[EM] Defection, nomination disincentive, MMPO

[Gervase Lam]

> Date: Thu, 16 Dec 2004 02:18:25 +0100 (CET)
> From: Kevin Venzke
> Subject: [EM] Defection, nomination disincentive, MMPO

> In "MinMax (Pairwise Opposition)" or "MMPO," a pairwise matrix
> is formed as in a Condorcet method, but the winners of pairwise
> contests are not determined. For each candidate, find how many
> ballots favored each other candidate over him, and record the
> largest number. Elect the candidate for whom this number is smallest.

For the situation I am thinking of using a voting method on, at the moment 
I would use MMPO.  This is because it is almost the same as MinMax 
(Winning votes) except for the fact that it is even easier to explain and 
present the results.  There is no need to mention about 
head-to-head/pairwise results.  You just need to mention how many voters 
ranked a candidate X above another candidate Y.

> This method has flaws: It fails Condorcet,

I initially thought MMPO did fail Condorcet.  However, I thought I worked 
out why it did not.  May be I am wrong.  Can you give an example of this?

> Majority, Plurality, and Clone Independence.

May be MMPO can be patched by Forest's de-cloning method before applying 
MMPO.

> But it does satisfy Later-no-Harm.

I think that sacrificing Condorcet in order to gain Later-no-Harm is most 
probably a worthwhile sacrifice.  For a while, I had been trying to think 
up of a reasonable pairwise method that fails Condorcet.  It looks like 
MMPO is one such method.

> 49 A, 24 B, 27 C>B or
> 49 A, 24 B>C, 27 C>B:
> A: score is 51 (number of B>A voters)
> B: score is 49 (number of A>B voters)
> C: score is 49 (number of A>C voters)
>
> In other words, a B-C tie, regardless of how many B or C voters
> defect from the other, so long as over 49 voters rank the same
> candidate above A. If at least 25 A voters pick a side between
> B and C, then that would break the tie, also.
>
> If a tie still remains, I suggest breaking it with Random Ballot
> or perhaps FPP, two methods which still satisfy Later-no-Harm.

Let's say that Candidate A was John Andrews, Candidate B was George Brown 
and Candidate C was Tim Charles.  The results of the above two scenarios 
could be presented as follows...

1st. Tim Charles  - 49:JA, 24:GB
2nd. George Brown - 49:JA, 27:TC
3rd. John Andrews - 51:GB, 27:TC

...and...

1st. Tim Charles  - 49:JA, 24:GB
2nd. George Brown - 49:JA, 27:TC
3rd. John Andrews - 51:GB, 51:TC

The above are basically whole pairwise matrices except that each line 
(i.e. candidate), the number of opposing votes is ordered in decreasing 
numeric order.  The lines are then ordered by number of opposing votes in 
the first "column" in increasing numeric order.

For example, according to the first line, 49 ballots ranked JA (John 
Andrews) above Tim Charles and 24 ballots ranked GB (George Brown) above 
Tim Charles.  As can be seen in the above,  Tim Charles and George Brown 
are initially tied with 49 opposing votes.

The most easily presentable tie-breaker I thought of using was to find the 
next highest opposing votes for each of the tied candidates.  The 
candidate with the lowest number of these opposing votes is the winner.

If there is still a tie here, then you go on to the next highest opposing 
votes for the tied candidates.  And so on, if required.

In the above situation, Tim Charles' next highest opposing votes is less 
than George Brown's.  Therefore, Tim Charles is the winner.  In this case, 
the tie-breaker also turns out to be a head-to-head contest between the 
Time Charles and George Brown, which Tim Charles wins.

In this situation, the number of voters is large enough to say that this 
was an extremely close election.  If George Brown had just got one less 
ballot that voted him below John Andrews, then George Brown would have 
won, regardless of the fact that George Brown loses the head-to-head 
against Tim Charles.

It can be seen that for an N candidate election, there can be up to (N-2) 
tie-breakers.  However, assuming that enough ballots are cast, it could be 
said any candidate who wins as a result of the tie-breaker is lucky.  
Therefore, it could be said that any tie-breaker could do.

Thanks,
Gervase.