On Sun, Jul 3, 2011 at 1:34 PM, Kristofer Munsterhjelm <[email protected]> wrote:
>>> Let me pull an old example again: >>> >>> 45: Left > Center > Right >>> 45: Right > Center > Left >>> 10: Center > Right > Left >>> >>> If there's one seat, Center is the CW; but if you want to elect two, it >>> seems most fair to elect Left and Right. If Center is elected, the wing >>> corresponding to the other winning candidate will have greater power. >> >> I disagree. In your example, clearly 55 prefer right to left, but only >> 45 prefer left to right. And center is the clear winner overall. >> Thus, if only two will be elected, it should be center and right. > > That's incompatible with the Droop proportionality criterion. The DPC says > that if there are k seats, and a fraction greater than 1/(k+1) of the > electorate all prefer a certain set of candidates to all others, then > someone in that set should be elected. In your example: 55: prefer Center to (Left or Right) 55: prefer Right to Left 45: prefer Left to Right and you say that the Droop criteria says that Right and Left must be elected! But by the rule you cite, then all three candidates must be elected, and thus the Droop quota requires electing k+1 candidates, which contradicts the statement of its own rule. Thus, QED, the Droop criteria doesn't elect k seats, and so must be abandoned, unless you are insisting on burying the 2nd choice candidates of all voters - like STV and IRV do to some voters, while considering the 2nd choice candidates of others. Thus, it seems logical, that only an IRV proponent, who insists on not looking at the 2nd choice votes of some voters would insist on using the Droop quota. I dare say, that to apply the Droop quota in many instances requires the unfair treatment of voters' votes (looking at only some voters 2nd and later choices, but not others) in most cases, or it would tend to elect more or less than the required number of seats otherwise. Hence, the Droop quota, not in this example, but in many others with the same number of voters and candidates, would require the unfair treatment of voters' rank choice votes or if not, it would elect the wrong number of seats. > > (Actually, the more general sense is that if more than p/(k+1) of the > electorate all prefer a set of q candidates to all others, then min(p, q) of > these candidates should win.) Throwing p in the expression, seems to make little sense. You mean if only p= 6/(k+1) = 2 voters prefers a set of 3 candidates to all others, in the case of k=2, then min(2,3) = 2 of these candidates should win! That's a funny rule. A more common sense rule IMO would be for k seats, elect the k candidates who are preferred above the rest of the candidates by more voters. -- Kathy Dopp http://electionmathematics.org Town of Colonie, NY 12304 "One of the best ways to keep any conversation civil is to support the discussion with true facts." Fundamentals of Verifiable Elections http://kathydopp.com/wordpress/?p=174 View some of my research on my SSRN Author page: http://ssrn.com/author=1451051 ---- Election-Methods mailing list - see http://electorama.com/em for list info
