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Condorcet is definitely an important property to have. But you also need a solution when there isn’t an outright Condorcet winner, and there is no good solution

Condorcet is definitely an important property to have. But you also need a solution when there isn’t an outright Condorcet winner, and there is no good solution to that. The tie-breaking mechanism of every Condorcet method is basically impossible to explain in simple terms, ranked pairs included. I often wonder if we should just use Condorcet and if there isn’t a clear winner, declare the result a tie. (That would likely be vulnerable to a kind of DoS attack, though.)

microcolonel 8y ago
Not to mention IRV/RCV usually produces a condorcet winner AFAIK.
dragonwriter 8y ago
> But you also need a solution when there isn’t an outright Condorcet winner, and there is no good solution to that. The tie-breaking mechanism of every Condorcet method is basically impossible to explain in simple terms, ranked pairs included. The Smith set is as easy to explain as the Condorcet winner, and random selection is trivial, so random selection from the Smith set seems to be a simple Condorcet tiebreaker.
- iainmerrick 8y ago
  So the tie-breaker is essentially to toss a coin? Hmm, there might be something in that...
  - dragonwriter 8y ago
    Limited to the Smith set, but sure. Same as lots of places do with FPTP and a tie; no Condorcet winner is effectively a “Condorcet tie” between the members of the Smith set, each of whom beats every candidate outside the set, but where victory within the set is undeterminable.
    - iainmerrick 8y ago
      Yes, UK elections use coin tosses on ties. It’s pretty rare to get an exact tie when thousands of people are voting, though, and it’s also hard to argue against the coin toss when literally nothing separates the two candidates. The Smith set tie would occur more often, and in complex scenarios where it really isn’t clear that it actually is a dead heat.
      - dragonwriter 8y ago
        You can do other tiebreakers in the Smith set, like plurality first-rank winner in the Smith set (proceeding down preference ranks, with random selection only if you fall off the bottom without resolving the tie.)
  - baddox 8y ago
    Indeed, if you allow randomness, you can get all the other desired attributes listed in Arrow’s impossibility theorem. Random ballot voting, for instance, has nearly every desirable property. There’s no reason to vote strategically (dishonestly). The race between candidates A and B can’t be affected by the presence or absence of any third candidate (there are no spoiler candidates). At least one voter is guaranteed to be happy with the election results (ok, that one is sort of a joke). But of course, random ballot has just one pesky undesirable property: the fact that it’s random.
xvedejas 8y ago
I kinda see range voting as the solution. To explain my intuition... Technically it's not condorcet, because a pair-wise winner doesn't always win, but if you relax the definition of "pair-wise winner" to allow for each voter to scale down the strength (weight) of their vote, then range voting does in fact always select the candidate which wins all weighted-pairwise competitions. I think whether you buy this intuition has to do with whether you think preference/utility is fundamentally ordinal or cardinal.
- bscphil 8y ago
  Your intuition is in fact correct. Balanski and Laraki have shown that a median based rating system, in the limit where all voters are rational and their utilities are maximized by electing their desired candidate, that the winner is always a Condorcet winner. This is one of the reasons that Majority Judgement (their name for this system) has so much going for it.
tunesmith 8y ago
Yeah, I've thought similarly. I actually consider multi-candidate Smith Sets a feature though, not a bug. Because I believe it means the electorate is honestly confused and not ready to make a decision yet. So one option is to hold a runoff between the candidates in the Smith Set. Another is crazier, but I haven't found a flaw yet. Basically introduce range/score voting on the ballot, but only use the scores if there is a multi-member Smith Set. That way you guarantee Condorcet compliance. And since the voters don't know if the ranges/scores will be used (and they probably won't), they don't have incentive to strategically score a less-desired candidate ahead of a more-desired candidate.
- iainmerrick 8y ago
  I like this idea the more I think about it, and I think it compares very well to methods being discussed elsewhere in this thread such as STAR (see https://news.ycombinator.com/item?id=18397839 https://news.ycombinator.com/item?id=18397839) I just thought of a way to possibly make it more intuitive. Use ranked voting, with the addition of a “nobody below this line” item that you can add to your ranking anywhere you like. Then at the runoff stage, you use approval voting, with “nobody below here” determining which candidates you approve. That gives you a nice easy-to-understand runoff, and is maybe even more understandable than normal ranked voting because you get to explicitly separate out the candidates you really can’t stomach. I suspect some people don’t like ranked voting because you’re expected to stack-rank lots of candidates you dislike.
  - quadrangle 8y ago
    I'm convinced that a "None of the Others" mark is pretty essential to making any ranked system good. But we're talking a lot of complexity to have both ranks and scores as separate marks. Super long ballot, will turn away most voters and seem really hard to process. It sounds like you might appreciate 3-2-1 voting. https://wiki.electorama.com/wiki/3-2-1_voting https://wiki.electorama.com/wiki/3-2-1_voting
    - tunesmith 8y ago
      It wouldn't have to be both ranks and scores. It could just be scores. But you then order the scores numerically to infer the rank. And you use the rank to find the Condorcet Winner. If there is no CW, then you use the scores from the candidates in the Smith Set to identify the winner from that Smith Set. (The point of the above was to respect the Condorcet Criterion, and 3-2-1 voting does not meet that criterion.)

microcolonel 8y ago

Not to mention IRV/RCV usually produces a condorcet winner AFAIK.

dragonwriter 8y ago

> But you also need a solution when there isn’t an outright Condorcet winner, and there is no good solution to that. The tie-breaking mechanism of every Condorcet method is basically impossible to explain in simple terms, ranked pairs included. The Smith set is as easy to explain as the Condorcet winner, and random selection is trivial, so random selection from the Smith set seems to be a simple Condorcet tiebreaker.

iainmerrick 8y ago
So the tie-breaker is essentially to toss a coin? Hmm, there might be something in that...
- dragonwriter 8y ago
  Limited to the Smith set, but sure. Same as lots of places do with FPTP and a tie; no Condorcet winner is effectively a “Condorcet tie” between the members of the Smith set, each of whom beats every candidate outside the set, but where victory within the set is undeterminable.
  - iainmerrick 8y ago
    Yes, UK elections use coin tosses on ties. It’s pretty rare to get an exact tie when thousands of people are voting, though, and it’s also hard to argue against the coin toss when literally nothing separates the two candidates. The Smith set tie would occur more often, and in complex scenarios where it really isn’t clear that it actually is a dead heat.
    - dragonwriter 8y ago
      You can do other tiebreakers in the Smith set, like plurality first-rank winner in the Smith set (proceeding down preference ranks, with random selection only if you fall off the bottom without resolving the tie.)
- baddox 8y ago
  Indeed, if you allow randomness, you can get all the other desired attributes listed in Arrow’s impossibility theorem. Random ballot voting, for instance, has nearly every desirable property. There’s no reason to vote strategically (dishonestly). The race between candidates A and B can’t be affected by the presence or absence of any third candidate (there are no spoiler candidates). At least one voter is guaranteed to be happy with the election results (ok, that one is sort of a joke). But of course, random ballot has just one pesky undesirable property: the fact that it’s random.

xvedejas 8y ago

I kinda see range voting as the solution. To explain my intuition... Technically it's not condorcet, because a pair-wise winner doesn't always win, but if you relax the definition of "pair-wise winner" to allow for each voter to scale down the strength (weight) of their vote, then range voting does in fact always select the candidate which wins all weighted-pairwise competitions. I think whether you buy this intuition has to do with whether you think preference/utility is fundamentally ordinal or cardinal.

bscphil 8y ago
Your intuition is in fact correct. Balanski and Laraki have shown that a median based rating system, in the limit where all voters are rational and their utilities are maximized by electing their desired candidate, that the winner is always a Condorcet winner. This is one of the reasons that Majority Judgement (their name for this system) has so much going for it.

tunesmith 8y ago

Yeah, I've thought similarly. I actually consider multi-candidate Smith Sets a feature though, not a bug. Because I believe it means the electorate is honestly confused and not ready to make a decision yet. So one option is to hold a runoff between the candidates in the Smith Set. Another is crazier, but I haven't found a flaw yet. Basically introduce range/score voting on the ballot, but only use the scores if there is a multi-member Smith Set. That way you guarantee Condorcet compliance. And since the voters don't know if the ranges/scores will be used (and they probably won't), they don't have incentive to strategically score a less-desired candidate ahead of a more-desired candidate.

iainmerrick 8y ago
I like this idea the more I think about it, and I think it compares very well to methods being discussed elsewhere in this thread such as STAR (see https://news.ycombinator.com/item?id=18397839 https://news.ycombinator.com/item?id=18397839) I just thought of a way to possibly make it more intuitive. Use ranked voting, with the addition of a “nobody below this line” item that you can add to your ranking anywhere you like. Then at the runoff stage, you use approval voting, with “nobody below here” determining which candidates you approve. That gives you a nice easy-to-understand runoff, and is maybe even more understandable than normal ranked voting because you get to explicitly separate out the candidates you really can’t stomach. I suspect some people don’t like ranked voting because you’re expected to stack-rank lots of candidates you dislike.
- quadrangle 8y ago
  I'm convinced that a "None of the Others" mark is pretty essential to making any ranked system good. But we're talking a lot of complexity to have both ranks and scores as separate marks. Super long ballot, will turn away most voters and seem really hard to process. It sounds like you might appreciate 3-2-1 voting. https://wiki.electorama.com/wiki/3-2-1_voting https://wiki.electorama.com/wiki/3-2-1_voting
  - tunesmith 8y ago
    It wouldn't have to be both ranks and scores. It could just be scores. But you then order the scores numerically to infer the rank. And you use the rank to find the Condorcet Winner. If there is no CW, then you use the scores from the candidates in the Smith Set to identify the winner from that Smith Set. (The point of the above was to respect the Condorcet Criterion, and 3-2-1 voting does not meet that criterion.)