[EM] non-deterministic methods

[Paul Kislanko]

Forest Simmons wrote:
> As Jobst recently pointed out, non-deterministic methods have 
> not been 
> adequately studied or promoted, considereing their potential 
> contribution 
> to fairness and to strategy free voting.
> 
> Consider, for example the following cycle of three:
> 
>    34 ABC
>    33 BCA
>    33 CAB
> 
> Though most methods would give the win to A, it seems like B 
> and C have 
> nearly as much claim as A.
> 
> What if we tossed two coins, and gave the win to B if they 
> both came up 
> heads, to C if they both came up tails, and to A otherwise.

If we did that, we would be assigning probability .25 that B or C would win
and .5 that A would win. So "B and C would have less claim" only because of
the way we assigned the "winner" based upon two coin-tosses.

How do you know to assign "HT or TH" to A, HH to B, and TT to C, except
you've already decided A "should" win?

> 
> This would give A a fifty percent chance of winning, and 
> divide the other 
> fifty percent equally between B and C.
> 
> Would this be fair?

If you're the dictactor, maybe. I think it loses information the same way
the US Electoral College system does. We went from a nearly .34:.33:.33 to
2:1:1 ratio. 

> 
> Does it give too much probability to A ?  or too little?
> 
> Some might say that A, B, and C should be more equal in 
> probability, since 
> the faction sizes are nearly equal.
> 
> But the determinists would say that they should be even less equal: A 
> should get one hundred percent of the probability.
> 
> Who is right?  And how should we assign the probabilities?

In the simple examples we use, this may not seem obvious, but we can use
probabilities if we must in a manner similar to the way we use the pairwise
matrix. The probability that A is preferred over B is .34, the probability
that A is preferred over C is .34, the probability that B is preferred over
A is .33, that B is preferred over C is .67, that C is preferred over A .66
that C is preferred over B .33

P(B>C|A) = .67, P(C>A|B) = .66, P(A>B|C) = .34 (where | means the best of
either)

Anything that gives A a 50 percent chance of winning would not be "fair" as
the word is defined in the laws of probability. A non-deterministic approach
should decide in favor of B or C at least 79 percent of the time (plug the
worst probability for B or C as p and the best probability for A as q into 

(p * (1-q))/(((p * (1-q))+(q * (1-p)))

to get that result). 

The "Though most methods would give the win to A," statement is what bugs me
(an ordinary voter) about "most methods". 2/3rds of the voters prefer
ANYBODY BUT A, so why should A be elected? If "most methods" select the FPP
winner A, why not just stick with plurality?