[EM] Condorcet divisor method proportional representation

[Kathy Dopp]

Thanks Kristofer.  I ignored the "all* in "all others".

I must say then, I simply do not like the Droop quota as a criteria
because it elects less popular candidates favored by fewer voters
overall and eliminates the Condorcet winners some times. The Droop
quota seems to go hand in hand with IRV and STV methods.

> The Droop proportionality criterion is thus a generalization of the mutual
> majority criterion, which says that if a majority prefers a certain set of
> candidates to everybody else, then someone from that set should win.

Yes.  Even in the case that a greater majority prefers a different set
of candidates over the ones elected by Droop.

Interesting.

Kathy

On Mon, Jul 4, 2011 at 9:22 AM, Kristofer Munsterhjelm
<km_elmet at lavabit.com> wrote:
> Kathy Dopp wrote:
>>
>> On Sun, Jul 3, 2011 at 1:34 PM, Kristofer Munsterhjelm
>> <km_elmet at lavabit.com> 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.
>
> That's not what it says. 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 other words, if more than a 1/(k+1) fraction voted A, B, and C, above all
> the other candidates, but not necessarily in the same order, then one of A,
> B, and C should win. Center is not preferred *to all others* by more than a
> Droop quota - only the 10 Center voters prefer Center to all others.
>
> If you want to be exhaustive, the count for all sets are:
>
> 100 voters prefer the set {Left, Center, Right} to all others
>  45 voters prefer the set {Left, Center} to all others
>  0 voters prefer the set {Left, Right} to all others
>  55 voters prefer the set {Right, Center} to all others
>  45 voters prefer the set {Left} to all others
>  45 voters prefer the set {Right} to all others
>  10 voters prefer the set {Center} to all others.
>
> A Droop quota is 1/(k+1), in this case, 100/3 = 33 + 1/3. That means that
> the DPC says:
>        Elect at least one from {Left, Center}  (since 45 > 33.3)
>        Elect at least one from {Right, Center} (since 55 > 33.3)
>        Elect at least one from {Left}          (since 45 > 33.3)
>        Elect at least one from {Right}         (since 45 > 33.3)
>
> and the only way to make that work is to elect Left and Right.
>
> The Droop proportionality criterion is thus a generalization of the mutual
> majority criterion, which says that if a majority prefers a certain set of
> candidates to everybody else, then someone from that set should win.
> (If the candidates in that set had agreed to run only one of the candidates,
> that candidate would, after all, have won by a straight majority.)
>
>>> (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.
>
> I forgot to say "fraction". If a fraction of more than p/(k+1) of the
> electorate all prefer a set of q candidates to everybody else, then min(p,
> q) should win. In the example above, with two winners, this works out as:
>
>  if more than 100 * 1/3 = 33 + 1/3 voters prefer a set to everybody else, at
> least one candidate from that set should win,
>  if more than 100 * 2/3 = 66 + 2/3 voters prefer a set to everybody else, at
> least two candidates from that set should win, unless the set has only one
> candidate in it (in which case it's impossible).
>
> For three winners:
>  if more than 100 * 1/4 = 25 voters prefer a set to everybody else, at least
> one should be elected from that set,
>  if more than 100 * 2/4 = 50 voters..., at least two candidates should be
> elected from that set (if possible),
>  if more than 100 * 3/4 = 75 voters..., at least three candidates,
>
> and so on.
>
>> 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.
>
> If you mean the k individual candidates, then consider an election of the
> form:
>
> 51: A1 > A2 > A3 > A4 > B1 > B2 > B3 > B4
> 49: B1 > B2 > B3 > B4 > A1 > A2 > A3 > A4
>
> four seats.
>
> The first candidate preferred by more voters is A1. The second is A2. The
> third is A3, and the fourth is A4. So the assembly turns out to have only
> A-candidates, even though only slightly more than half of the electorate
> ranked the A-candidates above the B-candidates. That's not very
> proportional.
>
> If you mean k candidates taken as a whole, then that's what the
> Condorcet-like multiwinner methods try to do. CPO-STV, for instance, tests
> different k-candidate "prospective councils" against each other as if they
> were individual candidates in a Condorcet election. How the pairwise
> matchups between councils are calculated depends on the method in question,
> and this calculation may lead a council to be the CW among prospective
> councils while still not including the single-winner CW.
>
>



-- 

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