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<p>Etjon,<br>
<br>
<blockquote type="cite"> 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.</blockquote>
<br>
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).<br>
And we don't need to know or care which it is.<br>
</p>
<p>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.<br>
<br>
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.<br>
<br>
Chris<br>
<br>
</p>
<div class="moz-cite-prefix">On 23/05/2025 2:05 am, Etjon Basha
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:CA+EJN6ROt8B2n3_n3rS=gjKoN+9xBbTmbhaRA044RQ2LyaXi2A@mail.gmail.com">
<meta http-equiv="content-type" content="text/html; charset=UTF-8">
<div dir="auto">Hi Chris,
<div dir="auto"><br>
</div>
<div dir="auto">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.</div>
<div dir="auto"><br>
</div>
<div dir="auto">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.</div>
<div dir="auto"><br>
</div>
<div dir="auto">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.</div>
<div dir="auto"><br>
</div>
<div dir="auto">Regards,</div>
<div dir="auto"><br>
</div>
<div dir="auto">Etjon </div>
</div>
<br>
<div class="gmail_quote gmail_quote_container">
<div dir="ltr" class="gmail_attr">On Thu, 22 May 2025, 11:59 pm
Chris Benham via Election-Methods, <<a
href="mailto:election-methods@lists.electorama.com"
moz-do-not-send="true" class="moz-txt-link-freetext">election-methods@lists.electorama.com</a>>
wrote:<br>
</div>
<blockquote class="gmail_quote"
style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Etjon,<br>
<br>
My favourite Condorcet method, Margins Sorted Approval, would
be <br>
relatively easy to hand count because it would only very
rarely need the <br>
full pairwise matrix.<br>
<br>
First just count the approvals to determine each candidate's
approval <br>
score. Those scores give us our initial order, from highest to
least <br>
approved. Now we are only interested in the pairwise results
between <br>
pairs of candidates which are adjacent to each other in this
order.<br>
<br>
(Our goal is to arrange the candidates in a chain where the
candidate at <br>
the head beats the candidate that is second who in turn beats
the <br>
candidate that is third, and so on. Ranked Pairs also does
that.)<br>
<br>
Next we do the pairwise comparison between the adjacent pair
of <br>
candidates with the smallest difference in their approval
scores. (If <br>
there is a tie for this, then the tied pair lowest in the
order.) If <br>
they are pairwise out of order (i.e. if the less approved of
the two <br>
pairwise beats the more approved) then the candidates change
places in <br>
the order to give us our new provisional ordering.<br>
<br>
We repeat this process to the end. (The order always
stabilises.) <br>
Then the candidate at the top of the final order is the
winner.<br>
<br>
There are two versions of this method, MSA (explicit) and MSA
<br>
(implicit). I prefer the more expressive (and more in the
Condorcet <br>
spirit) explicit version which allows voters to rank among
candidates <br>
they don't want to approve, versus the somewhat simpler (and
possibly a <br>
bit higher SU) implicit version which asks the voters to rank
only those <br>
candidates they approve.<br>
<br>
Benham meets Unburiable Mutual Dominant Third and I think this
doesn't, <br>
but Benham does need the full pairwise matrix (just the
win-loss-tie <br>
results) and overall isn't as good. So why put up with
relative <br>
"shortcomings" ?<br>
<br>
Chris<br>
<br>
On 22/05/2025 8:10 pm, Etjon Basha via Election-Methods wrote:<br>
> Good evening gentlemen,<br>
><br>
> I've been pondering the above issue, and already
consulted Gemini who <br>
> disagrees with me on the practicality of pairwise
matrices, so <br>
> couldn't help a lot.<br>
><br>
> I suspect that compiling pairwise matrices in the context
of a hand <br>
> counted election would be very time consuming, and quite
prone to <br>
> errors and challenges from all parties.<br>
><br>
> Assuming we agree on this (which you might not) is there
any practical <br>
> Condorcet method can can be hand counted?<br>
><br>
> I suspect Nanson is a reasonable candidate. Yes, it still
requires <br>
> log(candidates,2) counting rounds, and each of those
rounds require <br>
> sending a matrix of how many times each candidate was
ranked in which <br>
> position to a central location, so quite the bother
indeed.<br>
><br>
> Yet, I suspect this task can at least be completed within
acceptable <br>
> timeframes with an acceptable error rate by most
volunteers.<br>
><br>
> (Interestingly, Gemini considers Copeland easier to hand
count than <br>
> Nanson, which I disagree with)<br>
><br>
> Are there any simpler methods I'm unaware off, despite
any other <br>
> shortcomings such a method might have?<br>
><br>
> Best regards,<br>
><br>
> Etjon<br>
><br>
><br>
> ----<br>
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</blockquote>
</div>
</blockquote>
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