[EM] Proportional multi-winner ranked voting methods - guidelines?

Kristofer Munsterhjelm km_elmet at t-online.de
Thu Feb 23 14:20:20 PST 2017

On 02/23/2017 09:59 PM, robert bristow-johnson wrote:
> i wasn't aware that any STV would be Condorcet compliant (except for
> bottom-two runoff STV).
> my only thought about a multi-winner method that *is* Condorcet
> compliant is to simply run the rank-choice election according to some
> Condorcet method (say Schulze or RP or MinMax or BTR-STV), pick the
> Condorcet winner and assign that candidate to the first available seat,
> decrement the number of available seats by one, remove this winner from
> the candidate pool, and then see who the next Condorcet winner is from
> the remaining candidates.  rinse and repeat until all available seats
> are assigned.
> what would be wrong with that approach to multi-winner elections?

That method fails the Droop proportionality criterion and amplifies 
majorities into unanimities.


51: A>B>C>D
49: E>F>G>H

Four to elect gives a council of {A, B, C, D}. If one wants a 
proportional outcome (which is what all the quota business in STV is 
intended to accomplish), then a majoritarian multiwinner Condorcet 
election doesn't help much. If you want a majoritarian method, then it's 
okay, but the subject says "Proportional" :-)

A more proportional Condorcet method could be accomplished this way -- I 
think that would be the most simple somewhat proportional Condorcet 
method. For n seats:

* Repeat lots of times:
  - Randomly divide the voters into n groups
  - Order the groups in random order.
  - Determine the first group's winner according to the Condorcet method.
  - Give the first seat to this winner and eliminate him from every 
ballot (of every group).
  - Determine the second group's winner, elect, and eliminate.
  - Do so until you have n candidate assignments.
* Choose the assembly that you saw most often.

In the 51/49 example above, it's basically a coin toss as to whether any 
given group will elect one of {A,B,C,D} or one of {E,F,G,H}, and so you 
get a 50-50 split.

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