[EM] simple question (I think)

[Chris Benham]

Rob,

rob brown wrote:

> For instance, say there is no Condorcet winner.  Candidates A, B and C 
> all have 8 pairwise wins.  D has 7.  Could D still be chosen as the 
> winner by any "reasonable" method?

Yes. The method that just counts the number of pairwise wins is called 
Copeland.  It  hopelessly fails Clone Independence (Clone-Loser) and 
"Rich Party".

Imagine that that there are three candidates, each with the same number 
of pairwise wins, and the Condorcet method elects X.
Say  that the top cycle is  X>Z>Y>X
Now say we add a clone of  Y, that every voter ranks directly below Y.
Now Y and Z will each have an extra pairwise win, one more than X and so 
now (by the "Copeland criterion") X must lose to Z or Y.

Adding a clone  of  a losing candidate (not to say adding a 
Pareto-dominated candidate)  has changed the winner.

Parties and factions  that run more candidates will have an absurd and 
unfair advantage.


Chris  Benham