[EM] Oops! Forgot to include Chris's text. Chris MMT reply, complete this time.

[C.Benham]

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.