[EM] Condorcet divisor method proportional representation

[Kristofer Munsterhjelm]

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.