[EM] Re: 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.

> I must admit, (2) and (3) aren't good measures of which of two or more
> tied medians is the better.  Really [Votes above median - Votes below
> median] should be used in step (2).  But I can't picture clearly what
> strategic consequences this could have.

> Thanks,
> Gervase.

The last paragraph contains an error.  Really [Votes above median + 0.5 * 
Votes on the median] should be used in step (2).  A high value implies a 
"higher" median.  Alternatively, [Votes below median + 0.5 * Votes on the 
median] could be used.  A low value implies a "higher" median.  Both 
produce equivalent results.

This is probably fairer as the tiebreaker as it would be influenced by 
both the votes above and below the median.  Steps (2) and (3) possibly 
concentrate too much on the votes above or below the median, not both at 
the same time.  I thought this fairer tiebreaker would prevent the 
multi-district inconsistency that is possible with MCA...

Example: Three candidates (A, B and C) are voted by two districts, each of 
which receive ten ballots.  AB indicates a ballot with A given the 
Favoured vote, B the Acceptable vote and C the Unacceptable vote.

District 1: AB, AB, AB, AB, AB, BC, BC, BC, CB [A wins with 5/9 Favoured].
District 2: AC, AC, AB, BC, BC, BA, BA, CA, CA

In District 2, all three candidates have a median of Acceptable.  However, 
using [Votes above median + 0.5 * Votes on the median], A gets 5, B gets 
4.5 and C gets 4.  Therefore, A is the winner again.

Combining both districts, all three candidates have a median of 
Acceptable.  Using [Votes above median + 0.5 * Votes on the median], A 
gets 10, B gets 10.5 and C gets 6.5.  Therefore, B is the overall winner, 
despite the fact that B has not won any of the districts singly.

Thanks,
Gervase.