[EM] ?MAM better than VoteFair? Steve's 3rd dialogue with Richard Fobes

[Richard Lung]

Hello election methods members, Vote Fair,

A while ago, on request by Kristofer, I introduced a method of Binomial 
STV, which, unless extensive testing prove otherwise, should be equally 
applicable to single as well as multiple seat elections, being a more 
general theory of choice, generalising on traditional STV and Meek 
method. It doesnt depend on Condorcet pairing, a points system 
(descendants of Borda method), or premature exclusion (IRV and 
conventional STV).

Yours sincerely,
Richard Lung.


On 15/10/2015 23:52, VoteFair wrote:
> 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!
>
> The situations in which these methods can produce different results 
> 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 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,
>>
>> I know you are trying to find the time to reply to my earlier email.
>> However, 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?
>> Regards,
>> Steve Bosworth
>> Dear Steve Eppley,
>>
>> Thank you for being so helpful in explaining both exactly how MAM works
>> and how it is superior to Kemeny-Young, i.e. KY’s vulnerability to clone
>> attacks and its requiring a ‘combinatorial explosion of possible orders
>> of finish’ when used to elect one of many candidates.Thank you for
>> giving priority to clarity over brevity.In this spirit, you may also
>> wish to consider some of the explanatory additions I have made [within
>> square brackets] in your text below.At least, these helped me to be
>> surer that I was understanding you correctly.As a result, for example, I
>> am now convinced that MAM should be used to elect the US President.I
>> know that this would require a US constitutional amendment.
>>
>>
>> On a different but related question, I would very much appreciate
>> receiving any criticisms or suggestions you might have regarding the
>> very different electoral method I propose for electing the 435
>> congresspersons, a change that would also require a suitable US
>> constitutional amendment. I call this ‘new’ electoral system
>> Associational Proportional Representation (APR) and its details are
>> explained in the attachment.
>>
>>
>> Briefly stated, APR would allow each citizen to guarantee that their one
>> vote will continue mathematically to count in the House of
>> Representatives through the ‘weighted vote’ given to the congressperson
>> they had helped to elect.Each weighted vote would be exactly equally to
>> the number of citizens in the whole country who had helped to elect each
>> rep.A modified single transferable voting (STV) method would firstly be
>> used during APR’s primary election to discover all the ‘electoral
>> associations’ (geographically defined or not) who would be allowed to
>> send congresspersons to the House.APR’s later general election would use
>> a similar STV method to discover how many ‘weighted votes’ each
>> ‘association’s’ elected candidate(s) would have in the House.APR also
>> makes a limited use of ‘Asset’ voting to guarantee that no citizen’s
>> vote need be wasted.
>>
>>
>> I am aware that because STV reduces to IRV, it is theoretically
>> vulnerable to strategic voting.However, I do not yet see how anyone
>> could acquire the necessary for knowledge of how other people in the
>> whole country will vote using APR to enable any strategizer to be
>> successful. Consequently, I assume that any such criticism of APR would
>> only an attempt to apply what you call ‘aesthetic consistency criteria’.
>>
>>
>> I look forward to your comments.
>>
>>
>> Best regards,
>>
>>
>> Steve
>>
>>
>> Stephen Bosworth
>>
>>
>> ------------------------------------------------------------------------
>>
>> Subject: Re: MAM: A Question from Steve Bosworth
>> To: stevebosworth at hotmail.com
>> From: SEppley at alumni.caltech.edu
>> Date: Sun, 4 Oct 2015 22:25:23 -0400
>>
>> Hi Steve Bosworth,
>>
>> MAM isn't the same as Kemeny-Young unless there are fewer than 4
>> candidates. (Many pairwise voting methods behave the same with fewer
>> than 4 candidates; another method that behaves the same with fewer than
>> 4 is Schulze's method.)
>>
>> A criterion that MAM satisfies but KY doesn't is Independence of Clone
>> Alternatives. (If you're unfamiliar with Independence of Clone
>> Alternatives, the pair of examples below will help to explain it.) A
>> criterion that KY satisfies but MAM doesn't is Weak Reinforcement: If
>> two collections of votes tallied separately produce the same order of
>> finish, then that same order of finish must be produced when the votes
>> are tallied together.[SB:Not important for electing a president.]
>>
>> Independence of Clone Alternatives is important because it's easy for a
>> small minority to nominate clones, whereas Weak Reinforcement is
>> unimportant because a simple rule can prevent a minority from dividing
>> the voters into separate groups (districts). Weak Reinforcement is an
>> example of criteria I call "aesthetic consistency" criteria; society
>> won't be harmed by voting methods that fail those criteria. Too much of
>> the academic literature of social choice theory is about unimportant
>> aesthetic consistency criteria.
>>
>> 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
>
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