[EM] Blake's margins arguments - CORRECTION

[AVnow]

I wrote:

> Let's take another example:
> 101: A
> 1: BAC
> 101: CBA
>
> In this case, B defeats A 102>101,  A defeats C 102>101, and C defeats 
> B 101>1 (with 101 abstaining).  B>A and A>C are victories by majority, 
> but very weak victories.  C>B is a non-majority win, but a resounding 
> victory.  Just to get to a three-way tie, we must assume that all 101 
> abstaining votes really meant to choose B over C (and that's one 
> possibility out of 2^101 or over 
> 2,000,000,000,000,000,000,000,000,000,000, if we force each one take a 
> preference).

Correction:  Just to get a three-way tie, we must assume that 100 of the 
101 abstaining votes really meant to choose B over C, with 1 choosing C 
over B.  That's still a far greater assumption than any voting system 
should be making, so my overall argument here is still very strong.

Tom McIntyre

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