Dear Rob! you wrote: > Well, if all voters use an optimal approval strategy, this is something of > a knife-edge case. If the electorate were > > 100:A>D>C>B > 100:A>B>D>C > 100:A>C>B>D > 199:D>C>A>B > 200:B>D>A>C > 200:C>B>A>D > > then the approval votes at equilibrium would be > > 100:A>>D>C>B > 100:A>>B>D>C > 100:A>>C>B>D > 199:D>C>A>>B > 200:B>D>>A>C > 200:C>B>>A>D > > and A wins.
I don't understand this. Forest's method would initially give A an approval of 300, B an approval of 500, and C and D an approval of 499, hence it would reinforce B as long as B's weight is more than three times the weight of A, C, and D. From that point on, it would give A an approval of 300, B an approval of 500, C an approval of 599, and D an approval of 499, so that it would reinforce C. I don't see at all how A would arise as "the" winner... ---- Election-methods mailing list - see http://electorama.com/em for list info
