The following was originally posted on 10/16/98. Apparently the fractional vote method I used was identical to Blake's rich party example after all. On the other hand, this has more to do with IRO's behavior than with the fractional method. For example, the voters who ranked (X=Y=B) equally could have gotten together and voted: 4 X Y B 5 X B Y 4 Y X B 5 Y B X 4 B X Y 5 B Y X This would have had the same effect as counting the equal votes as fractional votes. Bart
What does MPV/IRO do when candidates are ranked equally? There seems to me to be three possibilities. 1. Give each of the alternatives a full vote. This is certainly what the voter who is giving the equal rankings would like. After all, there is always the danger that by the time your first choice is eliminated, your compromise second choice will be gone too. If you rank them both equally, this means you can help them both at the same time. But what that means is that each voter must weigh their desire to get their first choice, versus their desire to get one of their most acceptable choice. So, for example, A=B>C is less likely to elect A then A>B>C, but more likely to elect one of A or B. The voter must weigh these considerations, as they would in a method like approval. To me, this seems out of place in a ranked method, but perhaps some MPV / IRO advocates could comment. 2. Give each of the alternatives an equal fraction of the vote. So, for example, once A=B=C reaches the top of the ballot (through elimination), each of A, B, and C will get 1/3 of a vote. Once one of them is eliminated, the each get 1/2. And finally when two are eliminated, 1. This doesn't appear to have the problem I mentioned above, but it does fail GITC. Candidates are A and B, which are not twins, X and Y, which are. 42 A B X Y 30 B X Y A 27 X=Y=B A 32 X Y B A 31 Y X B A A 42 B 39 - eliminated X 41 Y 40 A 42 - eliminated X 75.5 Y 44.5 X 117.5 - winner Y 44.5 - eliminated Now without Y 42 A B X 30 B X A 27 X=B A 63 X B A A 42 - eliminated B 43.5 X 76.5 B 85.5 - winner X 76.5 - eliminated So, having a twin caused X to win. This is called the rich party problem because it means that parties that can afford to run more candidates will have an unfair advantage. 3. Just don't allow equal rankings, except by leaving candidates unranked. This is the most obvious solution. It is possible that the electorate wouldn't understand, and use, equal rankings anyway. And it passes GITC. Unfortunately, it passes GITC for the same kind of technical reasons that make plurality pass GITC. That is, because voters are forced to distinguish between candidates randomly, even if they have no preference, they will break up what based on their true preferences, are twins. However, the rich party problem remains. Consider the above example. Imagine that each voter randomly distinguished between candidates they consider equal. On average each equally ranked candidate would get an equal share of the vote, just like happened in the example above. So, the banning solution is equivalent in rich party effect to the fractional solution. --- To me, this argument suggests that solution 1 is the best. It may not be perfect, but at least it doesn't have the rich party problem. The main danger is that voters won't understand how to equally rank candidates, and that the results will be more like solutions 2 and 3. -----== Sent via Deja News, The Discussion Network ==----- http://www.dejanews.com/ Easy access to 50,000+ discussion forums
