FW: (4) MAM better than VoteFair: Steve's 4th dialogue with Richard Fobes
From: stevebosworth@hotmail.com
To: electionmethods@votefair.org
Subject: (4) MAM better than VoteFair: Steve's 4th dialogue with Richard Fobes
Date: Sat, 17 Oct 2015 17:34:28 +0000
Re: MAM better than VoteFair: Steve's 4th dialogue with Richard Fobes
Date: Thu, 15 Oct 2015 15:52:00 -0700
Subject: Re: ?MAM better than VoteFair? Steve's 3rd dialogue with Richard
Fobes
Dear
Richard,
Putting
aside our dialogue about APR which we have not yet completed, we agree that some
form of Condorcet method would be best for electing the US President, e.g.
either Kemeny-Young (VoteFair) or MAM. Perhaps we
also agree that in contrast to MAM, KY can produce flawed results because of
cloning, even if only ‘rarely’ as you put it.
This is confirmed by Steve Eppley’s KY example below in which the
adding of clone D to C resulted in the election of B rather than A.
At the same time, do you agree that this vulnerability of KY also gives
us the following related reasons for seeing MAM as superior:
Because KY is vulnerable to clone
attacks, this can provide an incentive for strategizers to nominate as many
clones as they can afford.
These additional candidates would both needlessly
make the ballots more complicated and increases the fear that these extra
candidates may produce unfair results.
These additional candidates would also
put an extra burden upon the KY computer program needed to calculate all the
possible sequences of winners. Current
computer technology might not even be able to cope with the resulting ‘combinational
explosion’ for a US presidential election.
Opponents of any electoral reform could
use all these arguments to undermine the credibility of KY. MAM’s relative simplicity would not be
vulnerable in these respects.
What do you think?
Best regards,
Steve Bosworth
See Steve Eppley’s example below:
On 10/14/2015 10:41 AM, steve bosworth wrote:
[...] below, Steve Eppley has just now relevantly replied to my
questions about MAM and VoteFair. As a result, do you see any reason
not
to prefer MAM over VoteFair popularity ranking given Eppley's
explanation of how VoteFair (Kemeny-Young) is both vulnerable to
clone
attacks and requires a ‘combinatorial explosion of possible orders of
finish’ when used to elect one of many candidates?
For the purposes of electing a US president, the Condorcet-Kemeny
method, the Maximize Affirmed Majorities (MAM) method, the
Condorcet-Schulze method, the Ranked Pairs method, and any of the
pairwise-counting (especially Condorcet) methods, would ALWAYS produce
the same result!
S: No, you seem to be
ignoring Eppley’s example in which the adding of clone D to C resulted in the
election of A rather than B.
R: > The situations in which these
methods can produce different results [are more likely to]
involve a much smaller number of voters.
When a large number of voters are involved, or when a very large number
of voters are involved (as in US presidential elections), the conditions
that can create differences between these methods [are much less likel] do
not occur.
Differences between these methods can arise when ballots have circular
ambiguity. This circular preference is analogous to the concept in the
game "rock paper scissors." In politics this means that about
one third
of the voters would mark a 1-2-3 ballot with something like Trump, Bush,
Clinton, and another third of the voters would mark their ballot Bush,
Clinton, Trump, and another third of the voters would mark their ballot
Clinton, Trump, Bush. Please read these sequences carefully and notice
that other specific ballot preferences -- namely Trump, Clinton, Bush,
and Bush, Trump, Clinton, and Clinton, Bush, Trump -- are missing. Such
preferences do not occur in political situations that involve a large
number of voters. Yet this is the unusual pattern that is presented in
the example below where the Condorcet-Kemeny method is being criticized.
As a further perspective about the number of ballots involved, in the
American idol polls that I host at VoteFair.org, circular ambiguity
sometimes shows up after a few people have voted, but circular ambiguity
always disappears when the number of ballots increases.
Of course from a mathematical perspective, if a majority of US voters
intentionally want to create a situation that has circular ambiguity --
that is, none of the candidates is a Condorcet winner (which means that
none of them is pairwise preferred over all the other candidates) --
then this situation can be achieved, but only if there is a high degree
of cooperation, such as marking ballots based on a central
recommendation that, in turn, is based on something like the first
letter of their last name.
To put this another way, keep in mind that vulnerability to clones only
affects the Condorcet-Kemeny method in rare circumstances. Emphasis on
the word rare, especially for political races.
The MAM method, the Condorcet-Schulze method, the Ranked Pairs method,
and other pairwise-counting methods also have vulnerabilities. But so
far no one has estimated the percentage of time in which these failures
occur. Expressed another way, current comparisons focus on pass-fail
criteria, without measuring how often, or how rare, those failures
occur. And new criteria are being added as voting becomes better studied.
To keep this discussion further in perspective, the gap between
pairwise-counting methods and instant-runoff voting is huge. This means
that if instant-runoff voting were used in a US presidential election,
the results could easily be quite different from the results of using
any of the pairwise-counting methods.
The fact that you want to use something like instant-runoff voting in
your APR method means that you are wise to recognize that a
single-winner election should use pairwise counting. The subtle
differences between pairwise-counting methods are far less important
within this perspective.
Regarding the computation time for the Condorcet-Kemeny method, if there
were 130 candidates in the same race (as happened when Arnold
Schwarzenegger became the governor of California), a cursory look at the
numbers would reveal which candidates have a possibility of winning, and
that would reduce the race to no more than ten candidates, and the
computations for ten candidates can be done within a few minutes using
just one computer. And that's using the slow approach of calculating a
ranking score for every possible sequence of those top candidates. When
the calculations are done efficiently, the elapsed time is just a few
seconds.
As long as you don't promote approval voting or instant-runoff voting
for use in US presidential elections I'm not going to argue about which
kind of pairwise-counting method should be used in US presidential
elections. (Here I'm talking about general elections. Approval voting
would be acceptable for primary elections.)
My bigger concern is using better methods for electing legislators and
parliament members and Congressmen. For that I've developed other
aspects of VoteFair ranking that go way beyond single-winner methods.
Again, I hope this feedback is helpful.
(I think I've caught up with answering your questions. If not, please
repeat the question.)
Richard Fobes
On 10/14/2015 10:41 AM, steve bosworth wrote:
Dear Richard,
[….]
To illustrate how Kemeny-Young and MAM work, here's a pair of
examples
that together also serve to illustrate the Independence of Clone
Alternatives criterion:
Example 1. Suppose there are 100 voters and 3 candidates A, B and C.
Suppose the voters' top-to-bottom rankings are:
40 35 25
A B C
B C A
C A B
Kemeny-Young defines the "best" order of finish as the
order of finish
that minimizes the sum of voters' pairwise preferences that the order
"disagrees" with.
There are three majorities:
75 rank B over C (with 25 opposed).
65 rank A over B (with 35 opposed).
60 rank C over A (with 40 opposed).
KY says the best order of finish is ABC. ABC disagrees with the 25
who
rank C over B and the 35 who rank B over A and the 60 who rank C over
A.
Thus the sum of pairwise preferences that ABC disagrees with is 120.
(All other possible orders of finish have a larger sum, and are
therefore worse, according to KY.) Since ABC is the order of finish,
the
winner is A.
Example 2: Same scenario as example 1, but suppose a clever voter who
favors B decides to also nominate a fourth alternative D that's
similar
to C (but slightly inferior to C). Now the voters' top-to-bottom
rankings are:
40 35 25
A B C
B C D
C D A
D A B
(Actually, let's not assume that every voter ranks C over D. Placing
D
below C in all the votes above made it easier for me to write the
votes.)
Now there are six majorities:
C is ranked over D by a majority of size x. (x doesn't matter but
let's
suppose it's largest.)
75 rank B over C.
75 rank B over D.
65 rank A over B.
60 rank Cover A.
60 rank D over A.
Now KY says the best order of finish is BCDA. BCDA disagrees with] D
over C and the 25 who rank C over B and the 25 who rank D over B and
the
65 who rank A over B and the 40 who rank A over C and the 40 who rank
A
over D, which sum to 295-x [i.e. 295 – 100]. ABCD is not the best
order
of finish because it disagrees with [35 who rank B over A] the 100-x
who
rank D over C and[ [the 25 who rank C over B] and the 25 who rank D
over
B and the [60 who rank C over A] 35 who rank B over A and the 60 who
rank D over A 60 who rank C over A, which sum to 305-x. The larger
sum
makes ABCD worse than BCDA, according to KY.
Now the KY winner is B. The clever voter who favors B changed the
winner
by nominating D. When small groups vote, typically the rules allow
one
or two people to nominate an alternative, which means it's easy for a
tiny minority to manipulate the outcome when a voting method isn't
independent of clones.
In example 1, MAM agrees that ABC is the best order of finish. In
example 2, MAM says the best order of finish is ABCD. (With MAM the
clever voter fails to change the winner.) Here's how MAM works:
MAM constructs the order of finish a piece at a time, by considering
the majorities one at a time, from largest majority to smallest majority:
For each majority, MAM places their more-preferred candidate ahead
of their less-preferred candidate in the order of finish, unless MAM
has already placed their less-preferred candidate ahead of their
more-preferred candidate (due to the "transitivity"
property of an
order of finish, as the following will explain).
In example 1, the largest majority rank B over C, so MAM places B
ahead
of C in the order of finish:
BC
The second largest majority is A over B, so MAM next places A ahead
of B
in the order of finish:
ABC
Because A finishes ahead of B and B finishes ahead of C, it follows
that
A finishes ahead of C too. (This is the transitivity property I
mentioned.)
The third largest majority is C over A, but MAM does not place C
ahead
of A because A is already [transitively] ahead of C.[A>B>C is
more
popular this sequence is only opposed by 60 preferences while
C>A>B is
opposed by 75 + 65 = 140 preferences.
At the same time, this 140 is the
‘Maximal Affirmed Majority’ for the finishing sequence of
A>B>C]
The completed order of finish is ABC, and MAM elects A.
In example 2, let's assume the largest majority is the voters who
rank C
over D (although this doesn't matter). Thus this is the first
majority
that MAM considers, and MAM places C ahead of D in the order of
finish:
CD
Next MAM considers the second largest majority, which is either the
75
who rank B over C or the 75 who rank B over D. Later I'll discuss how
MAM chooses between majorities that are the same size; for now let's
assume MAM decides the second largest majority is the 75 who rank B
over
D and the third largest majority is the 75 who rank B over C.
Since the second largest majority rank B over D, MAM now places B
ahead
of D in the order of finish:
CD
BD
Since the third largest majority rank B over C, MAM next places B
ahead
of C in the order of finish:
BCD
[Again, for simplicity, I have picked the follow majority as the 4^th
one even though it is also tied with the 65 who ranked D>A, the 65
who
ranked B>C, and the 65 who ranked B>D.]
The fourth largest majority [I have similarly chosen for simplicity]
is
the 65 who rank A over B, so MAM places A ahead of B in the order of
finish:
ABCD
Note that because A is ahead of B and B is ahead of C & D, this
means A
is ahead of C & D too. (The transitivity property.)
(The order of finish is now a linear ordering, ABCD, so if we wished
we
could stop here and elect A. But let's keep going.)
The fifth largest majority is either the 60 who rank C over A or the
60
who rank D over A. Again I'll defer an explanation of this
"tiebreaking"
of same-size majorities; for now let's assume MAM considers the fifth
largest majority to be the 60 who rank C over A and the sixth largest
majority to be the 60 who rank D over A.
The fifth largest majority rank C over A, but since A is already ahead
of C in the order of finish, MAM does not place C ahead of A.
The sixth largest majority rank D over A, but since A is already
ahead
of D in the order of finish, MAM does not place D ahead of A.
The completed order of finish is ABCD, so MAM elects A. Nominations
of
similar alternatives (D, which is similar to C) don't change the
outcome
with MAM.
There are other criteria satisfied by MAM but not by KY, but as far
as I
know the only criterion satisfied by KY but not by MAM is the Weak
Reinforcement criterion I mentioned above. When Peyton Young promoted
KY
in his papers and at least one book (Equity in Theory and Practice),
he
touted a criterion he called Local Independence of Irrelevant
Alternatives... which is satisfied by MAM.
(There's another criterion called Reinforcement by Donald Saari,
which I
call Strong Reinforcement, that neither KY nor MAM satisfy: If two
collections of votes separately elect X, then X must be elected if
the
votes are combined. The Borda method satisfies it, and so does
Plurality
Rule. Like Weak Reinforcement, I believe Strong Reinforcement is
unimportant, because it's simple to have a rule that prevents a
minority
from dividing the voters into groups.)
Another problem with KY is that it takes a long time to find its best
order of finish when there are many candidates. There's a
"combinatorial
explosion" of possible orders of finish, and each one needs to
be
checked. The difficulty in checking all possible orders of finish, to
calculate each one's sum, is why I didn't include KY in software I
wrote
years ago that compares voting methods head-to-head to see which
methods' winners are preferred by more voters. (My software generates
a
series of random votes, tallies the votes using two voting methods
and
accumulates statistics. I presume James Green-Armytage used similar
software to conduct his analysis.)
Regarding tiebreaking in MAM:
There are two places in MAM's algorithm where tiebreaking can be
required. One is when majorities are the same size, as mentioned
above.
The other can result from a tied pairing; the simplest example is a
two-candidate election in which the voters split 50/50. In this case,
consideration of all the majorities may not suffice to place one
candidate ahead of the other. (Transitivity may do it [as it might be
as
illustrated by the following example:]. For example: suppose the
"A
versus B" pairing is tied, and [at the same time there is] a
majority
rank A over C, and a majority rank C over B. [These additional
pairings
produce: A>C>B.] The order of finish after MAM considers
the[se] two
majorities is ACB, and no tiebreaking is needed to resolve A versus
B.)
In an election with many voters, it will be very rare that two
[pairing]
majorities are the same size, and [thus] it will be very rare that a
pairing is tied. So the simple definition of MAM provided above would
typically suffice for public elections.
The tiebreak procedure has been carefully chosen so that MAM
completely
satisfies criteria such as Independence of Clone Alternatives, Strong
Pareto, etc. In elections with many voters, tiebreaking would be
extremely rare so the [complete satisfaction of these criteria would
not
be in doubt’ completeness of the criteria satisfaction would not
matter,
but in voting by small groups it's important to [provide a tiebreaking
method that would] completely satisfy Independence of Clones. With
any
voting method that doesn't completely satisfy Independence of Clones,
elections in small groups can degenerate into farces, because each
faction will always have an incentive to nominate a huge number of
clones because there's a chance it can help and it can't hurt.[SB:It
can
help by effectively multiplying the size of the ‘majority’ of the
most
favored clone over the other clones.??????The more supports of the
favored clone explicitly favor her over the all of her slightly
inferior
clones, the greater will be her majority.This enables each
strategizer’s
vote to count (have more weight) than each other voter.]
The tiebreak procedure begins the same way for both same-size
majorities
and resolving tied pairings. A tiebreak ordering of the candidates is
constructed by randomly picking [one ballot]at a time, and including
into the tiebreak ordering all of [that] ballot's preferences that
don't
conflict with the preferences already included from the ballots [that
may have been] previously picked [in order to break any earlier
ties.This procedure is repeated] until the tiebreak ordering is
complete. Here's a simple example: Suppose there are 3 voters and 4
candidates A,B,C,D, and suppose the voters' top-to-bottom rankings
are
as follows:
A B C
B=C C D
D A
B
The middle vote [Voter 2 has] omitted A & D. MAM treats it the
same as
if the voter had [explicitly] ranked A & D equally at the bottom:
A B C
B=C C D
D A=D A
B
MAM picks one of the votes at random. Suppose it's the middle vote.
MAM
includes its preferences into the [1^st suggested] tiebreak ordering,
which becomes BC(A=D), identical to the [ordering of this picked
voter].
The tiebreak ordering isn't yet complete because it's not a linear
ordering (it ranks A & D the same) so MAM picks another vote at
random.
Suppose it's the left vote. The left vote ranks A over D, so MAM
includes the A over D preference into the tiebreak ordering, which
[now]
becomes BCAD. Now the tiebreak ordering is a linear ordering, so
there's
no need to pick any more votes [at random].
Let's return to example 2, which has two pairs of same-size
majorities:
two majorities are 75 voters, and two majorities are 60 voters. When
MAM
reaches the point at which it wants to consider the second largest
majority, it sees that two majorities have 75 voters. So it checks
the
sizes of their opposing minorities. If one of the same-size
majorities
is opposed by a smaller minority, then that would be the majority
that
MAM considers next. In example 2, though, the opposing minorities are
both the same size (25). So MAM picks a vote at random to start
constructing the tiebreak ordering. Suppose the vote that MAM picks
is
one of the 35 votes that rank B over C over D over A. All of this
vote's
preferences are included into the tiebreak ordering, which becomes
BCDA
(identical to the vote that was picked). The tiebreak ordering is now
a
linear ordering, so no more votes need to be picked. Next MAM
compares
the two same-size majorities by checking the positions of their
less-preferred candidates in the tiebreak ordering: The
less-preferred
candidate of the 75 who rank B over C is C, and the less-preferred
candidate of the 75 who rank B over D is D. Since D is behind C in
the
BCDA tiebreak ordering, MAM will consider [and give priority to] the
75
who rank B over D before considering the 75 who rank B over C. In
other
words, the second largest majority is the 75 who rank B over D, and
the
third largest majority is the 75 who rank B over C.
Later, when MAM reaches the point where it wants to consider the
fifth
largest majority, it finds 60 who rank C over A and 60 who rank D
over
A. In both of these same-size majorities, the less-preferred
candidate
is A. So MAM checks the positions of their more-preferred candidates
in
the tiebreak ordering. The more-preferred candidate of the 60 who
rank C
over A is C, and the more-preferred candidate of the 60 who rank D
over
A is D. Since C is ahead of D in the BCDA tiebreak ordering, MAM will
consider [and give priority to] the 60 who rank C over A before
considering the 60 who rank D over A. In other words, the fifth
largest
majority is the 60 who rank C over A, and the sixth largest majority
is
the 60 who rank D over A.
In the case where a pairing is tied and not resolved by transitive
majorities, MAM uses the [??? following]and give priority tiebreak
ordering to break the tie(s) in the order of finish. Suppose there's
a
5-candidate election and that after all the majorities have been
considered, the order of finish is (A=B)C(D=E). In other words, A
& B
are tied for first place, and D & E are tied for last place. MAM
will
pick votes at random to construct a linear tiebreak ordering as
described above. Suppose the tiebreak ordering is CEBDA. MAM breaks
the
ties beginning with the most significant tie, in this case A versus
B.
Since B is ahead of A in the tiebreak ordering CEBDA, MAM places B
ahead
of A, and the order of finish becomes BAC(D=E). The next tie to break
is
D versus E. Since E is ahead of D in the tiebreak ordering CEBDA, MAM
places E ahead of D, and the order of finish becomes BACED. MAM
always
constructs a linear order of finish, resolving all ties.
Because voters can express indifference between candidates, it's
theoretically possible that every vote will be indifferent between
two
candidates. (Identical twins?) In other words, the procedure of
randomly
picking votes may fail to produce a linear tiebreak ordering even
after
every vote has been picked. In this case, the incomplete tiebreak
ordering is completed (made linear) by randomly completing it. Then
any
ties can be broken.
There's a shortcut in the tiebreak procedure that can save labor in
some
cases, but it's not worth discussing the shortcut here. (Here's a
hint:
the tiebreaking ordering only needs to be as linear as necessary.)
Did I answer all your questions?
Would it help if I provide source code for MAM written in the Ruby
programming language? I tried to write it in a way that will make it
as
easy as possible for people unfamiliar with Ruby to understand it.
It's
not a complete program; it doesn't have the routines needed to input
votes or parse the votes, but those routines aren't unique to MAM;
they're needed in any pairwise voting method.
Also, you can freely use the online MAM server at:
It has a data entry box into which you can paste all the votes.
Clicking
a button underneath the data entry box causes the server to tally the
votes and construct the order of finish. You can paste all the votes
in
a single copy/paste operation. Here's example 2 in a format ready to
be
copy/pasted:
40: A B C D
35: B C D A
25: C D A B
There are a few interesting examples at the bottom of the webpage,
that
can be copied/pasted. And of course you can invent your own. Or use
real
votes from real groups of people.
There's at least one other website that supposedly provides the
ability
to tally votes using MAM, but I haven't seen their software and they
never asked to consult with me on the details. Mike Ossipoff once
showed
me an order of finish that had a tie in it, which means it could not
be
a correct implementation of MAM.
Best wishes,
Steve Eppley