<div dir="auto">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.)<div dir="auto"><br></div><div dir="auto">So chain building can ascend or descend ad libitum ... as they say, "Just follow your nose" </div><div dir="auto"><br></div><div dir="auto">Also it simplifies everything conceptually and computationally if we replace and shuffle the ballots between all consecutive drawings.<div dir="auto"><div dir="auto"><br></div><div dir="auto">Remember that the proto example was the candidate cycle ABCA with respective first place probabilities of p=fpA, q=fpB, and r=fpC.</div><div dir="auto"><br></div><div dir="auto">The possible maximal chains are ...</div><div dir="auto">A>B, B>C, and C>A,</div><div dir="auto">with probabilities proportional to ...</div><div dir="auto">pq, qr, and rp, respectively,</div><div dir="auto">when chains are built ad libitum.</div><div dir="auto"><br></div><div dir="auto">The condition for A winning is</div><div dir="auto">pq>max(qr,rp),</div><div dir="auto">which (turns out to be) equivalent to</div><div dir="auto">(p-r)>max(q-p, r-q),</div><div dir="auto">within this probability context of p+q+r=1.</div><div dir="auto"><br></div><div dir="auto">This last inequality is our fpA-fpC condition for A winning in our proto example.</div><div dir="auto"><br></div><div dir="auto">Next, a simple introduction to the method suitable for a voters' pamphlet ...</div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto"><br></div></div></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Thu, Oct 20, 2022, 1:24 PM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" rel="noreferrer noreferrer noreferrer" target="_blank">forest.simmons21@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="auto">One important (but easy) correction:<div dir="auto"><br></div><div dir="auto">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!</div><div dir="auto"><br></div><div dir="auto">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!</div><div dir="auto"><br></div><div dir="auto">I only mentioned it at all because of its earlier mention in related EM list threads.</div><div dir="auto"><br></div><div dir="auto">-Forest</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Thu, Oct 20, 2022, 11:48 AM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" rel="noreferrer noreferrer noreferrer noreferrer" target="_blank">forest.simmons21@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="auto"><div dir="auto">First a preliminary procedure to make sure no single candidate defeats every member of the support of the random ballot favorite:</div><div dir="auto">As long as there is such a candidate, retain only candidates of this índole, recalibrating between elimination steps.</div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto">Next: a non-deterministic lottery method ... Random Ballot Favorite Chain Climbing (RBFCC):</div><div dir="auto"><br></div><div dir="auto">Shuffle the ballots into some random order B1, B2, B3, ... and let ListF be a list of the candidates in the order induced by the first choices of the respective ballots in their order ... i.e. according to the order of their first appearance as a first choice on a ballot in the sequence B1, B2, B3, ...</div><div dir="auto"><br></div><div dir="auto">Now, Chain climb the list ListF by initializing the set variable CHAIN as the empty set, and then .... </div><div dir="auto">While some member of ListF defeats every member of CHAIN, add the first such candidate into CHAIN. EndWhile</div><div dir="auto"><br></div><div dir="auto">The head of the completed chain is the RBFCC (random trial) winner.</div><div dir="auto"><br></div><div dir="auto">Next, for each candidate X, let RBFCC(X) be the winning probability for X under this lottery.</div><div dir="auto"><br></div><div dir="auto">Finally, elect argmax RBFCC(X).</div><div dir="auto"><br></div><div dir="auto">Note that this method is Banks efficient, and obviously reduces to "fpA-SumfpC" in the eponymous three candidate case.</div><div dir="auto"><br></div><div dir="auto">On a practical note, should the computation of the RBFCC probabilities be intractable for some ballot set, then repeated trials in a MonteCarlo simulation of the lottery can be used to determine argmax RBFCC(X) with arbitrarily low error probability epsilon.</div><div dir="auto"><br></div><div dir="auto">Is this the simplest formulation of what we've been looking for?</div><div dir="auto"><br></div><div dir="auto">It doesn't seem like an easy method to "game".</div><div dir="auto"><br></div><div dir="auto">Other comments? Questions?</div><div dir="auto"><br></div><div dir="auto">Who can write this up in a way that Joe Q Public can easily relate to?</div><div dir="auto"><br></div><div dir="auto">-Forest</div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto"></div><div dir="auto"><br></div></div>
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