[EM] RCIPE version 2

[Richard Lung]

 A few comments from Richard Lung (not the VoteFair guy, (who is not to be confused, if I remember rightly, with Santucci, the vote guy!). 

As pointed out to Susan Simmons, which she acknowledged, eliminating candidates, during the count, loses voting information, before the count is over. It is not necessary with a binomial count, unlike all existing methods (uninomial counts) which employ elimination as an afterthought to an essentially uninomial election count. 
I repeat, in case you missed it, science demands one truth (to aspire to) not two. Therefore an election count and an exclusion/elimination count must be symmetrical. Call it symmetrical count requirement. But that is a binomial count.
And a binomial count indeed does imply higher order counts, governed by the binomial theorem. But a simple coherent first order binomial count should be sufficient for democratic representation. 
Whereas, FAB STV is the whole caboodle perhaps relevant to data mining. This system is monotonic, not vulnerable to strategic shuffling the preference orders. It avoids premature exclusion, and indeed premature election! It entirely avoids later harm, not just for transfer of surplus preferences.
It meets the Laplace condition of weighting a whole range of preferences in order of importance, unlike Condorcet pairing, whether or not the pairs are weighted in relative importance.
FAB STV recognises elections as statistical estimates of representation, and employs up to four averages to maximise accuracy. It accepts the "Impossibility" of a deterministic election result, and moves on. I beseech you all to do the same!

Yours sincerely,
Richard Lung.




On 25 Jul 2021, at 4:30 pm, Richard, the VoteFair guy <electionmethods at votefair.org> wrote:

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 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
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