[EM] Simplest Condorcet method to hand count?

[Chris Benham]

Etjon,

> If the seeding order is A>B>C****"", and we check pairwise between A 
> and B with A winning, and between B and C with B winning, we might be 
> confident in electing A right there and then.

If there are just 3 candidates then it is over and A has won. Both the 
other candidates each have one or two pairwise defeats, so either A is 
the Condorcet winner (pairwise beating C as well as B) or there is no 
Condorcet winner (the candidates are in a A>B>C>A cycle).
And we don't need to know or care which it is.

But if (as I suspect your aterixes are supposed to mean) there are more 
candidates,  then no it isn't safe to stop and announce A is the winner 
(especially in the "explicit" version).   But nonetheless we will 
usually need to use (and therefore know) much less than the full 
pairwise matrix.

This would be more true with extra candidates. With just three 
candidates we've only saved ourselves the trouble of looking at one 
extra pairwise comparison.

Chris

On 23/05/2025 2:05 am, Etjon Basha wrote:
> Hi Chris,
>
> am I correct in understanding that, under this count, we might need as 
> few as two passes (after the initial count) to finalise? If the 
> seeding order is A>B>C****"", and we check pairwise between A and B 
> with A winning, and between B and C with B winning, we might be 
> confident in electing A right there and then.
>
> Yes, in theory the very next candidate could overtake C, and then B 
> and then A but how likely could this be? Unlikely enough to stop the 
> count at the second count, Id say.
>
> If so, this does appear to be the simplest hand count so far between 
> Nanson and IRV-BTR. Probably not so if it requires as many passes as 
> there are candidates less one though.
>
> Regards,
>
> Etjon
>
> On Thu, 22 May 2025, 11:59 pm Chris Benham via Election-Methods, 
> <election-methods at lists.electorama.com> wrote:
>
>     Etjon,
>
>     My favourite Condorcet method,  Margins Sorted Approval, would be
>     relatively easy to hand count because it would only very rarely
>     need the
>     full pairwise matrix.
>
>     First just count the approvals to determine each candidate's approval
>     score. Those scores give us our initial order, from highest to least
>     approved.  Now we are only interested in the pairwise results between
>     pairs of candidates which are adjacent to each other in this order.
>
>     (Our goal is to arrange the candidates in a chain where the
>     candidate at
>     the head beats the candidate that is second who in turn beats the
>     candidate that is third, and so on. Ranked Pairs also does that.)
>
>     Next we do the pairwise comparison between the adjacent pair of
>     candidates with the smallest difference in their approval scores. (If
>     there is a tie for this, then the tied pair lowest in the order.)  If
>     they are pairwise out of order (i.e. if the less approved of the two
>     pairwise beats the more approved) then the candidates change
>     places in
>     the order to give us our new provisional ordering.
>
>     We repeat this process to the end.  (The order always stabilises.)
>     Then the candidate at the top of the final order is the winner.
>
>     There are two versions of this method, MSA (explicit) and MSA
>     (implicit).  I prefer the more expressive (and more in the Condorcet
>     spirit) explicit version which allows voters to rank among candidates
>     they don't want to approve, versus the somewhat simpler (and
>     possibly a
>     bit higher SU) implicit version which asks the voters to rank only
>     those
>     candidates they approve.
>
>     Benham meets Unburiable Mutual Dominant Third and I think this
>     doesn't,
>     but Benham does need the full pairwise matrix (just the win-loss-tie
>     results) and overall isn't as good. So why put up with relative
>     "shortcomings" ?
>
>     Chris
>
>     On 22/05/2025 8:10 pm, 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.
>     >
>     > 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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