[EM] Is "Inverse Nanson" better than standard Nanson?

[Forest Simmons]

Michael,

what you call "inverse Nanson" I would call second order iterated Borda
elimination.

These elimination methods can be iterated arbitrarily many times.

Even IRV gets the Condorcet Winner when there is one, if iterated
sufficiently many times.

This seems to be true for just about any (rankings based) method that
picks the correct winner in a two way contest.

As you note, Nanson and "inverse Nanson" get rid of what is perceived by
many students to be Borda's worst problem; without iteration Borda doesn't
always pick the majority winner in a three way contest. 

Other worse problems (in my opinion) that Borda suffers from are
sensitivity to clones and stategic incentives for insincere ranking.

It would be interesting to see how well inverse Nanson ameliorates these
two most serious problems of Borda.

Forest

On Sun, 1 Jul 2001, Michael Rouse wrote:

> Nanson's method -- which calculates the Borda score for each candidate, 
> drops the lowest, then recalculates -- can be proven to always find the 
> Condorcet winner if one exist. It is one of few non-Condorcet methods with 
> this property.
> 
> We can also invert Nanson's method like this:
> 1. Calculate the Borda count for all candidate.
> 2. Drop the highest-scoring candidate.
> 3. Re-order the remaining candidates.
> 4. As long as there are two or more candidates remaining, return to step 1.
> 
> This method can be shown to always pick the Condorcet loser in a set, if 
> there is one available. We can take this Condorcet loser and drop it from 
> the original list, then re-order the candidates and recalculate. In this 
> way we can generate a sorted candidate list from the bottom up, one that 
> will use Condorcet order if possible but will still find a single candidate 
> for each available position even if there are one or more circular ties.
> 
> The revised "inverse" Nanson method would add the following steps:
> ...
> 5. Put the final "loser" candidate in the appropriate position on a second 
> list, and remove from the original list.
> 6. Re-order the remaining candidates.
> 7. As long as there are two or more candidates remaining on the original 
> list, return to step 1.
> 
> Just as a small example of where the inverse Nanson would come up with a 
> Condorcet order and standard Nanson would not, consider the following:
> 2     3
> A    B
> B    C
> C    A
> 
> Borda would give the following count: A=4, B=8, C=3. With Nanson, we would 
> drop C, then check B(3) vs. A(2). You would end with the order BAC. In the 
> inverse Nanson, we would drop B since it is the highest, then compare A vs. 
> C. A would lose. Comparing C vs. B, C would lose. We would then end up with 
> the order BCA, which agrees with the Condorcet order.
> 
> Anyway, I think it has some advantages over the standard Nanson method. I's 
> be interested in any thoughts on it, especially examples where it breaks 
> the monotonicity criteria (Nanson does and I just want to get a feel for 
> how often that would happen in comparison). Thanks!
> 
> Michael Rouse
> mrouse at cdsnet.net
> 
>