On Fri, 23 May 2025, 12:27 am Steve Eppley via Election-Methods, election-methods@lists.electorama.com wrote:

On 2025-05-22 12:40, Etjon Basha via Election-Methods wrote:

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

Hi Etjon,

This is an interesting question. I agree that the hand-countability of
ballots, at least in the case of an audit, is an important practical
feature of an election.

I wonder if the ballot design itself could be modified to suit Condorcet
methods. So, you rank your candidates on the touch screen voting machine.
Then, the voting machine prints out your ballot, as is the case now.
However, rather than simply printing a piece of paper with your ranking, it
prints out each pairwise preference separately. So, if your ranking was A >
B > C > D, it would print out 6 ballots:

A>B
A>C
A>D
B>C
B>D
C>D

Ballots can then be sorted by type. That way, it is easy to tally the
ballots into the Condorcet matrix, and any entry into the matrix is easy to
double check. And, we can audit the count easily, as ballots should sum up
to the total number of voters, i.e., (A>C + C>A + A=C) should equal the
total number of voters, which should also equal (A>D + D>A + A=D), and so
on. And, as is the case now, you would also have a computer count to check
against.

On Thu, May 22, 2025 at 6:41 AM Etjon Basha via Election-Methods <
election-methods@lists.electorama.com> wrote:

election-methods@lists.electorama.com> wrote:

see tallied as a result. As long as there is some way to connect the two,
an electronic count is as foolproof as a hand count. If not, it's not.

counting would do it, and I've seen it done.

It's what any state that uses optical-scan technology does.  The actual
ballot that the voter marked is the "paper backup".  Unlike the stupid
computer card ("butterfly ballots") in which the physical instrument can be
misaligned in the jig that is in the voter booth (then it's possible that
the hole that is punched out is not for the candidate that the voter
intended) and the true intent of the voter is lost to the record, with
optical-scan, the true intent of the voter can always be discerned by
looking directly at the ballot that the voter marked.

an after-the-fact audit is impossible.

Why is that?

It's not the individual ballots, but we can link a particular sealed and
numbered ballot bag with a particular tally of the ballots (like with a
particular voting machine that voters insert their ballots into).  We can
confirm that the pairwise totals from the machine tally match the pairwise
totals in the recount.  In the U.S. the right to vote by secret ballot is a
big deal and we want no serial number, no identifying marks at all, on
the voted ballot that can possibly be traced back to the checklist and
identify the voter.

election-methods@lists.electorama.com> wrote:

keep an initially foolproof election system foolproof.

which a copy of the ballots verified by independent or multi-partisan
observers is published online in a downloadable format. I'm assuming the
tallying software is open source, available for free installation on
smartphones, and has been audited by some public interest groups you trust.
If you're really paranoid, you could shuffle the downloaded ballots and
globally replace the candidate IDs with dummy IDs, to check whether this
changes the result. People you trust could publish examples and their
expected results, to test your software.

the accuracy of a copy of the ballots, I don't understand how could you
have more trust in a hand-count.

believe is best (Maximize Affirmed Majorities) on the criterion I think is
most important (create a strong incentive for politicians to support
majority-preferred policies) seems simple enough to explain with the aid of
two simple examples: "Count all the head-to-head majorities. Then process
the head-to-head majorities one at a time, from largest majority to
smallest majority, placing each majority's more-preferred candidate ahead
of their less-preferred candidate in the order of finish." (To my eye,
Nanson isn't simpler.) The first example would have 3 candidates (perhaps
named Left, Center and Right) and a Condorcet winner (Center). The second
example would have 3 candidates (perhaps named Rock, Scissors and Paper)
and a majority cycle.

election-methods@lists.electorama.com> wrote:

your goal is to allow an election to be counted by a society that can't
even afford a cheap smartphone, I don't think this cost is a show-stopping
barrier, since smartphones are ubiquitous. So why settle for an inferior
tallying algorithm?

--

r b-j . _ . _ . _ . _ rbj@audioimagination.com

"Imagination is more important than knowledge."

.
.
.

Election-Methods mailing list - see https://electorama.com/em for list
info

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@munsterhjelm.no> wrote:

One other thing that is somewhat apropos to the subject is that Condorcet
methods are Precinct Summable.  So a recount can be done in a distributed
fashion.  The ballot bags never need to be transported dozens or hundred
kilometers to a central tallying location for the recount.

So if a recount, even a Condorcet recount (where there are N(N-1)/2 times
the pile of ballots must be processed, but for IRV that number could be
N-1, where N is the number of candidates), this laborious but simple work
can be divided up among many separate groups of people, done in a single
day, and the results are fully summable by anyone (the press, the
competing campaigns, and the general public, as well as the officials that
certify the outcome of the election).

It's distributed work.  It's quick, even for a state-wide or nationwide
recount.  It's decentralized.  It's summable.  It's transparent in
process.  Very hard to be dishonestly fixed.

This is one of two big reasons why I am such a hard-core Condorcet
advocate.  The other is, of course, electing anyone other than the
Condorcet winner is electing the wrong candidate and will violate Majority
Rule and violating that causes our votes to not be counted equally.

--

r b-j . _ . _ . _ . _ rbj@audioimagination.com

"Imagination is more important than knowledge."

.
.
.

between the ballots and individual machine counted and the actual bag.

scan method, but to the method where you vote on a machine and get a
printout to cast as a backup. You'd get this once you're out of the booth,
and would only be able to verify in full view of the volunteers. The font
one ID in the back would be too small to take a picture of anyway in this
limited time, and once the ballot is cast would make linking to the actual
voter impossible (assume a properly random ID).

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info

On 2025-05-24 07:22, Etjon Basha via Election-Methods wrote:

The only way I see to break this barrier is to use randomness. You could
randomly sample ballots if the complexity is prohibitive. E.g. you could
count a random pairwise contest from each ballot. But then you need a
trusted way to generate randomness.

Theoretically, you could even do this on the front-end: ask each voter
his preference for a randomly chosen pairwise contest X>Y. This is a
very simple ballot and would leave the voter to only have to consider
two candidates.[1] However, it would be unfamiliar and probably wouldn't
be received well: even in the best case, you've replaced voters having
to trust a machine count with voters having to trust your source of
randomness, and if the race has a few very clear frontrunners, the
majority of the voters who don't get asked about any frontrunner contest
might feel cheated out of their opinion.

But deterministically for Condorcet, I think v voters times n candidates
is the best you can do (at least without seriously thinking outside the
box). Because suppose otherwise, that you only check (n-1) candidates.
Then it's possible that the candidate that you didn't check is the
actual CW. There may be heuristics - e.g. if someone has a majority of
the first preferences, that candidate is the CW - but they won't always
hold.

On a related note, I have been thinking about how to formally prove that
a given method has a minimum degree of summability, or that for, say,
thre candidates, it must know the value of (e.g.) five variables. Such
an approach could show that, yes, you need n pairwise counts for every
known method. (It could also disprove it, but that would be more
surprising.)

For concrete methods that are always decisive, this becomes a linear
algebra problem.[2] I haven't found a way to solve this problem,
however. There's an obvious upper bound using separating hyperplanes,
but to my knowledge it's just that - an upper bound.

The problem for methods that pass some criteria is even harder. For
instance, I suspect that determining the innermost mutual majority set -
the set of candidates that methods passing mutual majority must elect
from - is not summable. But there are lots of methods that do pass
mutual majority and are summable. I would be surprised if my intuitive
hunch for Condorcet were to be disproven, though.

-km

[1] A fun theoretical system for people who consider rational ignorance
a problem would be to assign every voter a defined pairwise contest a
sufficiently long time before the election. Then the voter would only
have to evaluate these two candidates, and could do so in more detail.
But in practice, I don't think it would be accepted, and would require
significant technology in its own right.

[2] Each election can be characterized by a "vote vector" which lists,
for each preference orde, how many voters voted that way. Then each
candidate has a win region which is a union of convex polytopes in
n!-dimensional space, and every point in A's win region has coordinates
equal to a voting vector where the method declares A a winner. Then you
have to find a minimal dimension projection so that no point in
different candidates' win regions overlap after projecting all the win
regions down to that dimension.