4 ms·

I've seen people argue for Schulze or Ranked pairs over it. https://en.wikipedia.org/wiki/Schulze_method https://en.wikipedia.org/wiki/Schulze_method https://e

I've seen people argue for Schulze or Ranked pairs over it. https://en.wikipedia.org/wiki/Schulze_method https://en.wikipedia.org/wiki/Schulze_method https://en.wikipedia.org/wiki/Ranked_pairs https://en.wikipedia.org/wiki/Ranked_pairs The main distinction is that Ranked Choice can elect someone which is not the best compromise of everyone's choices, while Schulze and Ranked pairs will always elect the best compromise of everyone's choices. If there is a candidate who is preferred over the other candidates, when compared in turn with each of the others, they guarantee that candidate will win, while Ranked Choice does not. The downside to Schulze or Ranked Pairs is that saying that someone is your second choice could hurt the chances of your first choice to be elected, because by putting forward a compromising candidate, and since Schulze or Ranked Pairs optimize for that, you lean the tally towards your potential compromises (which are your second, third, etc. choices) in the case where the election are a "close call". If I understand correctly. I personally would prefer a system that optimizes for the compromise, as I think it's more important to get the candidate that the least people dislike, than it is to get the candidate that the most people adore. So if I understand correctly, it means something like: If you have A, B and C. 100% think B is second best. 40% think A is best. 19% think B is best. 41% think C is best. With Schulze or Ranked Pairs B will win, but with Ranked Choice C will win. That's because C is most people's first choice, so they win. But if you asked people to pick between C or B, B would technically win, because 19% would pick B (as it is their best), and 40% that think A is best but B is second best would also pick B, thus B vs C would get 59% votes for B and only 41% votes for C, that means that in a vote for B against C, B would win, but with Ranked Choice B loses, and C is elected.

mjevans 5y ago
The fallacy in the argument is that, first choice only, NO candidate has even a slim majority, let alone a strong one. If you decay the hypothetical 3 party set of options by eliminating the least popular candidate then you're back to first past the post which is the current status quo in the US. Schulze and similar methods use the ranked lists to evaluate pairs of candidates and eliminate the candidates that are universally the worst first / retain candidates that rank well in isolation. This is more likely to result in a compromise that works best for the most people.

mjevans 5y ago

The fallacy in the argument is that, first choice only, NO candidate has even a slim majority, let alone a strong one. If you decay the hypothetical 3 party set of options by eliminating the least popular candidate then you're back to first past the post which is the current status quo in the US. Schulze and similar methods use the ranked lists to evaluate pairs of candidates and eliminate the candidates that are universally the worst first / retain candidates that rank well in isolation. This is more likely to result in a compromise that works best for the most people.