[EM] Simplest Condorcet method to hand count?

[Etjon Basha]

Thank you all, gentlemen

I greatly enjoyed the discussion and FWIW come out of it with a greater
appreciation of Condorcet.

A summary for my own benefit: it seems like a hand-count makes one chose
between two counts for races with n candidates.

The first, more complex count involves  n (n-1) /2 passes through all
ballots to populate the pairwise matrix.
Though laborious, each pass is simple and does not require any coordination
with central, allowing any precinct to start their count as soon as polls
close, and complete as soon as they can. Once done, the best Condorcet
algorithms can be used.

The second, less laborious version involves a seeing count which ranks all
candidates, and then n-1 pairwise passes through all ballots to establish
which among the bottom two beats the other, then iterating against the
third from last, and so on. Using an approval seed allows the most approved
candidate to only have to win a single contest, giving decent chances of
emerging from a cycle.

Nanson, my original favourite, involves the least counts by far but
requires a lot more labour for each. Moreover, and I had missed this
earlier, once candidates start getting eliminated, each count becomes
harder and harder to compute, making errors more likely. Even a full
pairwise count might be easier that this.

Finally, IRV-BTR may involve fewer parwise passes than pseudo-C/A as per
above (the winner may gain a majority before all n-2 candidates are
eliminated), but it also requires additional secondary counts to transfer
the votes of the candidates eliminated, which would be quite a few ballots
at the top end.

And all of Nanson, pseudo-C/A and IRV-BTR require constant input to and
from central, making the count as fast as the slowest precinct allows.

In all, there seems to be no faster and more practical way than a pairwise
count to compute a Condorcet winner manually.


On Thu, 22 May 2025, 8:40 pm Etjon Basha, <etjonbasha at gmail.com> 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.
>
> Yet, I suspect this task can at least be completed within acceptable
> timeframes with an acceptable error rate by most volunteers.
>
> (Interestingly, Gemini considers Copeland easier to hand count than
> Nanson, which I disagree with)
>
> Are there any simpler methods I'm unaware off, despite any other
> shortcomings such a method might have?
>
> Best regards,
>
> Etjon
>
>
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