| > Is it just this simple? You sum their score, then instant runoff the top 2? Yes. First round counts stars, second round just counts preference. If I remember correctly, the main purpose of the second round is to discourage strategic voting -- if you give all other candidates the minimum score, to give your favorite the best chance, then you don't get to express a preference in the runoff, in the event that your favorite is not top 2. So it's only worth doing this if you really like your favorite so much that everyone else is the same to you. > It still definitely doesn't elect the condorcet winner... which you can see from 3 voters and 12 candidates My understanding (from Wikipedia) is that the condorcet winner is "one candidate who beats every other candidate pairwise". Since often no candidate fulfills this criteria, all condorcet systems that expect to survive contact with the real world need to specify a fallback way to choose a winner. I don't know much about the different strategies they use to do this. If my understanding is correct, then it seems like there is no condorcet winner in this scenario (rather, a 3-way tie). So I'm not sure how you expect a condorcet system to handle it. |
I'm new to these terms, but when I look up "condorcet winner" in on wikipedia, it's defined as: "the candidate that wins a majority of the vote in every head-to-head election"
So, I'm assuming "every head-to-head election" would be 10 factorial elections in our case (one for each pair).
Let's look at a couple of these elections and then extrapolate on the rest of them:
J vs A:
Voters 2 and 3 didn't put A on the list, while they did indicate that they liked J, so they would vote for J. Only Voter 1 would vote for A.
D vs A:
Again, two voters didn't put D on the list (1 & 3), but did indicate that they liked J, while only one voter (voter 2) likes D over A.
You can see that, for each candidate besides J, only one voters like them, while all voters at least like J a little. This means that J will win "every head-to-head election" and thus be the condorcet winner.
The only problem with this proof currently is that STAR doesn't explicitly elect a winner, so therefore it is not "choosing the wrong winner" as it isn't choosing any candidate. It is eliminating the condorcet winner though, which is concerning, and I believe a more complicated example would show STAR choosing a non-condorcet winner.
I like STAR more than RCV after reading up on STAR more. I believe that approval voting might be better though, as it completely elimintates the spoiler effect I believe. Though it may have other problems.
Thinking about this more, I believe a condorcet winner can be chosen with RCV (and STAR with some assumptions) just by changing the tallying system. I was able to dicern from my example who the condorcet winner was, so why not do the same in an election? Just simulate all head-to-head elections to find the condorcet winner. I must be missing something.
Here's a scenario where STAR elects a candidate that is not the condorcet winner:
I've marked important votes with a '!'. All others are unique votes for candidates with no support.In this scenario, B and G get the top two scores with 9 and 8. B wins the election with 3 votes (voters 1, 3, and 4) over G's two votes (voters 2 and 5). A is the condorcet winner though as they win every head-to-head contest (with close elections (3-to-2), against B and G). A was eliminated though, as they only scored 6 points.
This was complicated and too much fun to write, so it probably isn't practical in a real world scenario.