On 05 Nov 2011 17:46:49 -0700, Forest Simmons wrote: > > Dear EM Folks, I’ve been very busy, while watching postings on the > EM list out of the corner of my eye. I was very happy to notice > Mike Ossipoff’s interesting contributions. In particular his > promotion of a variant of IRV where equal rank counts whole strikes > me as promising in the context of the “chicken problem.” > > On a related note I have been thinking about how to make a monotone > variant of IRV. Perhaps the two ideas could be combined without > sacrificing all of the other nice features that IRV(=whole) seems to > have. > > It is well known that certain kinds of elimination methods cannot > satisfy the monotonicity criterion. The basic variant of IRV is of > that type, namely what I call the “restart with each step type” of > elimination method. This means that when one candidate is > eliminated, the next stage starts all over again without learning > from or memory of the eliminated candidate. Range elimination > methods that renormalize all of the ballots at each stage are of > this type, too, since the renormalization is an attempt to eliminate > the effect of the eliminated candidates on the remaining stages of > the process. > > But some methods of elimination that do not suffer from the “restart > problem” turn out to be monotone. For example, approval elimination > where the original approvals are kept throughout the whole > elimination process; trivially the highest approval candidate is the > last one left. > > Now here’s what I propose for an IRV variant: > > 1. Use the ranked ballots to find the pairwise win/loss/tie matrix > M. This matrix stays the same throughout the process. > > 2.Initialize a variable U (for Underdog) with the name of the > candidate ranked first on the fewest number of ballots, and > eliminate U from the ballots. > > 3.While more than one candidate remains, eliminate candidate X that > is ranked first on the fewest number of ballots after the previously > eliminated candidates’ names have been wiped from the ballots (as in > IRV elimination) and then replace U with X, unless U defeats X, in > which case leave the value of U unchanged. > > 4. Elect the pairwise winner between the last value of U and the > remaining candidate. > > Note that a simplified version of this where you just eliminate the > pairwise loser of the two candidates ranked first on the fewest > number of ballots in NOT monotone. We have to remember the previous > survivor and carry him/her along as "underdog challenger" to make > this method monotone. > > Note also that this method satisfies the Condorcet Criterion. So we > gain monotonicity and CC, but what desireable criteria do we lose? > It still works great on the scenario > > 49 C > 27 A>B > 24 B > > Candidate A starts out as underdog, survives B, and is beaten by C, > so C wins. But if B supporters really prefer A to C they can make A > win. On the other hand if the A supporters believe that the B > supporters are indifferent between A abd C, they can vote A=B, so > that B wins. > > When I have more time, I'll sketch a proof of the monotonicity. > > Comments? > > Thanks, > > Forest > ---- > Election-Methods mailing list - see http://electorama.com/em for list info
It strikes me as needlessly complex. Equal Rank whole-vote IRV, with some kind of Borda or score based elimination, would be much simpler. However, any quota-based ranked system (and any form of single-winner IRV has a quota of 50%) will be subject to one of the Woodall mono-* failure modes. So even if you manage to satisfy one of the monotonicity criteria, you will fail participation in some respect. Ted -- araucaria dot araucana at gmail dot com ---- Election-Methods mailing list - see http://electorama.com/em for list info
