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> Gibbard’s theorem assumes that each participant has a set of preferences about the entire outcome of the process. In this case, the outcome of the process is

> Gibbard’s theorem assumes that each participant has a set of preferences about the entire outcome of the process. In this case, the outcome of the process is the matching of all candidates to all positions. Doesn't this mean it's actually not very applicable to voting systems? Voters generally only care whether their preferred candidate comes first, not about the relative ordering of the other candidates. I suppose a voter might care how close their preferred candidate came to winning (or their margin of victory), but that's still a slightly weaker requirement than having a total ordering of all candidates relative to each other.

bo1024 5y ago
There are versions of Gibbard's theorem for outputting a single winner too. The famous one is Gibbard-Satterthwaite: https://en.wikipedia.org/wiki/Gibbard%E2%80%93Satterthwaite_theorem https://en.wikipedia.org/wiki/Gibbard%E2%80%93Satterthwaite_...
svcphr 5y ago
You can have weak inequalities in the preferences over alternatives -- e.g. you're indifferent between all rankings where your preferred candidate is number 1
cdsmith 5y ago
Author here. No, it doesn't mean that. If the election is held to determine who wins, then the outcome is who wins, and everyone has opinions about that same outcome. Contrast this to the matching problem, where the outcome is where ALL the candidates are assigned. You're allowed to have an opinion about where you are assigned. I'm allowed to have an opinion about where I am assigned. These opinions might conflict indirectly because of limited spots, but they don't conflict directly. This is very different from the election case, where if you and I have different opinions about who we want to win, those opinions are always contradictory.
- dane-pgp 5y ago
  What if an election is held to determine a set of two "finalists"? People might disagree about who the top two candidates are, but they might be satisfied as long as one of their top preferences makes it through to the second round. Obviously once there are two candidates remaining, Gibbard's Theorem doesn't apply, so I'm guessing that any procedure which reduces the set to two outcomes must itself be subject to strategy, but it would be interesting if allowing people to cast a separate vote in a run-off election was enough to make the first round no longer require strategy.
  - cdsmith 5y ago
    If the result of the election is the choice of two finalists, then as long as there are more than two candidates, there are more than two possible outcomes. For example, if the candidates are A, B, and C, then the possible outcomes are: (A and B), (A and C), or (B and C), so 3 in all. In general, for n candidates, there are n * (n - 1) / 2 outcomes. By Gibbard's Theorem, then, either there is a dictator, or there is strategy. Another way to see this: the entire process of choosing two finalists and then having a runoff to choose an ultimate winner counts as a collective decision-making process. That there are two separate votes doesn't actually matter. By Gibbard's theorem, then, if there isn't a dictator, then the entire process is strategic. Since the chance to use strategy doesn't occur in the simple-majority final runoff, by process of elimination, it must occur in choosing the candidates who qualify for the runoff.

bo1024 5y ago

There are versions of Gibbard's theorem for outputting a single winner too. The famous one is Gibbard-Satterthwaite: https://en.wikipedia.org/wiki/Gibbard%E2%80%93Satterthwaite_theorem https://en.wikipedia.org/wiki/Gibbard%E2%80%93Satterthwaite_...

svcphr 5y ago

You can have weak inequalities in the preferences over alternatives -- e.g. you're indifferent between all rankings where your preferred candidate is number 1

cdsmith 5y ago

Author here. No, it doesn't mean that. If the election is held to determine who wins, then the outcome is who wins, and everyone has opinions about that same outcome. Contrast this to the matching problem, where the outcome is where ALL the candidates are assigned. You're allowed to have an opinion about where you are assigned. I'm allowed to have an opinion about where I am assigned. These opinions might conflict indirectly because of limited spots, but they don't conflict directly. This is very different from the election case, where if you and I have different opinions about who we want to win, those opinions are always contradictory.

dane-pgp 5y ago
What if an election is held to determine a set of two "finalists"? People might disagree about who the top two candidates are, but they might be satisfied as long as one of their top preferences makes it through to the second round. Obviously once there are two candidates remaining, Gibbard's Theorem doesn't apply, so I'm guessing that any procedure which reduces the set to two outcomes must itself be subject to strategy, but it would be interesting if allowing people to cast a separate vote in a run-off election was enough to make the first round no longer require strategy.
- cdsmith 5y ago
  If the result of the election is the choice of two finalists, then as long as there are more than two candidates, there are more than two possible outcomes. For example, if the candidates are A, B, and C, then the possible outcomes are: (A and B), (A and C), or (B and C), so 3 in all. In general, for n candidates, there are n * (n - 1) / 2 outcomes. By Gibbard's Theorem, then, either there is a dictator, or there is strategy. Another way to see this: the entire process of choosing two finalists and then having a runoff to choose an ultimate winner counts as a collective decision-making process. That there are two separate votes doesn't actually matter. By Gibbard's theorem, then, if there isn't a dictator, then the entire process is strategic. Since the chance to use strategy doesn't occur in the simple-majority final runoff, by process of elimination, it must occur in choosing the candidates who qualify for the runoff.