One important (but easy) correction:

In order to make this method Monotone, we have to start the chain from the
bottom of the list ListF. That's what puts the "Climbing" in Random Ballot
Favorite Chain Climbing!

A comment on exposition for public consumption ... no mention of Condorcet
Smith, Landau, or Banks should be included in the method description, any
more than a brief introduction to IRV needs to explain what to do if the
last three remaining candidates have the same first place transferred vote
totals.  Every generic public ballot set will have a Banks member with at
least one first place vote, so no need to get people worried right off the
bat about what to do in the impossibly rare contrary case.  Stick with
generic conditions in the voters' pamphlets ... just make sure that the
rare exceptional possibilities are covered in the published official legal
definition, as well as the RAQ's (Rarely Asked Questions) if not the FAQ's!

I only mentioned it at all because of its earlier mention in related EM
list threads.

-Forest

On Thu, Oct 20, 2022, 11:48 AM Forest Simmons forest.simmons21@gmail.com
wrote:

It turns out that what we want is chain building by fitting each newly
drawn ballot favorite into the chain where-ever possible without destroying
its total order relation. (A chain is a totally ordered set.)

So chain building can ascend or descend ad libitum ... as they say, "Just
follow your nose"

Also it simplifies everything conceptually and computationally if we
replace and shuffle the ballots between all consecutive drawings.

Remember that the proto example was the candidate cycle ABCA with
respective first place probabilities of p=fpA, q=fpB, and r=fpC.

The possible maximal chains are ...
A>B, B>C, and C>A,
with probabilities proportional to ...
pq, qr, and rp, respectively,
when chains are built ad libitum.

The condition for A winning is
pq>max(qr,rp),
which (turns out to be) equivalent to
(p-r)>max(q-p, r-q),
within this probability context of p+q+r=1.

This last inequality is our fpA-fpC condition for A winning in our proto
example.

Next, a simple introduction to the method suitable for a voters' pamphlet
...

On Thu, Oct 20, 2022, 1:24 PM Forest Simmons forest.simmons21@gmail.com
wrote:

In some informal settings it is convenient and appropriate to use a chance
event to make a decision ... flipping a coin to see which team is first at
bat ... drawing straws to see who is point man on patrol ... rock paper
scissors to see who washes the dishes.

In 1965 when I turned 18, a selective service lottery picked me to be one
of the lucky guys eligible for conscription into Uncle Sam's VietNam war.

In the context of voting methods the "benchmark" lottery draws one or more
ballots (depending on the number of recruits needed) at random to see whom
is to be elected/selected.

Some times it is appropriate to turn a chance method into a deterministic
method by simply selecting/electing the nominated candidate(s) with the
greatest chance of winning the election lottery ... e.g. the one(s) with
the most votes.

That brings us to the "Random Ballot Favorite Chain Building" RBFCB voting
method.

First, we will give the chance version, and then the associated
deterministic version ... both based on the voter submitted, non-random,
ranked choice ballots.

As usual, we start by tallying the ballot preferences to find out all of
the head-head matchup winners.

Naturally if X outranks Y on more ballots than Y outranks X, then X is the
matchup winner for the X vs Y match.

Then we start to build up our "chain" by drawing a ballot B at random and
initializing a list named FavoredChain (FC for short) with the name of the
candidate most favored by ballot B.

We continue building up the FC chain by drawing additiomal ballots and
inserting the names of their respective favorites into their appropriate
positions relative to the other FC chain members ... above all of the ones
they defeated in the matchup tally, and below all of the ones by whom they
were defeated.

If some particular candidate cannit be fit into the FC list in this way,
then it is left out of list as we turn our attention to the next randomly
drawn ballot.

When no additional candidate can be fit into the FC chain order in a way
that respects the relevant matchup wins, then the FC chain is complete,
i.e. "maximal."

The candidate that ended up at the head of the list is the winner of this
round of the Random Ballot Favorite Chain Building lottery.

The winner of the corresponding deterministic method is simply the
candidate with the best chance of winning additional rounds of the RBFCB
lottery.

How is it determined from the ballots which candidate has the best chance
of winning the RBFCB lottery?

Usually, it can be determined without running the lottery simulation even
once.

For example, if (as per usual) there is one candidate that wins all of its
matchups in the matchup tally, she will go straight to the head of the FC
chain as soon as any ballot is drawn on which it is the most favored
candidate. Thereafter all additional favorites will be incorporated in the
FC list below this undefeated matchup winner.

In the rare case when no candidate is undefeated in the matchups, and the
winning probabilities cannot be easily calculated from the table of matchup
winners in conjunction with the first place vote totals information ... in
that very rare case the lottery can be run as many times as necessary to
get the winning probabilities with any desired degree of statistical
certainty.

I hope that this exposition will be clearer than mud to most EM list
readers ...perhaps even most Scientific American readers could catch the
drift.  Math Monthly readers might be miffed at the lack of algebra.

Still a ways to go for the voters' pamphlet version:-)

-Forest

On Fri, Oct 21, 2022, 6:56 AM Forest Simmons forest.simmons21@gmail.com
wrote:

In the quoted message I did a sloppy job of hand waving to convince myself
that Random Ballot Favorite Chain Building (RBFCB) gives the same result as
fpA-fpC in the prototypical example.

I sloppily conjured a formula out of the clear blue for the RBFCB winning
probabilities, and then asserted without proof that it turned out to be
equivalent to the max(fpA-fpC) criterion.

I was depending on mental graphs of the two formulae to coincide for a
quick confirmation of my intuition, but it turns out that the online
Wolfram Alpha plotter is very persnickety about 3D graphics ... it just
gives up way too soon!

So let's take a more careful run at it, and let the chips fall where they
may ... if it turns out that these two methods are not equivalent, then we
have two nice voting methods instead of only one:-)

Let's approach the problem from a Markov process point of view. The
possible states of the chain are the initial empty set {}, the singletons
A, B, and C; and the possible completed maximal chains A>B, B>C, and C>A.

We need to construct a transition matrix of the transition probabilities
from the various states to the new states as the process evolves. One
should draw a directed graph with arrows showing the state transitions.
Those arrows should be labeled with the respective transition
probabilities.
Once the digraph is properly labeled, the transition matrix can be read off
directly from the digraph.

In this case the digraph is a tree, with the initial empty state {}  as the
root node.

The three arrows emanating from it enter the three singleton states A, B,
and C. The corresponding arrows are labeled with the transition
probabilities p, q, and r, respectively.

These singleton states have arrows connecting them to the maximal states:

arrows from A to A>B and from A to C>A, as well as arrows from B to both
B>C and  A>B, and arrows from C to both C>A and B>C.

What are the respective transition probabilities that should serve as
labels for these arrows from the singleton states to the final states?

This is where we have to be careful ... sloppy handwaving will not do here!

Why is it so tricky? Because the transitions from the single states to the
final states are "conditional probabilities."

For example, the transition from A to A>B has a conditional probability of
q/(q+r), rather than the straight probability q.

Why is that? Because (at this stage) when we draw a ballot whose favorite
candidate is A, we must ignore it and draw again until we get a ballot with
a B or C favorite, since we already have an A in our current Markov state.

So the respective transitions from A to A>B and from A to C>A, are the only
possibilities, which means their probabilities must sum to unity.

Imagine that there are 20 A ballots, 35 B ballots and 45 C ballots, and we
must ignore all of the A ballots if drawn, because only one A can be
incorporated into the chain, which it already has, since our transition is
from state A to A>B or to C>A.

There are 80 currently active ballots still in play at this juncture... 35
B's and 45 C's, so the respective conditional transition probabilities are
evidently 35/80 and 45/80.  In general, they would be q/(q+r) and r/(q+r).

Check that they add up to unity, just as 35/80 and 45/80 do.

Now that we have the transition arrows labeled, how do we figure the
probabilities of the final states, the leaves of the tree representing the
completed chain?

First we multiply the transition probabilities along each path to get the
probability of the leaf state the end of the path.

For A the path from A>B through singleton B has a value of qp/(p+r), while
the path through the singleton A state has a value of pq/(q+r).

We add these two path values together to get the probability of the chain
A>B.
This sum, qp/(p+r)+pq/(q+r),
is A's chance of winning the RBFCB election lottery :

Similar expressions give the winning probabilities for B and C.

To simplify the expressions we make use of the grand common denominator
(q+r)(r+p)(p+q). This common denominator can be employed to clear the
fractions on the rationale that when all denominators are equal, the
largest fraction is the one with the largest numerator.

So A's numerator will be
pq(p+q)(q+r+r+p), which simplifies to
pq(1-r)(1+r), since q+r+p is just 1.

If we divide every such expression by pqr, we maintain the proportionality,
and we see that the winning probability for A is proportional to (1-rr)/ r,
or 1/r -r. In other words, the bigger 1/r-r, the better  A's chance of
winning. But this expression is monotone decreasing in r, so the smaller r,
the greater A's chance of winning.

Can we conclude that A wins iff and only if r<min(p,q)?

At least we can safely conclude that A wins if (1/r-r)>max(1/p-p,1/q-q).

This makes sense intuitively because r is the probability of drawing a
ballot favoring C, which is the only candidate that defeats A.

One should compare graphically the above condition(1/r-r)>max(1/p-p,1/q-q)
with the corresponding fpA-fpC condition
(p-q)>max(q-r,r-p)
to see how close they are.

I think we can see how useless it is for the A supporters to bury C. It
does not change A's chance of winning the RBFCB lottery at all!

On the other hand, increasing A's first place support can help A only
indirectly ...by competing for C's first place support.

Check my algebra and Markov tree reasoning.

It would be easy to check these results independently with MonteCarlo
simulations.

-Forest

On Fri, Oct 21, 2022, 6:56 AM Forest Simmons forest.simmons21@gmail.com
wrote:

Good news ... the monotone decreasing property of (1/r)-r does indeed show
that in our prototypical three candidate cycle that the winner is A, B, or
C, depending on which of r, p, or q is smaller. In other words, depending
on which of fpC, fpA, or fpB is smaller.

In other words, un the three candidate Smith case the RBFCB method (the
deterministic version) is equivalent to Ranked Pairs or Beatpath CSSD, or
River with defeat strength gauged by bow few (not how many!) first place
votes the losing candidate has in opposition to the winner.

For all practical purposes RP(lack of losing first place opposition) is
equivalent to RBFPCB. The public proposal could describe it either way ...
you could just call it "lack of first place votes" or LFPV Condorcet.

For clarity, in the ABCA cycle if C has fewer first place votes against A,
than A has against B, or B against C, then the cycle is broken at the
weakest link ... the CA link, so the ABCA cycle is replaced with the
(strongest) beatpath ABC, making A the winner.

Thanks for your patience!

On Fri, Oct 21, 2022, 6:03 PM Forest Simmons forest.simmons21@gmail.com
wrote: