[EM] IRV's "majority winner". What if we let the people choose?

[James Green-Armytage]

<jgilmour at globalnet.co.uk> writes:
>Now consider:
>49  A<C<B
>48  B<C<A
>  3  C<B<A
>IRV winner = B;  CW winner = C.
>I doubt very much whether most electors would accept C as the "winner" if
>this were an election for
>Sate Governor, much less for a directly elected President of the USA.  If
>anyone has evidence to the
>contrary I'd like very much to see it.
>James Gilmour

	Well, if the votes were sincere to begin with, then it is axiomatic that
C will win a runoff election against B. That is, 51 prefer C to B, and 49
prefer B to C. If I prefer B to C, and they are in a runoff election
against each other, of course I will vote for B, even if I would have
rather had A.
	I'd agree with Mike to the extent that whenever there is a sincere
Condorcet winner (wins all pairwise comparisons) who differs from the
sincere IRV winner, the Condorcet winner should always be expected to win
a runoff election against the IRV winner, because s/he necessarily has a
pairwise win against that candidate.
	However, if there is no clear Condorcet winner, I mean if there is a
cycle and a completion method needs to be used, then the completion method
winner will not necessarily have a pairwise beat against the IRV winner,
and hence will not be expected to win in a runoff. Just for example:

18: A>C>B
15: A>B>C
18: B>A>C
14: B>C>A
20: C>B>A
15: C>A>B
100 total

IRV
A		B		C
33		32		35
+18		eliminated	+14
51				49
A wins

pairwise comparisons
A:B = 48:52
A:C = 51:49
B:C = 47:53

	No Condorcet winner, weakest defeat is A:C 51:49, therefore C wins
minimax etc.
	So the IRV winner is A, the minimax winner is C, and A beats C in
pairwise comparison.

	Now, I don't mean to imply by any means that IRV is better than minimax,
or that Condorcet should be completed by IRV. I'm just responding to what
people have said.

my best,
James