[EM] Acquiescing majority MMT (MIKE OSSIPOFF)

[fsimmons at pcc.edu]

Mike,

yes it is the same as the one you repeated at the end of your reply below.

But notice that in the ballot set

49 C
27 A>B
24 B

there are two Acquiescing Majorities, namely both {A, B} and {B, C}, and that C has more top votes than B.

Forest


> From: MIKE OSSIPOFF 
> To: 
> Subject: [EM] Acquiescing majority MMT
> Message-ID: 
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> 
> Forest--
> 
> Let's find out what its properties are.
> 
> Just preliminarily, it sounds like one of the MMT ideas that I 
> considered. 
> 
> But it seemed to me, at the time (if it's the same method I was 
> considering) that, if a ballot can be counted in that majority 
> merely by rating each candidate
> in the set equal to or over every candidate outside the set, 
> then, in the ABE, the B votes could
> rate A at bottom, with C, and still be part of the relevant 
> majority. So there there is
> the set required by the acquiescing rule, and there is one 
> candidate rated above bottom
> by everyone in that set: Candidate B.
> 
> So, the method that I'd considered wouldn't pass in the ABE. I 
> don't know if the method
> you describe is the same one, but, preliminarily, it sounds similar.
> 
> But maybe not. Any possibility could yield improvement. 
> 
> My definition of that set was something like this:
> 
> A set of candidates rated equal to or over everyone outside the 
> set by each member of the
> same majority of the voters.
> 
> Mike Ossipoff
> 
> 
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