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I've always thought that descriptions of Condorcet methods are harder than they need to be. Even the wikipedia page describes them badly. The Condorcet criter

I've always thought that descriptions of Condorcet methods are harder than they need to be. Even the wikipedia page describes them badly. The Condorcet criteria itself is easy. If a candidate would beat every other candidate in a one-on-one matchup, that candidate should be the winner. There, that's it. That's really how it should be described. And then as a next step, we can then describe "tiebreakers". Instead, wikipedia (and others) describe Condorcet only in terms of their tiebreakers, making Condorcet methods themselves appear flawed[1]. You see, loops are possible. In a vote, it's possible that A would always beat B, B would always beat C, and C would always beat A. And the various Condorcet tiebreakers (like Schulz) are ways to resolve those situations. But the point is, if that happens, the vote really is accurately uncovering an indecisiveness that actually exists in the voting population. So it makes sense that a "tiebreaker" would be non-ideal in some sense. And they all are in some way. The problem is not with the tiebreaking methods. The problem is with the voting population. I've always thought a good voting system would be the following: First, rank the votes. If a Condorcet winner exists, the election is over. If not, restrict the candidate pool to the Schwartz set, and start a new round of campaigning so the voting population can research more and better educate themselves. [1]. Yes, we can argue that the Condorcet criteria is "perfect" if a Condorcet Winner exists, with one caveat: it implies the value that all votes have equal value. If you have that value, it's a logical implication that the Condorcet Winner (if it exists) should be the winner. But other values are possible, such as a utilitarian mindset of wanting to meet greatest social utility, such as some voters being far more passionate or educated than others. Following those implications bring up other tricky problems, though.

eli 13y ago
That's still pretty tough to explain. I had a friend who worked advocating for simple runoff voting initiatives, but gave up because it was too much work having to explain it to be people.
- tunesmith 13y ago
  I dunno - Instant Runoff got a lot of traction, and that one is even harder to explain/count and can come up with crazier results, too. I've always thought Condorcet could get more traction if it were rebranded as "Instant Round Robin" or something. People understand round robin tournaments.
  - dragonwriter 13y ago
    IRV is very simple to explain to people familiar with Majority/Runoff, since its basically a way of automating majority/runoff.
chongli 13y ago
Why not ditch the ordinal voting and go with a cardinal system? People have seen and understood the system used in judging competitions such as figure skating, diving and gymnastics for many years. Wouldn't it be completely straightforward for them to see it on a ballot? Range voting is the name of that system and, unlike Condorcet methods, it is not subject to Arrow's impossibility theorem.
- bradbeattie 13y ago
  Because range voting devolves to approval voting. Why rate a candidate 4/10 if you don't want them to win? You insincerely rate them 0/10. Why rate a candidate 8/10 if you want them to win? You insincerely bump them to 10/10. On top of that, you then have to consider the potentially lower popularity of your favourite choice against that of the current lead contenders. Do you reduce your vote for the better of two evils so that your sincere preference has a better chance? Ugh. I'm not trying to suggest tactical voting doesn't exist in other systems, but to suggest that it doesn't in Range voting is either misinformed or disingenuous.
  - DennisP 13y ago
    Now suppose there's a candidate I like pretty well, not as much as my favorite, but better than several others. Do I vote 10/10 or 0/10? Which is more tactical? It's a little hard to say. With approval voting I have to make that decision, setting some kind of threshold of approval. With range voting it might be easier to just vote honestly. At worst, everybody votes tactically and you get approval voting, which is still a good system. But sometimes the tactical decision is difficult, in which case range voting gives you the option of just voting honestly.
  - chongli 13y ago
    but to suggest that it doesn't in Range voting is either misinformed or disingenuous. It's a good thing I didn't suggest it, then.
- YokoZar 13y ago
  Because range voting embraces the property of valuing votes differently per voter. Implicitly, range voting gives more weight to voters who are aware of the top two candidates and can strategically rate one much lower than the other, even if their opinions of them are approximately the same. Valuing one voter over the others like that is a dangerous road to travel down.
baddox 13y ago
I agree. For many of the attributes of voting systems, you can actually remove the voting analogy altogether and think of any competition between individuals, like tennis. In a tennis tournament, you could easily have non-transitive results, where A defeats B, B defeats C, and C defeats A. So how do you organize a tournament to determine the "best" tennis player? Round robin tournaments (where every player plays every other player) can have non-transitive results. Of course, there's usually not enough time to do that anyway, so tournaments go with a bracket system. Brackets have their own problems, namely that it's very possible that the bracket winner would lose to a player who was knocked out of the bracket earlier by some third player.
kybernetikos 13y ago
I've written a system that allows condorcet voting with small numbers of voters (we used it to decide a HackDay winner recently), and with small numbers of voters you have to allow ties. None of the discussions of condorcet methods on wikipedia seem to talk much about allowing ties. The other thing is, that I sometimes want to allow people not to rank an option at all (e.g. if they participated in a HackDay team, they shouldn't rank their own team), once you allow options to be unranked, you open up the possibility of multiple roots to the graph (e.g. one person votes that A > C and leaves B unranked, and another person votes that B > C and leaves A unranked), so you need to do something reasonable in that case too. In my first implementation of ranked pairs, I would lock in results based on their victory margin, but if a number of results had the same victory margin, they would be locked in based on their order in the system - clearly wrong. My favourite theory now is just to find all strongly connected components and collapse them into a group which is equally tied (not just the Schwartz set either - cycles can occur further down the graph and relative rankings are still interesting to competitors). The joint winners are all options in all groups that have no losses. http://kybernetikos.github.io/graff/demo/ http://kybernetikos.github.io/graff/demo/ was me playing around with the ranked pairs way of removing cycles compared to just grouping them.
- tunesmith 13y ago
  Shouldn't it be possible to remove the Schwartz Set candidates and then re-run the count to find out the remaining relative rankings?
  - kybernetikos 13y ago
    That's probably a nice way of doing it - iteratively work out all schwartz sets (if there are multiple roots, there can be multiple schwartz sets) and remove them until there are none left to work out the rankings. Now I have to go back to the code!
    - tunesmith 13y ago
      I did something semi-related with beatpaths.com when it was active. I created a cyclic graph of NFL team wins. There would obviously be loops (such as two teams in a division splitting the series). So I'd simply remove the loops from the graph; smallest loops first. Eventually you'd end up with a DAG (perhaps with multiple roots), and I'd run a modified tsort (with various tiebreakers) to come up with a power ranking. It was kind of a silly website but I wanted to see how accurate NFL picks could be using only wins and losses. (Answer: somewhat competitive with other methods, but not very - but still, fun, and with pretty pictures!) Anyway, yes - you can come up with a full ranking of candidates in a Condorcet method by finding the condorcet winner (or schwartz set), removing, and running again, and keep going until you've fully ranked all candidates.

eli 13y ago

That's still pretty tough to explain. I had a friend who worked advocating for simple runoff voting initiatives, but gave up because it was too much work having to explain it to be people.

tunesmith 13y ago
I dunno - Instant Runoff got a lot of traction, and that one is even harder to explain/count and can come up with crazier results, too. I've always thought Condorcet could get more traction if it were rebranded as "Instant Round Robin" or something. People understand round robin tournaments.
- dragonwriter 13y ago
  IRV is very simple to explain to people familiar with Majority/Runoff, since its basically a way of automating majority/runoff.

chongli 13y ago

Why not ditch the ordinal voting and go with a cardinal system? People have seen and understood the system used in judging competitions such as figure skating, diving and gymnastics for many years. Wouldn't it be completely straightforward for them to see it on a ballot? Range voting is the name of that system and, unlike Condorcet methods, it is not subject to Arrow's impossibility theorem.

bradbeattie 13y ago
Because range voting devolves to approval voting. Why rate a candidate 4/10 if you don't want them to win? You insincerely rate them 0/10. Why rate a candidate 8/10 if you want them to win? You insincerely bump them to 10/10. On top of that, you then have to consider the potentially lower popularity of your favourite choice against that of the current lead contenders. Do you reduce your vote for the better of two evils so that your sincere preference has a better chance? Ugh. I'm not trying to suggest tactical voting doesn't exist in other systems, but to suggest that it doesn't in Range voting is either misinformed or disingenuous.
- DennisP 13y ago
  Now suppose there's a candidate I like pretty well, not as much as my favorite, but better than several others. Do I vote 10/10 or 0/10? Which is more tactical? It's a little hard to say. With approval voting I have to make that decision, setting some kind of threshold of approval. With range voting it might be easier to just vote honestly. At worst, everybody votes tactically and you get approval voting, which is still a good system. But sometimes the tactical decision is difficult, in which case range voting gives you the option of just voting honestly.
- chongli 13y ago
  but to suggest that it doesn't in Range voting is either misinformed or disingenuous. It's a good thing I didn't suggest it, then.
YokoZar 13y ago
Because range voting embraces the property of valuing votes differently per voter. Implicitly, range voting gives more weight to voters who are aware of the top two candidates and can strategically rate one much lower than the other, even if their opinions of them are approximately the same. Valuing one voter over the others like that is a dangerous road to travel down.

baddox 13y ago

I agree. For many of the attributes of voting systems, you can actually remove the voting analogy altogether and think of any competition between individuals, like tennis. In a tennis tournament, you could easily have non-transitive results, where A defeats B, B defeats C, and C defeats A. So how do you organize a tournament to determine the "best" tennis player? Round robin tournaments (where every player plays every other player) can have non-transitive results. Of course, there's usually not enough time to do that anyway, so tournaments go with a bracket system. Brackets have their own problems, namely that it's very possible that the bracket winner would lose to a player who was knocked out of the bracket earlier by some third player.

kybernetikos 13y ago

I've written a system that allows condorcet voting with small numbers of voters (we used it to decide a HackDay winner recently), and with small numbers of voters you have to allow ties. None of the discussions of condorcet methods on wikipedia seem to talk much about allowing ties. The other thing is, that I sometimes want to allow people not to rank an option at all (e.g. if they participated in a HackDay team, they shouldn't rank their own team), once you allow options to be unranked, you open up the possibility of multiple roots to the graph (e.g. one person votes that A > C and leaves B unranked, and another person votes that B > C and leaves A unranked), so you need to do something reasonable in that case too. In my first implementation of ranked pairs, I would lock in results based on their victory margin, but if a number of results had the same victory margin, they would be locked in based on their order in the system - clearly wrong. My favourite theory now is just to find all strongly connected components and collapse them into a group which is equally tied (not just the Schwartz set either - cycles can occur further down the graph and relative rankings are still interesting to competitors). The joint winners are all options in all groups that have no losses. http://kybernetikos.github.io/graff/demo/ http://kybernetikos.github.io/graff/demo/ was me playing around with the ranked pairs way of removing cycles compared to just grouping them.

tunesmith 13y ago
Shouldn't it be possible to remove the Schwartz Set candidates and then re-run the count to find out the remaining relative rankings?
- kybernetikos 13y ago
  That's probably a nice way of doing it - iteratively work out all schwartz sets (if there are multiple roots, there can be multiple schwartz sets) and remove them until there are none left to work out the rankings. Now I have to go back to the code!
  - tunesmith 13y ago
    I did something semi-related with beatpaths.com when it was active. I created a cyclic graph of NFL team wins. There would obviously be loops (such as two teams in a division splitting the series). So I'd simply remove the loops from the graph; smallest loops first. Eventually you'd end up with a DAG (perhaps with multiple roots), and I'd run a modified tsort (with various tiebreakers) to come up with a power ranking. It was kind of a silly website but I wanted to see how accurate NFL picks could be using only wins and losses. (Answer: somewhat competitive with other methods, but not very - but still, fun, and with pretty pictures!) Anyway, yes - you can come up with a full ranking of candidates in a Condorcet method by finding the condorcet winner (or schwartz set), removing, and running again, and keep going until you've fully ranked all candidates.