[Election-Methods] PR favoring racial minorities (Gilmour/Dopp?)

[Kristofer Munsterhjelm]

Juho wrote:
> On Jul 31, 2008, at 0:50 , Kristofer Munsterhjelm wrote:
> 
>> In any case, the left-right problem would still be a limitation to 
>> RRV, where ballots are set so that RV (and any sensible method) would 
>> elect Center first, but where electing an assembly of two should elect 
>> Left and Right. For some reason, that problem doesn't appear in party 
>> list, but I don't know why.
> 
> I'm not sure if I understood you correctly, but isn't that the same 
> difference as between highest average based methods 
> (http://en.wikipedia.org/wiki/Highest_averages_method) and largest 
> reminder based methods 
> (http://en.wikipedia.org/wiki/Largest_remainder_method) that both can be 
> used with party lists?

To be more clear, this is an example I devised for Range:

52: Left: 1.0 Center: 0.5 Right: 0.0 (leftists)
50: Left: 0.0 Center: 0.5 Right: 1.0 (right-wingers)
13: Left: 0.0 Center: 1.0 Right: 0.0 (moderates)

In pure Range, Left has a score of 52, Center has a score of 64, and 
Right has a score of 50. Obviously, in the single-winner case with Range 
rules, Center should win.

However, in the assembly of size two, the Left and Right voters have a 
Droop quota each, and thus, if the method is to fulfill the Droop 
proportionality criterion, it must elect Left and Right.

> I guess any method that picks the winners one by one until sufficient 
> number of representatives has been elected fails to elect both the 
> centrist (1 representative) and left+right (2 representatives) in the 
> example above. On the other hand if we follow the example that means 
> that Alabama paradox is what we want.

It would appear as such, since a candidate can lose by having the 
assembly size increase (if that candidate is a centrist). If we use the 
example above and say that Left and Right contribute to the left and 
right side of the spectrum equally, but the centrist contribute to both, 
then electing Center and (either Left or Right) is undesirable, because 
it tips the balance of power to either left or right, respectively.

But party list PR doesn't have a problem with the Alabama paradox, yet 
it doesn't fall to this problem, either. Why is that? I think the reason 
is that in party list PR, you only have one vote. Therefore, the Left 
and Right voters must vote Left only and Right only. Because of party 
proportionality, there's no vote splitting problem*, so one can vote for 
the favorite party.

With a divisor/average method like Sainte-Laguë, since it's protected 
against the Alabama paradox, that means that it'll fail when the 
assembly is small enough. In the case above, it would elect Left with an 
assembly size of one.

Thus it seems that in order for the method to be able to compromise 
effectively, it must technically break the Alabama paradox. I think that 
hybrids are possible, though. Although I haven't checked it in detail, 
QPQ (which I mentioned earlier) seems to be such a hybrid, because it 
reduces, or is claimed to reduce, to D'Hondt (or Sainte-Laguë) when 
voters vote strictly on party lines. More research would be needed...

* Well, significantly less of a vote splitting problem; it depends on 
the size of the districts and the nation. Here in Norway, we have 
additional measures to compensate district disproportionality with 
nation level seats, kind of like MMP but with parties only.