[EM] fpA - fpC (typos galore, bordering on thinkos)

Forest Simmons fsimmons at pcc.edu
Tue Mar 16 18:02:41 PDT 2021


Too much hurry ...starting at *** below,  typos abound ... x-y should be
x-z, then y-z should be y-x, and z-x should be z-y.

On Tuesday, March 16, 2021, Forest Simmons <fsimmons at pcc.edu> wrote:

> In his book, The Geometry of Voting, Saari approaches the problem of a
> resolving a Condorcet cycle in the three candidate case by first canceling
> ABC  ballots against CBA ballots, BCA ballots against ACB ballots, and CAB
> ballots against BAC ballots. This first step cannot eliminate the Condorcet
> cycle, but it reduces the number of factions to three: either ABC, BCA, &
> CAB or else
> CBA, BAC, & ACB.
>
> Without loss in generality suppose the three factions are
>
> x: ABC
> y: BCA &
> z: CAB
>
>
> Now suppose that y = min(x,y,z). Saari now removes y ballots from each
> faction, which eliminates the middle faction entirely:
>
> (x-y): ABC
> (z-y): CAB
>
> With only two factions there can no longer be a Condorcet cycle.
> Majoritarians would say that A or C must win depending on which of the two
> remaining factions is larger. A should win if (x-y) is larger than (z-y),
> which is true iff x - z is positive.
>

*** x-y---> x-z, etc.  I'm going to make the changes...

>
> Saari continues his analysis to argue in general (no matter which of the
> three cyclical factions is smallest) the largest difference among (x -z),
> (y -x), (z-y) should determine whether A, B, or C wins, respectively.
>
> [Note x-z is fpA - fpC]
>
> Saari goes on to show that this result is equivalent to the Borda Count,
> and rests his case that Borda is the best election method (to make a long
> story short).
>
> To be continued ....
>
>
>
>
>
> ************************************************
>>
>
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