[EM] (4) MAM better than VoteFair: Steve's 4th dialogue with Richard Fobes

[VoteFair]

On 10/17/2015 10:34 AM, steve bosworth wrote:
 > ...
 > 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 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.

In a U.S. presidential election the Condorcet-Kemeny method would NEVER 
produce a result that would be different from any other (reasonable) 
pairwise-counting method!

Please re-read my previous message.  As I tried to explain, the example 
that Steve Eppley wrote is purely mathematical, and would NEVER occur in 
any REAL election -- of any type, with any number of voters.

 > At the same time, do you agree that this vulnerability of KY also gives
 > us the following related reasons for seeing MAM as superior:

This vulnerability is very, very tiny, and is of about the same 
magnitude as the vulnerabilities of other good voting methods.

 > 1)Because KY is vulnerable to clone attacks, this can provide an
 > incentive for strategizers to nominate as many clones as they can afford.

If voters strategize, they would not try to create a situation of 
circular ambiguity.  Instead they would simply mark their ballots in a 
way that produces the result they want.  They would intentionally AVOID 
circular ambiguity because it makes the outcome far less predictable. 
Circular ambiguity is a fringe/edge effect where the differences between 
pairwise-counting methods occur.  (BTW, the above comments about 
strategy also apply to other pairwise-counting methods.)

 > 2)These additional candidates would both needlessly make the ballots
 > more complicated and increases the fear that these extra candidates may
 > produce unfair results.

The full VoteFair ranking method simplifies the ballot.  It eliminates 
candidates who have almost no chance of winning (based on 
political-party popularity determined from earlier elections).

 > 3)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.

VoteFair ranking software calculates even a "complex" election in less 
than a couple of minutes (often within seconds).  The complexity does 
not depend upon the number of candidates.  If someone says that this is 
not so, then they are referring to blindly calculating a sequence score 
for every possible sequence without first looking at the ballots to see 
that some candidates have absolutely no chance of winning.

 > 4)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.

Not true.  The Condorcet-Kemeny method has lots of credibility because 
it calculates very(!) fair results.

Actually the Condorcet-Kemeny method is easier to explain than, say, the 
Condorcet-Schulze method (which is "cloneproof").

I have not studied the MAM method beyond its use of pairwise counting 
because it is a new method, and I don't recall seeing it in Wikipedia.

The usual arguments on this forum discuss whether Condorcet-Kemeny or 
Condorcet-Schulze or Ranked Pairs is a better method.  If you are 
ignoring those methods in favor of the newcomer MAM then you have 
wandered into the middle of a conflict that involves far more complexity 
than your understanding of a single difference between Condorcet-Kemeny 
and MAM.

Again, for use in U.S. presidential elections, any pairwise-counting 
method will ALWAYS produce the same results!

In haste,

Richard Fobes


On 10/17/2015 10:34 AM, steve bosworth wrote:
> To-Richard Fobes-V-F4
>
> Re: MAM better than VoteFair: Steve's 4th dialogue with Richard Fobes
>
>>  Date: Thu, 15 Oct 2015 15:52:00 -0700
>>  From: ElectionMethods at VoteFair.org
>>  To: election-methods at lists.electorama.com
>>  CC: stevebosworth at hotmail.com
>>  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 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:
>
> 1)Because KY is vulnerable to clone attacks, this can provide an
> incentive for strategizers to nominate as many clones as they can afford.
>
> 2)These additional candidates would both needlessly make the ballots
> more complicated and increases the fear that these extra candidates may
> produce unfair results.
>
> 3)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.
>
> 4)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:
> http://MAM.hostei.com
>  > > <http://mam.hostei.com/>
>  > > 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
>  >
>