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

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

Forest Simmons
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 <[hidden email]> 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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