Mike,
As I pointed out in my last message, I made a mistake with the example I
gave. There should have been only 10 B>A votes.
45: C
06: D>A
39: A>B
10: B>A
So there are a hundred voters and no what you call "mutual-majority
candidate set".
But if it weren’t big enough, and if the D voters wanted to add
themselves to it, then they’d have only to vote D=A>B. By MMT2’s
definition of a mutual majority candidate set.
I see. It seems that contrary to what I claimed, this method does meet
the FBC as you say.
But overall IMO it pays far too high a price for "no defection
incentive" and FBC compliance. It has random-fill and Burial incentives
and fails Mono-add-Plump.
Chris Benham
Mike Ossipoff wrote (9 Dec 2011):
Chris said:
> As far as I can see the examples I gave apply equally well to "MMT2".
> I've pasted them in at the bottom.
He was referring to his posting copied and replied to below:
>
> I think this (MMT2) fails the FBC. Say sincere is:
>
> 45: C
> 06: D>A
> 39: A>B
> 20: B>A
>
> There is no "mutual majority set" (by your latest definition)
My latest MMT version is still MMT2. It’s my latest, final,
and best MMT version.
By its definition of a mutual-majority candidate set, in
your example, {A,B} is a mutual-majority candidate set.
But if it weren’t big enough, and if the D voters wanted to
add themselves to it, then they’d have only to vote
D=A>B. By MMT2’s
definition of a mutual majority candidate set.
Therefore, there would be no violation of FBC in your
example.
Your example illustrates a general fact: It’s possible to be
counted in support of any mutual majority candidate set without voting
anyone
over your favorite. MMT2 meets FBC.
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