[EM] IRV vs RCV
Kristofer Munsterhjelm
km_elmet at t-online.de
Wed Dec 18 04:36:51 PST 2019
On 18/12/2019 06.15, Rob Lanphier wrote:
> Kristofer, I stand corrected on both of the points I was trying to
> make (thank you!). More inline:
>
> On Tue, Dec 17, 2019 at 4:02 PM Kristofer Munsterhjelm
> <km_elmet at t-online.de> wrote:
>> 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:
>>> 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.
>
> Oh, that's delightfully simple! Your informal proof of BTR-IRV
> passing Smith seems correct to me.
>
> I'm now struggling to figure out what the practical benefits of the
> other Condorcet methods over BTR-IRV. Given that BTR-IRV is
> reasonably simple to explain, it has an intuitive connection to IRV,
> it's hard to understand what the practical benefit is to advocating
> for other Condorcet-winner compliant systems.
As Chris Benham pointed out, it fails clone independence. It also
inherits a lot of the compliance failures of IRV (e.g. nonmonotonicity,
no reversal symmetry), and it's not summable. Contrasted with Schulze,
it doesn't tend to elect Minmax winners, and with Ranked Pairs, it fails
LIIA.
I vaguely seem to recall that Warren found out it does worse than
Schulze/RP/etc on Bayesian regret. But I'm not entirely sure of this.
I'm going through the messages of the Condorcet Yahoo group, and it
seems like Alex Small proved something along the lines of BTR-IRV giving
the same winner as Benham with three candidates. Let's see if I can redo
it and make it clearer.
Either we have a Condorcet cycle or there's a CW. If there's a CW, the
equivalence obviously holds. So suppose without loss of generality that
there's an A>B>C>A cycle and that A is the IRV winner. (It's always
possible to relabel the candidates so that this holds.)
When there are two candidates left in IRV, the ultimate IRV winner must
beat the loser pairwise. So given the assumptions above, the full order
must be A>B>C (otherwise, A and C would be in the final round, which
contradicts that A is the IRV winner).
BTR-IRV will first hold a runoff between B and C, and since B beats C
pairwise, B wins. In the next runoff, A wins against B.
Benham will directly eliminate C. In the next round, we have A vs B and
A wins. So the winners are the same.
That seems to show that if you think Smith cycles above three are rare,
then it doesn't matter whether you use Benham or BTR-IRV, because
they'll give the same winner: the CW if there is one, and the IRV winner
if there's a cycle. (Or it almost does: I have to show that the Smith vs
non-Smith runoff rounds don't alter anything.)
But since I'm always adding caveats, BTR-IRV's failure of clone
independence could lead parties to puff up the size of the Smith set in
an attempt to benefit. So if you want a good Smith-IRV hybrid *in
general*, Benham is better.
>>> 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.
>
> Based on what I learned about Copeland back in 1996 when I was first
> learning this stuff, I somehow dismissed the usefulness of the
> Copeland set, and exhalted the use of the Smith set (since
> Smith//Minmax(wv) seemed to be the preferred method discussed on EM
> back in 1996, as I recall). That paper looks like something I should
> spend more time reading.
The Copeland set is interesting: for instance, it's a subset of the
uncovered set[1]. However, there don't seem to be any good ways to turn
the set into a proper method that satisfies good properties (clone
independence etc).
[1] E.g. proposition 3.2. of
https://researchportal.port.ac.uk/portal/files/1894394/SELCUK_2010_cright_EL_A_characterization_of_the_Copeland_solution.pdf
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