8

Following this question on the Gibbard-Satterthwaite (GB) theorem, I was wondering how the Majority Judgment (MJ) voting system fits in.

Quick summary of how the MJ works: you attribute each candidate with a mention. The candidate with the highest median mention wins.

The GB theorem states that, for three or more candidates:

  1. The rule is dictatorial
  2. There is some candidate who can never win, under the rule, or
  3. The rule is susceptible to tactical voting.

The MJ, according to its creators, does not follow this theorem. The first two points are obvious, and for the third, they claim it is system affected the least by strategical voting.

What does the Gibbard-Satterthwaite theorem mean applied to the Majority Judgment voting system?

Improve this question
0
8

The correct answer is: assertion 3.

In all my answer, I will assume that there are 3 voters or more. (With 1 voter, the system is obviously dictatorial. And with 2 voters, the answer to your question depends on the exact tie-breaking rule that is used.)

Remark that a subset consisting of more than half the voters can always choose the winning candidate by giving her the best grade and attributing the worst grade to all the other candidates.

  • This precludes the existence of a dictatorial voter, so assertion 1 is false.
  • This proves that any candidate can be elected, so assertion 2 is false.

Hence (still with the assumption of having 3 voters or more), Gibbard's theorem implies that as soon as there are 3 candidates or more, MJ is susceptible to tactical voting.

But there is worse. For example, consider the following situation. I use grades for clarity, but the example can be immediately translated with appreciations instead.

Voter 1: A 10, B 0.

Voter 2: B 7, A 0.

Voter 3: B 9, A 8.

Candidate A's median: 8. Candidate B's median: 7. So, A wins. But voter 3 can manipulate toward the following situation.

Voter 1: A 10, B 0.

Voter 2: B 7, A 0.

Voter 3: B 9, A 0.

Candidate A's median: 0. Candidate B's median: 7. So candidate B wins.

Conclusion: MJ is manipulable even when there are only 2 candidates!

I never understood why Balinski and Laraki base a large part of their argumentation in favor of MJ on the issue of manipulability. Although MJ does have some interesting features, immunity to manipulation is certainly not one of them.

Improve this answer
4
  • once you go out beyond the trivial numbers aka 100 or more the issue of single voter manipulation goes completely away. – SoylentGray Jan 9 '17 at 19:14
  • 6
    Yes, that's true, like for any "reasonable" voting rule. Cf. for example On asymptotic strategy-proofness of classical social choice rules, Slinko, 2002. However, there is the issue of manipulation by a coalition, i.e. a subset of voters (by the way, it is my main research interest, precisely because of the reason you mention). You can immediately translate my example with 3 groups of 100 voters each, for instance. – François Durand Jan 10 '17 at 8:48
  • 1
    Is that really manipulation though? Or is that combined will of 300 people – SoylentGray Jan 10 '17 at 15:54
  • 3
    What it shows is that in MJ, when you prefer B to A, no ballot defends your opinion best than (B 10, A 0). Ok, no problem, let us say that it is the "standard" ballot when you prefer B to A (whatever your intensity of preference). But with this new frameword, MJ does not meet the Independence of Irrelevant Alternatives (because it should elect the Condorcet winner in any situation, and there does not always exist one). So, MJ, does not escape Arrow's theorem anymore... And while it is now not manipulable for 2 candidates, it still is for 3 candidates or more, because of Gibbard's theorem. – François Durand Jan 11 '17 at 9:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .