[EM] Simplest Condorcet method to hand count?

[Etjon Basha]

Ah, no, the CW mustn't always have a majority of approvals given that at
least one candidate has such a majority.

Sad.

On Fri, 23 May 2025, 8:13 am Etjon Basha, <etjonbasha at gmail.com> wrote:

> 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
>>
>
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