[EM] MCA, 3RC, Pair Approval, Manipulability

[Kevin Venzke]

Forest,

 --- Forest Simmons <fsimmons at pcc.edu> a écrit :
> > Question: What advantage does MCA have over three-rank Condorcet?
> >
> 
> Good Question!
> 
> I think it's easier to explain to the lay voter: "If nobody's favorite
> gets more than fifty percent of the votes, then the most acceptable
> candidate wins."
> 
> We need to compare the two methods on more examples to clarify the other
> advantages and disadvantages.

I thought a bit about my question after posting it, and these reasons
occur to me:
1. MCA is so close to Approval that it's almost a two-rank method.  Thus if
it's agreed that two-rank Condorcet (Approval) is better than three-rank 
Condorcet, then MCA should be better for those same reasons, whatever they are.
2. It's easier to explain, as you said.
2. Three-rank Condorcet requires a cycle-breaker of some kind.  That raises
the question of which it should be, and makes it even harder to explain.

Advantages of 3RC over MCA:
1. I'm worried that less savvy voters will fill out an MCA ballot like
it was a 3RC ballot.  In other words, that they will put unduly lousy
candidates in the second rank.
2. The expectation from ordering two candidates is the same no matter which
two ranks you place them in.  In MCA, A>B>_, A>_>B, and _>A>B all have
different values.
3. Strategy is simpler, because MCA has two ways to win, and more odds to
estimate.

Regarding manipulability by false information, I would say we should
suspect right off the bat that 3RC will outperform MCA, because MCA
strategy requires more information (as from polls).


> The Strong FBC method holds up well on this example because the B and C
> supporters have no incentive to approve any pair other than {B,C} and the
> A supporters have no incentive to approve any pair other than {A,B}. The
> finalist pair then turns out to be {A,B} and A wins.

Is this so?  I would think that:
A voters approve (A,B) as you say, because although A is destined to lose
based on the poll, B is better than C;
B voters approve (A,B) and (B,C) because they can win both (they think they
are the CW, so they're fine as long as they get in the winning pair);
C voters approve (A,C) expecting to win, and/or (B,C) to ensure a decent
result.

(It also occurs to me that some A voters might want to approve (B,C), in
order to avoid (A,C) winning, resulting in a C victory.  Possible?  The
strategy A uses in the first place is "defend the CW"...)

Expected results:
(A,B): 6000
(A,C): 0-4000
(B,C): 2500-6500
Either (A,B) or (B,C) wins, B wins in either case

Actual results:
(A,B): 7000
(A,C): 0-3000
(B,C): 1000-4000
So (A,B) wins, and A beats B, as hoped.

(But if enough A supporters actually approved (B,C), B could still take
it.  In that case, this method is not so good, but at least it would
be clear from the results that the A supporters were duped.)


Kevin Venzke
stepjak at yahoo.fr


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