[EM] IRV variants

[C.Benham]

Forest Simmons wrote (5 Nov 2011):


> 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?


Yes. I don't get it. (I am confused by your explanation of the algorithm).

How do you think this is better than your latest version of Enhanced DMC?

I think a good method is the IRV-Condorcet hybrid that differs from IRV 
only by before any and each elimination
checks for an uneliminated candidate X that pairwise beats all the other 
uneliminated candidates and elects the
first such X to appear.

That of course gains Condorcet, and it keeps IRV's Mutual Dominant Third 
Burial Resistance property.
So if a candidate X pairwise beats all the other candidates and is 
ranked above all the other candidates on more than
a third of the ballots then (as with IRV) X must win and a rival 
candidate Y's supporters can't get Y elected (assuming
they can somehow change their ballots) by Burying X.

Does your method share that property?

> 49 C
> 27 A>B
> 24 B
>
> Candidate A starts out as underdog, survives B, and is beaten by C, so 
> C wins.


 From what I think I do understand of your algorithm description, 
doesn't candidate B start out as "underdog"?

Chris Benham