[EM] FBC survey & Simmons latest lottery method

[Simmons, Forest]

I wrote ...

**We have shown that when preferences of the type C>F or C>>F are replaced with C=F (without lowering approval of C), the probability of the winner coming from the set  {C,F} is not decreased.

But I overlooked one possibility:  the case where before the change C is the approval winner and F majority defeats C, while after the change F becomes the approval winner (and still pairwise defeats C) but doesn't pairwise defeat all of the other alternatives.   Then before the change {C,F} is the winning pair, while after the change {F,D} is the winning pair, which would decrease the probability of the set  {C,F}.
 
In other words, suppose that the approval order is  F>C>D, and that F majority defeats C but does not majority defeat D.
 
Then the winning pair is {F,D}.  
 
Now suppose that we can lower the approval of F to just below that of C, without causing C to pairwise defeat F.
 
If we have enough control, we can change the approval order to  C>F>D without C majority defeating F.
 
The the winning pair is  {C,F}, an improvement from the point of view of those F supporters that like C more than D.
 
So this method does not satisfy that (**)  version of the FBC.
 
Here's an example:
 
40 F>C>>D
25 C>D>>F
35 D>F>>C
 
The approval order is  F>C>D, and the winning pair is  {F, D}.
 
Let N be a number between 11 and 24.  If  N of the  F>C>>D camp change their ballots to C>>F>D,  then the approval order becomes C>F>D, but F still majority defeats C, so the winning pair becomes  {C,F}, an improvement from the point of view of the F supporters. 
 
Well, back to the drawing board.
 
What if we give all of the probability to the approval winner except when the second place approval alternative is not pairwise defeated by the approval winner, in which case the top two approval alternatives share the probability, either by random ballot or fifty/fifty ?
 
Forest
 
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