[EM] IRV variants

[fsimmons at pcc.edu]

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