[EM] Weighted voting systems for proportional representation

[Kathy Dopp]

The system you describe *is* still precinct summable in the sense of
reporting the sums for each possible slate of candidates for each
precinct or polling location - this is at least a whole lot fewer sums
than the number of possible ballot choice permutations including
partially filled out ballots that IRV/STV would require to be reported
and sampled to be precinct summable (reporting all individual ballots'
choices would be less to report in most cases).

To be summable, this system would require reporting (N choose S) sums
where N is the number of total candidates in the contest and S is the
number of seats being elected.  This is a lot of sums - but could, I
imagine, be mathematically sampled and audited to limit the risk of
certifying the wrong slate much more easily than IRV methods could be
- but I'm not certain about that until I have the time to think about
it more (not any time soon).

On Sat, Jul 23, 2011 at 12:00 PM, Kristofer Munsterhjelm
<km_elmet at lavabit.com> wrote:
> Kathy Dopp wrote:
>>
>> On Sat, Jul 23, 2011 at 9:33 AM, Kristofer Munsterhjelm
>> <km_elmet at lavabit.com> wrote:
>>>
>>> Kathy Dopp wrote:
>>>>>
>>>>> From: Kristofer Munsterhjelm <km_elmet at lavabit.com>
>>>>> 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.
>
>



-- 

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