[EM] More easily hand-counting three-slot Condorcet

[Jobst Heitzig]

Dear Kevin!

Let us recall: you suggest to use only three slots and to interpret
candidates in slots 1 and 2 as approved of and candidates in slot 3 as
not approved of. Then you came up with the topic of how to measure
defeat strength best without having to count all winning votes, and
suggested to use approval scores. I pointed out that when using approval
of A to measure the strength of A>B, you count some people towards that
strength who actually prefer B to A, and that this possibility will be
 counter-productive when trying to convince people to go voting. I
suggested to try either the number of voters who approve of A but not of
B (which is a special case of weighted pairwise), or at least to use the
number of voters who don't approve of B instead of those who approve of A.

All of this discussion was based upon the assumption that there are
*many* candidates (not just 3) and, of course, that the voters would be
told that slots 1 and 2 meant approval while slot 3 did not. So, your
recent example
	9 A>B>C	
	8 B>C>A
	7 C>A>B
doesn't fit into that setting at all. It only shows that it is not very
sensitive to interpret slots 1 and 2 as approval when there's only 3
candidates! And perhaps neither when there are more candidates...

It is always the same: When we restrict voters ability to express
preferences unnecessarily (such as using only rankings or even only 3
slots) and impose all kinds of strange assumptions (such as the one
about approval in slots 1 and 2), we must not wonder when getting
strange results...

Jobst


PS: I think my recent "grand compromise" proposal should be a good
common starting point for further discussion since it incorporates many
of the important ideas we had this year. It would be nice if you could
take it as serious as I took your proposal serious.