[EM] RCIPE version 2

[Richard, the VoteFair guy]

On 7/24/2021 2:19 PM, Kristofer Munsterhjelm wrote:
 > On 7/22/21 5:31 PM, VoteFair wrote:
 >> On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote:
 >>  > How about this?
 >>  >
 >>  > - Eliminate the candidate with the least number of winning subgroups.
 >>  > - If there is a tie, break that tie by IRV.
 >>  > ...
 >>
 >> Isn't the first step basically Copeland's method?
 >
 > No, because there's no elimination in Copeland (and it doesn't pass
 > LIIA). It would just elect the candidate/s with the most
 > winning subgroups.

I see you're right, of course.

I admit your suggestion is clever because it includes Condorcet loser 
elimination.

Yet I'm sure lots of non-math-savvy voters will not trust that the 
candidate with the least number of wins is not always the least popular. 
  I too share that lack of trust.

Keep in mind that lots of voter don't trust the idea that the winner of 
all the pairwise contests is always the most popular.

 > But then clone independence is not important after all because the
 > methods are ugly. I can't quite determine whether clone independence is
 > important or not.

It's important that the failure rate is small. But it doesn't need to be 
zero.

 > That's true. You implicitly need some kind of valuation of the different
 > failure rates. For instance, if you want LNHarm and LNHelp, you have to
 > give up either monotonicity or mutual majority. Which it's going to be
 > depends on what values you place on the different criteria.

I'm not concerned about monotonicity, LNHarm, LNHelp or any other 
failures that are difficult to exploit. I'm much more concerned about 
exploitable failures.

Admittedly, as a fan of Condorcet-Kemeny, I favor looking deep into the 
ballots, and I favor ways of "sorting" that basically move the biggest 
pairwise counts into one half of the usual matrix while moving the 
smallest pairwise counts into the other half, where the dividing line is 
the diagonal where candidates are paired with themselves.

 > Or to put it differently: if the method insists on a zero failure rate
 > for Condorcet loser, why shouldn't it insist on a zero failure rate for
 > Condorcet winner, say? And, equivalently, if "merely a low rate of
 > failure" is good enough for the Condorcet criterion (or say, clone
 > independence), why is it not good enough for Condorcet loser?

I admit I'm intentionally avoiding a zero failure rate for Condorcet 
winner because that makes the method into a Condorcet method, and those 
have been vilified (portrayed as evil) by the FairVote organization, and 
to some extent by STAR fans.

Plus, just as a voter is not likely to trust that the candidate with the 
fewest wins is least popular, they aren't likely to trust that the 
candidate who wins all the pairwise matches is most popular.

So at this point I'm still happy with eliminating the Condorcet loser as 
the top priority and otherwise eliminating the candidate who has the 
smallest pairwise support count (which basically counts how many 
remaining candidates are ranked below the candidate being scored).

At this point I continue to be open to suggestions for something better, 
but that window of time is closing very soon.

Again, thank you Kristofer for your wise feedback!

Richard Fobes
The VoteFair guy


On 7/24/2021 2:19 PM, Kristofer Munsterhjelm wrote:
> On 7/22/21 5:31 PM, VoteFair wrote:
>> On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote:
>>  > How about this?
>>  >
>>  > - Eliminate the candidate with the least number of winning subgroups.
>>  > - If there is a tie, break that tie by IRV.
>>  > ...
>>
>> Isn't the first step basically Copeland's method?
>
> No, because there's no elimination in Copeland (and it doesn't pass
> LIIA). It would just elect the candidate/s with the most winning subgroups.
>
>> That's an ugly "method" that fails to look beneath the surface.
>>
>> IRV also fails to look beneath the surface, which is why it too is an
>> "ugly" method.
>
> That leads me to wonder which is the case.
>
> You said you couldn't replace the IRV tiebreaker with minmax elimination
> because IRV is cloneproof and minmax is not -- that clone independence
> was important because it "protects against money-based vote splitting
> tactics". So I found something that invokes IRV's clone independence
> more often.
>
> But then clone independence is not important after all because the
> methods are ugly. I can't quite determine whether clone independence is
> important or not.
>
>>  > But again, the ungrouped mechanic is not cloneproof.
>>
>> Being cloneproof is not a goal. The goal is to have a very small
>> failure rate for clone independence.
>
> Then you could check the alternatives by that metric. A method seeming
> ugly may not necessarily have any bearing on the rates of failure.
>
>> Also, electing the Condorcet winner is not a goal. The goal is to have
>> a very small Condorcet criteria failure rate.
>>
>> To repeat my concern, attempting to get a zero failure rate will cause
>> other kinds of failure rates to increase.
>
> That's true. You implicitly need some kind of valuation of the different
> failure rates. For instance, if you want LNHarm and LNHelp, you have to
> give up either monotonicity or mutual majority. Which it's going to be
> depends on what values you place on the different criteria.
>
> The same would hold for rates. Say you want to find the method that
> minimizes w * x, where x is the rates of each failure type
> (monotonicity, vote splitting, teaming, crowding, favorite betrayal...).
> Then the weights of the w vector provide a measure of indifference: how
> much of failure type 1 is an acceptable trade for one unit of failure
> type 2?
>
> Or to put it differently: if the method insists on a zero failure rate
> for Condorcet loser, why shouldn't it insist on a zero failure rate for
> Condorcet winner, say? And, equivalently, if "merely a low rate of
> failure" is good enough for the Condorcet criterion (or say, clone
> independence), why is it not good enough for Condorcet loser?
>
>> I'm still willing to consider improvements, but it needs to find a
>> balance between what voters can understand -- both through an animated
>> video and through words -- and what yields low failure rates.
>>
>> Again, thank you Kristofer for applying your clear understanding to
>> this revision from RCIPE 1 to RCIPE 2.
>
> You're welcome :-)
>
> -km