving IIA.

Arrow's IIA does not imply an IIA similar to those different from this at the top of this article nor conversely.[7]

In the first edition of his book, Arrow misinterpreted IIA by considering the removal of a choice from the consideration set. Among the objects of choice, he distinguished those that by hypothesis are specified as feasible and infeasible. Consider two possible sets of voter orderings (${\displaystyle R_{1}}$, ...,${\displaystyle R_{n}}$ ) and (${\displaystyle R_{1}'}$, ...,${\displaystyle R_{n}'}$) such that the ranking of X and Y for each voter i is the same for ${\displaystyle R_{i}}$ and ${\displaystyle R_{i}'}$. The voting rule generates corresponding social orderings R and R'. Now suppose that X and Y are feasible but Z is infeasible (say, the candidate is not on the ballot or the social state is outside the production possibility curve). Arrow required that the voting rule that R and R' select the same (top-ranked) social choice from the feasible set (X, Y), and that this requirement holds no matter what the ranking is of infeasible Z relative to X and Y in the two sets of orderings. IIA does not allow "removing" an alternative from the available set (a candidate from the ballot), and it says nothing about what would happen in such a case: all options are assumed to be "feasible."

In a Borda count election, 5 voters rank 5 alternatives [A, B, C, D, E].

3 voters rank [A>B>C>D>E]. 1 voter ranks [C>D>E>B>A]. 1 voter ranks [E>C>D>B>A].

Borda count (a=0, b=1): C=13, A=12, B=11, D=8, E=6. C wins.

Now, the voter who ranks [C>D>E>B>A] instead ranks [C>B>E>D>A]; and the voter who ranks [E>C>D>B>A] instead ranks [E>C>B>D>A]. They change their preferences only over the pairs [B, D], [B, E] and [D, E].

The new Borda count: B=14, C=13, A=12, E=6, D=5. B wins.

The social choice has changed the ranking of [B, A] and [B, C]. The changes in the social choice ranking are dependent on irrelevant changes in the preference profile. In particular, B now wins instead of C, even though no voter changed their preference over [B, C].

Consider an election in which there are three candidates, A, B, and C, and only two voters. Each voter ranks the candidates in order of preference. The highest ranked candidate in a voter's preference is given 2 points, the second highest 1, and the lowest ranked 0; the overall ranking of a candidate is determined by the total score it gets; the highest ranked candidate wins.

Considering two profiles:

• In profiles 1 and 2, the first voter casts his votes in the order BAC, so B receives 2 points, A receives 1, and C receives 0 from this voter.
• In profile 1, the second voter votes ACB, so A will win outright (the total scores: A 3, B 2, C 1).
• In profile 2, the second voter votes ABC, so A and B will tie (the total scores: A 3, B 3, C 0).

Thus, if the second voter wishes A to be elected, he had better vote ACB regardless of his actual opinion of C and B. This violates the idea of "independence from irrelevant alternatives" because the voter's comparative opinion of C and B affects whether A is elected or not. In both profiles, the rankings of A relative to B are the same for each voter, but the social rankings of A relative to B are different.

This example shows that Copeland's method violates IIA. Assume four candidates A, B, C and D with 6 voters with the following preferences:

The results would be tabulated as follows:

• [X] indicates voters who preferred the candidate in the column caption to the one in the row caption
• [Y] indicates voters who preferred the candidate in the row caption to the one in the column caption

Result: A has two wins and one defeat, while no other candidate has more wins than defeats. Thus, A is elected Copeland winner.

Now, assume all voters would raise D over B and C without changing the order of A and D. The preferences of the voters would now be:

The results would be tabulated as follows:

Result: D wins against all three opponents. Thus, D is elected Copeland winner.

The voters changed only their preference orders over B, C and D. As a result, the outcome order of D and A changed. A turned from winner to loser without any change of the voters' preferences regarding A. Thus, Copeland's method fails the IIA criterion.

In an instant-runoff election, 5 voters rank 3 alternatives [A, B, C].

2 voters rank [A>B>C]. 2 voters rank [C>B>A]. 1 voter ranks [B>A>C].

Round 1: A=2, B=1, C=2; B eliminated. Round 2: A=3, C=2; A wins.

Now, the two voters who rank [C>B>A] instead rank [B>C>A]. They change only their preferences over B and C.

Round 1: A=2, B=3, C=0; B wins with a majority of the vote.

The social choice ranking of [A, B] is dependent on preferences over the irrelevant alternatives [B, C].

This example shows that the Kemeny–Young method violates the IIA criterion. Assume three candidates A, B and C with 7 voters and the following preferences:

The Kemeny–Young method arranges the pairwise comparison counts in the following tally table:

The ranking scores of all possible rankings are:

Result: The ranking A > B > C has the highest ranking score. Thus, A wins ahead of B and C.

Now, assume the two voters (marked bold) with preferences B > C > A would change their preferences over the pair B and C. The preferences of the voters would then be in total:

The Kemeny–Young method arranges the pairwise comparison counts in the following tally table:

The ranking scores of all possible rankings are:

Result: The ranking C > A > B has the highest ranking score. Thus, C wins ahead of A and B.

The two voters changed only their preferences over B and C, but this resulted in a change of the order of A and C in the result, turning A from winner to loser without any change of the voters' preferences regarding A. Thus, the Kemeny-Young method fails the IIA criterion.

This example shows that the Minimax method violates the IIA criterion. Assume four candidates A, B and C and 13 voters with the following preferences:

Since all preferences are strict rankings (no equals are present), all three Minimax methods (winning votes, margins and pairwise opposite) elect the same winners.

The results would be tabulated as follows:

• [X] indicates voters who preferred the candidate in the column caption to the one in the row caption
• [Y] indicates voters who preferred the candidate in the row caption to the one in the column caption

Result: A has the closest biggest defeat. Thus, A is elected Minimax winner.

Now, assume the two voters (marked bold) with preferences B > A > C change the preferences over the pair A and C. The preferences of the voters would then be in total:

The results would be tabulated as follows:

Result: Now, B has the closest biggest defeat. Thus, B is elected Minimax winner.

So, by changing the order of A and C in the preferences of some voters, the order of A and B in the result changed. B is turned from loser to winner without any change of the voters' preferences regarding B. Thus, the Minimax method fails the IIA criterion.

In a plurality voting system 7 voters rank 3 alternatives (A, B, C).

• 3 voters rank (A>B>C)
• 2 voters rank (B>A>C)
• 2 voters rank (C>B>A)

In an election, initially only A and B run: B wins with 4 votes to A's 3, but the entry of C into the race makes A the new winner.

The relative positions of A and B are reversed by the introduction of C, an "irrelevant" alternative.

This example shows that the Ranked pairs method violates the IIA criterion. Assume three candidates A, B and C and 7 voters with the following preferences:

The results would be tabulated as follows:

The sorted list of victories would be:

Result: A > B and B > C are locked in (and C > A cannot be locked in after that), so the full ranking is A > B > C. Thus, A is elected Ranked pairs winner.

Now, assume the two voters (marked bold) with preferences B > C > A change their preferences over the pair B and C. The preferences of the voters would then be in total:

The results would be tabulated as follows:

The sorted list of victories would be:

Result: All three duels are locked in, so the full ranking is C > A > B. Thus, the Condorcet winner C is elected Ranked pairs winner.

So, by changing their preferences over B and C, the two voters changed the order of A and C in the result, turning A from winner to loser without any change of the voters' preferences regarding A. Thus, the Ranked pairs method fails the IIA criterion.

This example shows that the Schulze method violates the IIA criterion. Assume four candidates A, B, C and D and 12 voters with the following preferences:

The pairwise preferences would be tabulated as follows:

Now, the strongest paths have to be identified, e.g. the path D > A > B is stronger than the direct path D > B (which is nullified, since it is a tie).

Result: The full ranking is C > D > A > B. Thus, C is elected Schulze winner and D is preferred over A.

Now, assume the two voters (marked bold) with preferences C > B > D > A change their preferences over the pair B and C. The preferences of the voters would then be in total:

Hence, the pairwise preferences would be tabulated as follows:

Now, the strongest paths have to be identified:

Result: Now, the full ranking is A > B > C > D. Thus, A is elected Schulze winner and is preferred over D.

So, by changing their preferences over B and C, the two voters changed the order of A and D in the result, turning A from loser to winner without any change of the voters' preferences regarding A. Thus, the Schulze method fails the IIA criterion.

A probable example of the two-round system failing this criterion was the 2002 French presidential election. Polls leading up to the election have suggested a runoff between centre-right candidate Jacques Chirac and centre-left candidate Lionel Jospin, in which Jospin has been expected to win. However, the first round was contested by an unprecedented 16 candidates, including left-wing candidates who intended to support Jospin in the runoff, eventually resulting in the far-right candidate, Jean-Marie Le Pen, finishing second and entering the runoff instead of Jospin, which Chirac won by a large margin. Thus, the presence of many candidates who did not intend to win in the election changed which of the candidates won.