Just another example vote set FYI. 2 AB 2 AC 1 B 1 C 4 D
Natural winners are maybe A and D. 100 AB 100 AC 1 B 1 C 4 D Natural winners are maybe B and C. Is it a problem that additional support to A (and B and C) meant that A was not elected? (A was top ranked by all the new voters. B and C were top ranked by 50% of the new voters.) Juho On 8.8.2011, at 22.40, [email protected] wrote: > It seems that if a PR method chose slate {X, Y} for a two winner election, > and only X or Y received > increased support in the rankings or ratings, then {X, Y} should still be > chosen by the method. > > But consider the following approval profile (for a two winner election): > > 3 X > 1 XY > 2 Y > 2 Z > > It seems pretty clear that the slate {X, Y} should be elected, and that is > the PAV decision. > > Now suppose that X gets additional approval on some ballots but the Y and Z > approvals stay the same: > > 2 X > 3 XY > 2 Z > > Now PAV elects {X, Z}, and this seems like the right choice, because this > slate completely covers the > electorate, unlike any other pair. Candidate Y has more approvals than Z, > but everybody that approves > Y also approves X, so given that X is part of the slate, Y would only > contribute half a satisfaction point > per ballot, while Z adds a full point per ballot. Since 2>1.5, Z wins over Y > for the remaining position on > the slate. > > This violates the strong monotonicity ideal of the first paragraph, but does > not violate a weaker version > that says if only one candidate X of the winning slate gets additional > support on some ballots (and all > other candidates have the same or less support as before on all ballots) then > that one candidate X > should be a part of the new slate. > > Now let's look at this example from the point of view of the Ultimate Lottery: > > In the before scenario, the Ultimate Lottery probabilities x, y, and z for > the respective candidates X, Y, > and Z are obtained by maximizing the product > > x^3*(x+y)*y^2*z^2 subject to the constraint x+y+z=1 > > The solution is exactly (x,y,z)=(45%, 30%, 25%) . > > After the increase in support for x the Ultimate Lottery probabilities are > obtained by maximizing the > product > > x^3*(x+y)^3*z^3 subject to the same constraint x+y+z=1. > > The solution is precisely (x, y, z) = (75%, 0, 25%) . > > Note that (in keeping with the strong ideal expressed at the beginning of > this message) the only > candidate to increase in probability was X, the one that received increased > support. It did so at the > expense of Y whose probability decreased to zero. So Z passed up Y without > any change in its > probability. That's basically why Z took Y's place on the slate without any > increased support on the > ballots. > > So this helps us understand (in the PR election) why the weaker member of the > two winner slate > changes from Y to Z, and why we cannot expect the strong monotone property > for a finite winner PR > election; the discretization in going from the ideal proportion of the > Ultimate Lottery to a finite slate > allows only a crude approximation to the ideal proportion. > > In other words, it is just one of the classical apportionment problems in > disguise. > > How do other PR methods stack up with regard to monotonicity? > > Since IRV is non-montone, automatically STV fails even the weak montonicity > sartisfied by PAV. > > How about the other common methods? > ---- > Election-Methods mailing list - see http://electorama.com/em for list info ---- Election-Methods mailing list - see http://electorama.com/em for list info
