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>>1. Pairwise tally, using the ranked ballots only. Elect the Condorcet
>>winner if one exists.
>
David wrote:
>
>So in the examples:
>45 A 100 > B 70 > C 0
>10 B 100 > A 70 > C 0
>5  B 100 > C 70> A 0
>40 C 100 > B 70 > A 0
>
>45 A 100 > B 10 > C 0
>10 B 100 > A 90 > C 0
>5  B 100 > C 90 > A 0
>40 C 100 > B 10 > A 0
>B wins in both cases even though in the second example B is disliked by
>85% of voters.
>David Gamble

I reply:
	Yes, that is correct. In both of those examples, B is a Condorcet winner,
and is therefore elected by my method. Hence yours is a criticism of
Condorcet methods in general, rather than my method in particular. This
criticism, i.e. the possibility of a low-utility Condorcet winner, has
been debated before. If memory serves, you prefer IRV to Condorcet, or did
at one point. 
	So, the standard pro-Condorcet counter-argument is to say that if we were
doing an IRV count that was repeated over and over again, you would
probably get to the point pretty quickly where B would win every time.
That is, in the first round, perhaps, people would give their sincere
rankings, and A would win the first IRV tally. But in subsequent rounds,
the C>B>A voters would be likely to realize that they could get a better
result for themselves (B instead of A) by voting B in first place. So C
would be eliminated first and B would win the tally. At this point, it
seems to me like the situation would be largely in equilibrium, in that
there wouldn't be a way for any of the voting blocs to get an immediately
preferable result.
	The fact that IRV, if repeated several times, is likely to settle on the
Condorcet winner, makes me think that Condorcet is an improvement on IRV.
That is, Condorcet is likely to reach an equilibrium outcome in one vote
that could take IRV several rounds to establish. And the problem with
taking several rounds is that public elections don't allow for this, and
so in a one or two round system there is necessarily a lot of
second-guessing that distorts preferences.
	
	So anyway, yes, the method I'm proposing is Condorcet efficient, and if
you aren't convinced that Condorcet efficiency is a good thing, then you
probably won't like my method. For me, Condorcet efficiency is of foremost
importance in single-winner voting, partly because of the reasoning stated
above.
	However, looking at your example, I thought of a criticism quite
different from the one you made. 
45 A 100 > B 10 > C 0
10 B 100 > A 90 > C 0
5  B 100 > C 90 > A 0
40 C 100 > B 10 > A 0
>
	I'm noticing that in my method as I've stated it so far, B is the initial
winner (because he's a CW), but the A voters can cause a cycle by
truncating, a cycle which A wins. Maybe you would be okay with A winning
here, David, but for me it's a disaster, first of all because B was the
CW, and in general because it creates a strong strategic incentive to
truncate, distorting preferences. So it's not good that my method allows
that to happen so easily. 
	Winning votes Condorcet methods do brilliantly at protecting the CW from
strategy in cases like these, and it's hard for my method to compete with
that.
	But one thing that I can offer is that perhaps my method can be amended
so that non-majority pairwise beats can be treated differently from
majority pairwise beats. One possibility is to treat non-majority defeats
as ties.

	I'll probably write more about this in a subsequent posting...

my best,
James