[EM] IRV vs RCV

Kristofer Munsterhjelm km_elmet at t-online.de
Tue Dec 17 16:02:49 PST 2019


On 17/12/2019 08.12, Rob Lanphier wrote:
> On Sat, Dec 14, 2019 at 7:21 AM robert bristow-johnson
> <rbj at audioimagination.com> wrote:
>>> On December 14, 2019 2:44 AM Rob Lanphier <robla at robla.net> wrote:
>>> I also like Ranked Pairs, but
>>> I suspect that all of the methods that pick a candidate out of the
>>> Smith set are indistinguishable in real-world conditions.  My hunch is
>>> that an analysis of the public elections that had ranked ballots would
>>> reveal that all of them had a single Condorcet winner,
>>
>> except, of course, Burlington Vermont 2009.
> 
> I'm pretty sure all of the Condorcet-winner compliant methods chose
> Andy Montroll, given the ballots from the Burlington 2009 election.
> Copeland, Schulze, Ranked Pairs,  etc.  Was there a discrepency
> between Condorcet methods, or just the well-documented discrepency
> between the Condorcet methods and IRV?
> 
>>> and therefore
>>> there would be no difference between the results of Ranked Pairs,
>>> Schulze, Tideman, Schulze, or even Copeland.
>>
>> this is the difficult point i have tried to say here.  i think that
>> Schulze is likely the most resistant to voting strategy, but it's
>> too difficult to explain to legislators and the public.>
> Yeah, I agree.  I'm willing to take it on faith that BTR-STV is more
> susceptible to strategy than methods that guarantee Smith set
> membership, but I suspect that Condorcet-compliant methods perform
> better at strategy resistance than standard IRV does.

Doesn't BTR-IRV pass Smith? Suppose that X is in the Smith set and Y is
not. Once X and Y meet in the bottom-two runoff, then by definition of
the Smith set, X beats Y pairwise, so Y is eliminated.

This holds no matter who X and Y are, so a Smith set member can never be
eliminated in a bottom-two runoff when facing a candidate outside the
Smith set. And since every candidate has been subjected to at least one
such runoff, every member outside the Smith set must necessarily have
been eliminated. Thus the method passes Smith.

Or am I missing something? :-)

>> BTR-STV is different.  Schulze, RP, MinMax (dunno about Copeland) all elect the same candidate in the case of an CW or a Smith set of 3.  i don't ever ever ever expect to see a Condorcet RCV ever have a Smith set larger than and i really don't expect to see one without a CW.
> 
> Copeland isn't guaranteed to pick a candidate out of the Smith set
> when the Smith set is bigger than one, so it's possible it'll pick a
> different winner than Schulze, RP, MinMax, etc when the Smith set is
> 3.

That also seems wrong. See theorem 1 of
http://dss.in.tum.de/files/brandt-research/choicesets.pdf.


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