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? ---- Election-methods mailing list - see http://electorama.com/em for list info