[EM] recommendations

[Kevin Lamoreau]

James G.-A., you wrote:

> 	Are you aware of Tideman's ranked pairs method? I believe that it is very
> similar to MAM. To be honest, I don't think about tiebreaking nuances,
> because to me, a tie should just be called a tie. (Not a majority rule
> cycle, but an actual tie given the basic rules of the method... I'd rather
> call it a tie than make a new rule to resolve it randomly.)

Yes, I am aware of Tideman’s ranked pairs method.  I supported it for a while, 
and I may be veering back almost to it.   I’m pretty easily swayed: since sometime 
during the fall semester of 2003, my favorite single-winner voting method has 
gone from IRV (not before I had thought about what turned out to be Condorcet 
voting methods but before I realized there was significant support behind them) to 
Shultze to ranked pairs to plurality (I was drawn to the fact that the most a voter 
could do voting-wise to help elect a candidate was to rank that candidate first => 
vote for that candidate, and that the only way that changed/added votes could 
result in one candidate being elected over another is if at least one vote was 
changed/added to have the first candidate ranked higher with respect to the second 
candidate - this of course is accomplished by only allowing a voter to rank one 
candidate first and requiring that voter to effectively rank all other candidates 
equally) to my “locked preferences” method and possibly now to the margins 
equivalent of locked preferences (more on that transition below).  Besides the 
margins v. winning votes divide, the only difference between Tideman’s method 
and MAM seems to be the tiebreaker used.  Steve Eppley did a good job pointing 
out weaknesses in two tiebreakers Tideman had proposed in his detailed 
definition of maximize affirmed majorities ( 
http://alumnus.caltech.edu/~seppley/MAM%20procedure%20definition.htm ), but 
Tideman and TN Zavist announced the latter of those two tiebreakers in 1989 and 
that may no longer be the tiebreaker that either of them are advocating for use in 
the ranked pairs method.

> 	You are advocating winning votes rather than margins, am I right? ...

You were right, at least, but I may be shifting my advocacy back margins.  Chris 
Benham has referred me to a page on Blake Cretney’s website on winning votes 
v. margins ( http://condorcet.org/rp/inc.shtml ), which I found very persuasive.  At 
least some of the potential benefits of winning votes over margins seem to 
become moot if voters have a good understanding of whatever method is being 
used.  The use of margins seems more intuitive to me than that of winning votes, 
and the effect of ranking two candidates equally seems to be more often and more 
thoroughly halfway between the effect of ranking either candidate ahead in 
margins than in winning votes.  If I do decide to shift to margins, I would support 
a ranked pairs-type method with pairwise ties counting as a defeat of each 
candidate involved over the other of strength 0 and resolved, if necessary, along 
with any other cases of defeats of equal strength (and in one fell swoop) by Steve 
Eppley’s random ballot tiebreaker.  This would require me to give up having all 
preferences sorted, unless I wanted to include “defeats” of negative strength, 
which would be silly.

> 	On the subject of single-winner methods, I try to plug my "weighted
> pairwise" method every chance I get. So here I am plugging it again: I'd
> like you to read the proposal and see what you think.
> http://fc.antioch.edu/~jarmyta@antioch-college.edu/voting_methods/weighted_pairwise.htm

I read your proposal (or at least most of it – I didn’t read everything in order and 
didn’t read every word, but I read enough to know exactly what you are 
proposing).  You made some good points but I’m afraid I was unconvinced.  One 
aspect of your plan seems disjointed to me.  The first of your “additional 
provisions” states, “If the winning side of one defeat constitutes a majority (of the 
valid vote), and the winning side of another defeat does not constitute a majority, 
then the majority defeat is necessarily considered to be stronger. Otherwise, the 
weighted magnitude is always the determining factor in relative defeat strength.”  
I think if weighted magnitude should be the determining factor in the relative 
defeat strength of a 53% to 44% pairwise win and a 51% to 46% pairwise win, 
then it should also be the determining factor in the relative defeat strength of a 
51% to 46% pairwise win and a 49% to 48% pairwise win.  I used those pairs of 
percentages, all of which leave 13% of the vote for equal rankings, so as to reach 
out to backers of both margins and winning votes while making my point.  That 
said, I don’t think weighted magnitude should be used at all.  I don’t have any 
mathematically-based reason why.  I think it may give voters a greater range of 
options then they are willing to take the time to make.  You would likely have 
some voters who rank one candidate ahead of another but give the second 
candidate a higher rating.  While the introduction of any new voting method 
would likely lead to the results of some elections being taken to court, as 
happened in Ann Arbor Michigan when IRV was introduced and the plurality 
winner in the three-way mayoral election sued after narrowly losing when the 
third candidate’s votes were redistributed, I think the introduction of your method 
would lead to multiple court cases and Florida 2000-type controversies.  I must 
admit, though, that some of my aversion to the use of ratings in Condorcet 
completion methods is ascetic, which probably stems from years of thinking 
mostly about purely ordinal voting methods.  Still, you deserve credit for 
exploring an area that few, if any, have explored in conjunction with Condorcet 
methods.

In some possible outcome-based Condorcet-flavored multiple-winner election 
methods, ratings could be useful in determining what outcome in a pairwise 
contest to attribute the voter’s votes to if it could not be determined based on the 
voter’s ordering of the candidates.  If, for example, in a method like CPO-STV 
except that it is non-proportional, if two outcomes varied by two candidates and a 
voter preferred one of the candidates in just one of the outcomes ahead of both of 
the candidates just in the other outcome, who the voter in turn preferred over the 
other candidate who was just in the first outcome, the ratings of the two 
mentioned candidates in each outcome could be added up and the voters vote in 
that pairwise contest attributed to the outcome whose candidates had the higher 
combined rating.  If the combined ratings of the two candidates were equal, then 
the voter would be deemed to have no preference between the two outcomes.  
Obviously, if the mth ranked candidate of those who are in one of the outcomes 
but not the other was ranked greater (and perhaps, even equal if at least one 
candidate was ranked greater) than his/her counterpart in the other outcome, than 
the voter’s vote would go to the first outcome.  Come to think of it, you could 
make this method arguably as proportional as CPO-STV by having votes that rank 
a candidate in both outcomes first (among the candidates in either or both 
outcomes) go to that candidate initially.  If any candidates had more votes than 
the quota, you would use your favorite surplus method to determine the portion of 
each vote that is transferred, and those portions would go to one outcome in the 
pairwise contest or the other or neither based on what I have described above.  

> >Moving on to multiple-winner elections, ideally I would 
> >support CPO-STV, in pure form except that equal rankings 
> >would be allowed (how that would work is a topic for 
> >another post).  
> 
> 	Yes, do tell. Probably have to divide votes into fractions, right?

In some cases yes, in some cases no.  When deciding how to handle equal 
rankings in CPO-STV, one should first think about how CPO-STV works.  In 
each comparison of outcomes, there are effectively only n + 2 candidates, with n 
being the number of candidates in both outcomes.  The candidates are those n 
candidates, the “candidate” representing each voter’s top preference among the 
candidates in one of the two outcomes but not the other, and the “candidate” 
representing each voter’s top preference among the candidates only in the other 
outcome.  Since the votes of candidates who are only in one of the two outcomes 
are never transferred in that comparison, I see no need to resolve equal rankings 
that occur below any candidates that are only in one outcome (and this includes, 
for any vote which ranked more than two candidates, any portion of that vote for 
which a candidate who is only in one outcome has been given a ranking above 
candidates who rank remains equal).  In fact, for since they are immaterial to the 
outcome, I propose that, in any comparison of outcomes, the preferences of any 
voter below the voter’s top-ranked candidate or candidates among those who are 
only in one outcome not be tabulated at all.  That would be more consistent with 
the decision not to resolve equal rankings below the top-ranked candidate or 
candidates in only one slate.  It could also save the need for additional resolution 
of equal rankings.  If it isn’t clear by now, I believe equal rankings should be 
resolved separately for each comparison of outcomes.  Now let’s move on to how 
for how to resolve the equal rankings the resolution of which can affect the 
outcome:

In a given comparison of outcomes, if two or more candidates are ranked mth 
place (among candidates in either or both outcomes) and all of them are in the 
same outcome and no other outcome, then it really doesn’t matter what fraction of 
the vote is counted as an mth place ranking for each candidate as long as all of the 
vote is counted as an mth place vote for one of the candidates who are just in that 
outcome.  What is most intuitive, however, is for the mth place rank to be split 
evenly among each candidate who is ranked tied for mth place, so that is what I 
propose.  If two or more candidates are ranked mth place and all of them are in 
only one outcome but there is at least one such candidate in each outcome, then 
that means the top-ranked candidate or candidates who are in each outcome other 
than the ones who are in both outcomes are ranked equally, since we have already 
established that these rankings would not even be tallied if any candidates who 
were only in one of the outcomes had been ranked higher.  Since votes which rank 
two candidates equally are not counted for either candidate in a pairwise contest 
between the two candidates in Condorcet, I propose that in the case in question no 
candidate be awarded with the mth place rank.  However, since the voter did give 
those candidates, who are all in only one of the two outcomes, the mth place rank 
(after candidates not in either outcome were excluded), and in order to be 
consistent, none of the voter’s preferences below mth place should be counted.

If two or more candidates who are all in both outcomes are ranked mth place, then 
each candidate should be ranked mth place on an equal portion of the vote.  If n 
candidates (n>=1) who are all in both outcomes and at least one candidate who is 
only in one of the two outcomes are ranked mth place, then each of the equally 
ranked candidates who are in both outcomes should be given the mth place 
ranking in 1/(n+1) of the vote, with the mth place rank for the remaining 1/(n+1) 
of the vote being given to the remaining candidate if there is only one such 
candidate, being divided equally among the remaining candidates if they are all in 
the same outcome, or being exhausted if the remaining candidates are in opposing 
outcomes.  Such a resolution is equidistant to how the mth place rank would be 
distributed if only one candidate who is in both outcomes was ranked mth place 
for all n such candidates who are ranked tied for mth place and how the mth place 
rank would be distributed if only a candidate or candidates who were in one 
outcome were ranked mth place or tied for mth place.  If any candidates are still 
ranked equally in any of the portions of the vote after the mth place rank has been 
decided, the (m+1)st place rank shall be decided separately for each portion of the 
vote (except any portion where the mth place rank went to a candidate in only one 
slate or where the vote was exhausted) using the appropriate formula.  This shall 
be continued until no candidates remain who are ranked equally (not counting 
those whose rank is not to be tabulated anyway), whence the STV process can 
begin.

If there are multiple instances of tied candidates neither of which occur after the 
top-ranked candidate or candidates who are only in one slate, the resolutions of 
those ties should be done independently of each other, and if more than one of 
such instances require division of the vote, the divisions resulting from separate 
instances should be done at 90 degree angles if you know what I mean.  I have not 
yet decided how to handle equal rankings in the preliminary STV count.  My first 
instinct is to apportion the appropriate ranks (like first and second in a two-way 
tie for first or third, forth and fifth in a three-way tie for third) equally and as 
independently as possible among the equally ranked candidates, but I am afraid 
that would lead to instances where a candidate was declared elected who might 
not be in the rightful winning outcome if no outcomes are excluded.  Perhaps also 
a candidate would be excluded from consideration under one of your (James) 
shortcuts who would actually be in the rightful winning outcome.  If someone 
could prove that such a resolution would never result in either of those two things 
happening, then I would support that resolution.  If that is not the case, however, 
it would be nice if someone could come up with a resolution that would never 
result in the elimination of the winning outcome with reasonable shortcuts but that 
will still result in most outcomes that can be proven not to be the winning 
outcome before you start comparing outcomes being excluded using reasonable 
shortcuts.

> 	Yes, slates are a good idea in any STV method. Note that I could rank a
> few individual candidates, then rank a whole party slate, then rank a
> couple more individual candidates, then another party slate after that,
> and so on. If you have already ranked a few candidates from a party ahead
> of the party slate, then of course they are not included the second time
> around. Hence, this can be a shortcut to bumping a candidate you like to
> the head of a slate. 
> 	Or, with a computer interface, you could insert a slate into your
> rankings and then move some of the individual candidates around as you
> like. Basically, slates can serve as a convenient shorthand without
> limiting voter options.

I hate to break it to you, but I was actually talking about voters voting for slates 
instead of candidates, rather than in addition.  I probably shouldn’t have called my 
proposal party-list CPO-STV but rather closed party list voting with seat 
allocation by CPO-STV.  The slates need not necessarily correspond to parties, 
though I should admit that in my mind they would except for independent 
candidates, which could form slates or not depending on their choosing.  From the 
time I first thought of it, which I believe was after reading your e-mail that started 
this recommendations thread, I had doubts about it, but I thought it would be neat 
for me to mention it on this listserv.  There are some things about party-list voting 
in general and closed party list voting in particular that attract me, but I don’t 
want to take the time to mention those things or get into a big debate about it so I 
won’t mention them.  Your suggestions regarding slates definitely have merit, but 
I feel that if the slates are not going to be rigid (where a voter has to rank 
successively each candidate in one slate before moving on to another slate except 
where a voter ranks all the candidates of more then one slate equally), then they 
should not exist at all.  Their existence could lead some voters to think that the 
slates paid more of a role in the election than they did.

One possible method I just thought of today, which I’m not really advocating for 
but I thought I’d mention, is open party list voting with seat allocation by CPO-
STV.  Voters would, if they chose, both rank the slates and rank the candidates in 
each slate, with the candidates in each slate being listed in a distinct section of the 
ballot.  “Party list CPO-STV” could be used to determine the number of seats 
each slate was entitled, and then a ranking of all the candidates could be 
determined by the voter’s votes (although there could be problems where voters 
ranked two or more slates equally but ranked the candidates in at least one of 
those slates), after which regular CPO-STV could be used to determine the 
winning outcome among all those with the right number of seats for each slate.

Equal rankings given by voters to slates in party list CPO-STV would be handled 
the same for the candidates (or positions, if you’re using the open party list 
variety) in the slates as equal rankings given by voters to candidates in regular 
CPO-STV, except that only the top as-yet unranked candidates in each slate 
which was ranked equally with another slate get a portion of that rank from that 
vote.  In each resulting portion of the vote, the next candidate in that slate, if there 
is one, shall the appropriate portion of the next rank out of that portion of the vote.

> >One 
> >might be able to use shortcuts in addition to those 
> >proposed by Mr. Green-Armytage without further 
> >compromising the integrity of the method.  
> 
> 	What did you have in mind?

It seems pretty clear to me, and could possibly be proven, that when you are 
allocating seats among the slates, if you take a set of outcomes that are alike 
except for the number of seats two slates received, then there will be no cycles 
within that set (and I count A>B, B>C but A=C as a cycle as I do A=B, B=C but 
A>C), there will be no cases of three or more outcomes in that set pairwise tying 
each other unless all outcomes in the set are pairwise tied, and if the outcomes in 
the set that differ by only one seat are placed next to each other on the x-axis on a 
graph and the reverse order of their rank out of the outcomes in the set is the y-
axis (so that the rank 1 is the highest), then the result will be either a mountain 
peak or straight line with the possible exception of a plateau at the top.  Thus, if 
you know that an outcome with two candidates elected from party A, two 
candidates elected from party B and one candidate elected from party C pairwise 
beats an outcome with two candidates elected from party A, one candidate elected 
from party B and two candidates elected from party C, then you can tell that each 
of those outcomes pairwise beats an outcome with two candidates elected from 
party A, no candidates elected from party B and three candidates elected from 
party C.  This could allow one to avoid doing some comparisons.  What I stated in 
the first sentence in this paragraph may be true of any straight line of outcomes, 
and if so that could allow one to avoid doing even more comparisons.

> my best,
> James

Thank you, and my best to you,

Kevin