Kathy Dopp wrote:
On Sat, Jul 23, 2011 at 9:33 AM, Kristofer Munsterhjelm
<[email protected]> wrote:
Kathy Dopp wrote:
From: Kristofer Munsterhjelm <[email protected]>
I don't think that passes DPC (since Borda doesn't pass Majority), but
it passes the weaker "force proportionality" criterion (in that an 1/n
faction can, by strategy, force their representative to be the one they
want). So it is at least better than SNTV, except for the whole bit
about not being summable :-)
It *is* a summable method because it would simply require reporting
and summing the sum from each precinct for every ballot's (1) 1st
choice candidate, (2) 2nd choice candidate, and so forth.
Is it? I may be wrong, but consider these alternate scenario snippets:
Scenario A:
1: X: 2, Y: 1, rest 0
1: A: 2, B: 1, rest 0
Scenario B:
1: X: 2, B: 1, rest 0
1: A: 2, Y: 1, rest 0
Say the current slate is {X,Y}. Then in the first scenario, you want to add
two points to the score from the first voter's ballot, and none from the
second ballot; but in the second scenario, you want to add two points from
the first ballot and one from the second.
Yet the per-candidate sums give X two points and Y one in both scenarios.
Even if you count the vote ranked style, you get:
First scenario: first place: one X and one A, second place: one Y and one B.
Second scenario: first place: one X and one A, second place: one B and one
Y.
I was reading what you wrote *literally* when you said: "Any given
slate has a score equal to the sum of, over all ballots, the
*highest* rated candidate on that ballot that is also in the given slate."
I'll try to be more precise. Consider the intersection of the slate in
consideration (call it S) and the vth voter's ballot where ballots are
b_{1}...b_{n}. Then f(S, b_{v}) is equal to the highest rated candidate
present in the intersection of S and the vth voter's ballot.
In other words, it is the highest rated candidate that is both in S (the
slate under consideration), and on v's ballot.
Then, define f(S), the score of slate S, to be equal to
SUM (k = 1...n) f(S, b_{k})
where there are n voters. Then we want to find the S for which f(S) is
maximized. Say that the maximum is reached for f(S_W). Then S_W is the
winning slate.
In the example above, in the first scenario, you have:
1: X: 2, Y: 1, A: 0, B: 0
1: A: 2, B: 1, X: 0, Y: 0
(expanding the zeroes).
S is here {X,Y}, so the intersections are:
1: {X: 2, Y: 1}
1: {X: 0, Y: 0}.
Thus the contribution from the first voter is 2, because that's the
highest rated candidate both on his ballot and in S. The contribution
from the second voter is 0, and the final score is 2.
Second scenario:
1: X: 2, B: 1, Y: 0, A: 0
1: A: 2, Y: 1, B: 0, X: 0
S is again {X,Y}, so the intersections are:
1: {X: 2, Y: 0}
1: {Y: 1, X: 0}
Then the contribution from the first voter is 2 and the contribution
from the second voter is 1, adding up to 3.
You also said "Tiebreaks are leximax (sum of, over all ballots, the
second highest rated candidate, etc)." Taken *literally* this means
that the votes of the 2nd highest candidate is only considered for
breaking ties among slates.
That is true. The 2nd highest rated candidate of the intersection of S
and v's ballot is only considered in a tiebreak.
The English language is so ambiguous. Your system would have to make
it clear to voters that ranking a candidate on the ballot *at all* is
likely to help that candidate win - even if ranked last.
In the Range variant, rating the candidate above bottom would help him,
even if you rated him at 0.001, because it's above the 0 (bottom value)
the system assumes if you don't rate him. In the Borda variant, ranking
someone last on a truncated ballot can help him unless he's also the
last candidate you rank. That is, if there are 8 candidates, and you put
someone below the 7 candidates already on your ballot, he won't be
helped by that, but if your ballot only had 3 other candidates, adding
that candidate last would help him.
I probably would *not* support a system that counted ballots in a way
that is not precinct summable.
Summable proportional multiwinner methods are not common, but I think a
summable Bucklin type method was discussed here at some point. I can't
remember which it was, though.
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