[EM] MCA and median

[Gervase Lam]

> > (1) The winning candidate is the candidate with the highest median
> > rank or score.
> > (2) If more than one candidate satisfies (1), then break the tie by
> > making the candidate with the least number of votes below the median
> > the winner.
> > (3) If more than one candidate satisfies (2), then break
> > the tie by making the candidate with the most number of votes above
> > the median the winner.

All I was trying to do here was to point out that if you applied the above 
rules to a 2 or 3 level ballot, you would respectively get Approval and 
MCA.

> I like this idea, but won't voters use Approval strategy?  I mean, as a
> voter, your ranking of a candidate only matters if you're the median
> voter for that candidate.  In that case, I'd expect to vote either
> "thumbs-up" or "thumbs-down" with no middle rankings.
>
> Kevin Venzke

Possibly.  With MCA, the theory is that the need for you to vote 
"thumbs-up" (i.e. the Favoured vote) for candidates you are indifferent 
about (i.e. the Acceptable vote) is lower.  The only time you would need 
to do this is if it is predicted that one of your "thumbs-down" candidates 
will win by tallying over 50% of the Favoured votes.  In other words, it 
is predicted that your "thumbs-down" candidate would have a median of 
Favoured.

See the following links for some interesting MCA posts:

<http://groups.yahoo.com/group/election-methods-list/message/9692>
<http://groups.yahoo.com/group/election-methods-list/message/9707>
<http://groups.yahoo.com/group/election-methods-list/message/9708>

Extending from the 3 levels of MCA to a larger number of levels, a similar 
rule could be applied.  You need to raise the level of a candidate on your 
ballot if the predicted median level of another candidate, who you rate 
less, is higher.  You would need to raise it to at least the level of the 
other candidate's predicted median.  Or something like that.

Hmmm.  The above paragraph makes it sound like the (strong) favourite 
betrayal criterion is not satisfied if more than 3 levels were used.  Is 
this correct?

Thanks,
Gervase.