[EM] A, B. C, D: all votes were wasted

[Brian Olson]

Instant Runoff Normalized Ratings (Proportional Representation mode) gets this right.

I fed it ratings ballots:
10x A=9 B=5
11x C=9 B=5
12x D=9

It correctly found B as the most popular and D as the second elected winner.
This is possible because it considers the weight expressed by the voter in the entire ballot, not just one choice at a time.

> On Jan 31, 2016, at 6:16 AM, Sennet Williams <sennetwilliams at yahoo.com> wrote:
> 
> K: > >> For something like
> > >> 
> > >> 10: A>B
> > >> 11: C>B
> > >> 12: D
> > >> 
> > >> and two seats, electing A and C wastes votes (12 of them to be exact),
> > >> but electing B and D doesn't.
> > > 
> > > S: No. In this case, APR would elect C with a ?weighted vote? of 11 and
> > > D with a weighted vote of 12. The 10 votes given to A would be wasted
> 
> In fact, all votes would be wasted.  The voters are left disenfranchised with two opposing paid office-holders (C & D)  when in fact the most voters supported B.  C & D would take office and negotiate with each other for what they want regardless of what the voters wanted.  
> In a more U.S. realistic scenario (one winner), the serious politicians would all have followed the priorities of the most voters, like "B", and the best looking candidate with B's priorities would win and take office and break their promises so that she could promise them again in the next election.  
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