[EM] Simplest Condorcet method to hand count?

[Etjon Basha]

Hi Kristofer,

I don't think running as many counts as we have candidates would be
feasible in practice. Would probably still be better than running one pass
by pairwise matrix, but still.

Now, if we use approval as an initial ordering count, I think one might
already be able to say something about who the CW would be if there was
one.

If any candidates are approved by a majority, would the CW be one of them?
I reckon.

If none are approved my a majority, would the CW be in the top three?

If either of these hold, we might be able to get by with only a few passes,
and I think the first condition at least would hold.

Regards,

Etjon

On Fri, 23 May 2025, 2:37 am Kristofer Munsterhjelm, <
km-elmet at munsterhjelm.no> wrote:

> On 2025-05-22 12:40, Etjon Basha via Election-Methods wrote:
> > Good evening gentlemen,
> >
> > I've been pondering the above issue, and already consulted Gemini who
> > disagrees with me on the practicality of pairwise matrices, so couldn't
> > help a lot.
> >
> > I suspect that compiling pairwise matrices in the context of a hand
> > counted election would be very time consuming, and quite prone to errors
> > and challenges from all parties.
> >
> > Assuming we agree on this (which you might not) is there any practical
> > Condorcet method can can be hand counted?
> >
> > I suspect Nanson is a reasonable candidate. Yes, it still requires
> > log(candidates,2) counting rounds, and each of those rounds require
> > sending a matrix of how many times each candidate was ranked in which
> > position to a central location, so quite the bother indeed.
>
> How about this method? Use some base method (e.g. FPTP or even just a
> random order) to order the candidates. Then repeatedly remove, from this
> order, the pairwise loser of the two candidates ranked last on it. (I.e.
> pit the two last ranked candidates against each other pairwise; pit the
> winner of that contest against the third-last ranked candidate, etc.)
> Last man standing wins.
>
> This has one initial count (if you don't use a random order), and n
> pairwise counts.
>
> The benefits vs Nanson are that it doesn't require any Borda counting,
> just whether X beats Y pairwise. In addition, just like Nanson, it's
> summable if you're okay with calculating the Condorcet matrix ahead of
> time. It passes Smith and is easy to do interactively (possibly using
> approval or something equally simple to create the initial agenda order).
>
> The disadvantages are that while the worst-case number of rounds is the
> same, Nanson probably has fewer rounds with realistic elections. It's
> also nonmonotone and probably worse in this respect than Nanson,
> although I haven't verified this.
>
> -km
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20250523/5b6600f8/attachment.htm>