<div dir="auto">I have used the fractional Top and Bottom counts as a basis for clone independent Friendly voting variants.<div dir="auto"><br></div><div dir="auto">A much better solution (imho) is to replace these fractional Top a d Bottom counts with my adaptation of the Martin Harper lotteries to the Implicit Approval and Bottom Disapproval counts:</div><div dir="auto"><br></div><div dir="auto">Let's start with the Bottom Disapproval counts. In our example ...</div><div dir="auto">48 C</div><div dir="auto">28 A>B</div><div dir="auto">24 B</div><div dir="auto"><br></div><div dir="auto">The respective bottom disapprovals for A, B, and C are</div><div dir="auto">48+24=72, 48, and 52.</div><div dir="auto"><br></div><div dir="auto">Since A's disapproval is greatest, according to Martin Harper's rule all of the ballots that disapprove of A contribute to A's bottom disapproval lottery probability of 72 percent. The other 28 percent of the probability goes to C.</div><div dir="auto"><br></div><div dir="auto">My version of the Implicit Approval Lottery probabilities are a little trickier because in my version only Top or Equal Top get full approval ... while rankings between Too and Bottom count for half approval.</div><div dir="auto"><br></div><div dir="auto">On this basis the respective approval counts are 28, 24+(1/2)28=38, and 48.</div><div dir="auto"><br></div><div dir="auto">C gets 48 percent of the Simmons/Harper lottery probability. The next highest approval value is 38 for B. So acording to the Harper rule, all of the B ballots contribute to B's probability. </div><div dir="auto"><br></div><div dir="auto">But what about the A ballots ... they supported both the 38 percent approval of B and the 28 percent approval of A.</div><div dir="auto"><br></div><div dir="auto">Well their approval of B was only fractional, so we compare (1/2)38=19 with full 28. Since 28>19, all of the A ballots go to A's lottery probability of 28 percent.</div><div dir="auto"><br></div><div dir="auto">Harper's original purpose was to appease the IRV vote transfer mindset by specifying which approval ballots transferred to which candidates in an approval election. I extended that to the general Range context, and in particular the implicit approval context where equal top gets 100 percent approval, ranked between bottom a d too 50 percent approval, and bottom gets zero.</div><div dir="auto"><br></div><div dir="auto">Any way ... the idea works great in this context ... each ballot ends up supporting exactly one candidate in each of the two lotteries ... the implicit approval lottery and the disapproval lottery.</div><div dir="auto"><br></div><div dir="auto">Back to our example:</div><div dir="auto"><br></div><div dir="auto">The approval lottery probabilities are</div><div dir="auto">28, 24, and 48, while disapproval probabilities are 72, 0, and 28.</div><div dir="auto"><br></div><div dir="auto">The method one scores for a candidate are the disapproval probabilities minus approval probabilities of the corresponding hostile candidates:</div><div dir="auto">S(A)=d(C)-a(C)=28-48=-20</div><div dir="auto">S(B)=dA-aA=72-28=44</div><div dir="auto">S(C)=dB-aB=0-24=-24</div><div dir="auto"><br></div><div dir="auto">The method 2 scores S' are the friendly approval probabilities minus the friendly disapproval probabilities:</div><div dir="auto">S'(A)=aA+aB-dA-dB=28+24-72-0=-20</div><div dir="auto">S'(B)=aB+aC-dB-dC=24+48-0-28=44</div><div dir="auto">S'(C)=aC+aA-dC-dA=48+28-28-72=-24</div><div dir="auto"><br></div><div dir="auto">So B wins again.</div><div dir="auto"><br></div><div dir="auto">The method 5 scores (to be minimized) are the friendly disapproval probabilities plus the hostile disapproval probabilities...</div><div dir="auto">A-->dA+dB+aC=72+0+24=96</div><div dir="auto">B-->dB+dC+aA=0+28+28=56</div><div dir="auto">C-->dC+dA+aB=28+72+24=124</div><div dir="auto"><br></div><div dir="auto">B minimizes the score.</div><div dir="auto"><br></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Mon, Feb 13, 2023, 4:23 PM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com">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>Let's add the Method 1&2 calculations for the same example that we did the Method 3,4&5 tallies in our previous message (copied below).</div><div dir="auto"><br></div><div dir="auto">Here's a summary of the data from that example:</div><div dir="auto"><br></div><div dir="auto"><div dir="auto" style="font-family:sans-serif">48 C</div><div dir="auto" style="font-family:sans-serif">28 A>B</div><div dir="auto" style="font-family:sans-serif">24 B</div><div dir="auto" style="font-family:sans-serif"><br></div><div dir="auto" style="font-family:sans-serif">For the different sums of differences S and S' we need the following Too and Bottom (fractional) counts:</div><div dir="auto" style="font-family:sans-serif">Top(A)=28, Top(B)=24, Top(C)=48</div><div dir="auto" style="font-family:sans-serif">Bot(A)=(48+24)/2=36,</div><div dir="auto" style="font-family:sans-serif">Bot(B)=48/2=24,</div><div dir="auto" style="font-family:sans-serif">Bot(C)=(24/2)+28=40</div><div dir="auto" style="font-family:sans-serif"><br></div><div dir="auto" style="font-family:sans-serif"><div dir="auto">Method 2 elects the candidate Y that maximizes S'(Y), the Sum (over the candidates X friendly to Y) of Top(X)-Bot(X):</div><div dir="auto">S'(A)=Too(A)-Bot(A)+ Top(B)-Bot(B)</div><div dir="auto">= 28-36+24-24=-8</div><div dir="auto">S'(B)=Top(B)-Bot(B)+Top(C)-Bot(C)</div><div dir="auto">=24-24+48-40=8</div><div dir="auto">S'(C)=TopC-BotC+TooA-BotA</div><div dir="auto">=48-40+28-36=0</div><div dir="auto">The winner is argmax S'(Y) = B</div></div></div><div dir="auto"><br></div><div dir="auto">Method 1 elects the candidate Y that maximizes S(Y), the Sum over the hostile candidates X of Bot(X)-Top(X):</div><div dir="auto">S(A)=Bot(C)-Too(C)=40-48=-8</div><div dir="auto">S(B)=Bot(A)-Top(A)=36-28=8</div><div dir="auto">S(C)=Bot(B)-Top(B)=24-24=0</div><div dir="auto">The winner is argmax S(Y) = B</div><div dir="auto"><br></div><div dir="auto">So B is the winner of all five methods.</div><div dir="auto"><br></div><div dir="auto">Why is this so?</div><div dir="auto"><br></div><div dir="auto">Well, the fractional counts are the same counts we would get by symmetric completion of the ballots ... and we know that all five methods are equivalent given complete ballots.</div><div dir="auto"><br></div><div dir="auto">-Forest</div><div dir="auto"><br><br><div class="gmail_quote" dir="auto"><div dir="ltr" class="gmail_attr">On Mon, Feb 13, 2023, 7:58 AM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" target="_blank" rel="noreferrer">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">Continuing in the same context for a little while we introduce Methods 4 & 5, respectively as maximizing and minimizing the mined and subtrahend, respectively, of the difference arising in method 3 ...<div dir="auto"><br></div><div dir="auto">Method 4: Maximize the minuend, which is the sum of the Bottom totals of the hostile candidates and the Top totals of the friendly candidates.</div><div dir="auto"><br></div><div dir="auto">Method 5: Minimize the subtrahend, which is the sum of the Top totals of the hostile (vexing) candidates and the Bottom totals of the Friendly candidates.</div><div dir="auto"><br></div><div dir="auto">Note the numerical and psychological advantages of Methods 4&5 due to all terms being positive: ... numerical stability and linguistic congruence in the sense tally=sum, as opposed to difference, in the public mind.</div><div dir="auto"><br></div><div dir="auto">That said, let's specialize to our Example 1 context again for Method 4: Here there is only one candidate (C) hostile to candidate A, to whom the other two candidates A and B are friendly ... so the tally for candidate A is ...</div><div dir="auto">Top(A)+Top(B)+Bottom (C), which is just fpA+fpB+fpA in terms of faction sizes.</div><div dir="auto"><br></div><div dir="auto">For comparison purposes only, let's subtract zero in the form of n-n from this tally, where n is the total number of ballots fpA+fpB+fox. After the dust settles we see that A's tally is numerically equal to ...</div><div dir="auto">fpA-fpC+n, the same score for A as in methods 1 and 2, up to a constant n.</div><div dir="auto"><br></div><div dir="auto">Similarly, Method 5 yields (for A) a tally of fpC-fpA+n to be minimized. </div><div dir="auto"><br></div><div dir="auto">So in the special context of Example 1, all five methods agree on the winner.</div><div dir="auto"><br></div><div dir="auto">But once we go beyond Example 1, it seems to me that Method 5 is better strategically than Method 4 ... because Method 4, with two terms of foA in its tally, gives too much control to the A faction ... the likely culprit of the burial that created the ABCA beat cycle.</div><div dir="auto"><br></div><div dir="auto">So I propose tentatively Method 5:</div><div dir="auto"><br></div><div dir="auto">Elect the candidate that minimizes the total tally TT given by: friendly bottom totals plus hostile top totals.</div><div dir="auto"><br></div><div dir="auto">Just one item of business to finish the five method specifications: when rankings are not complete Top and Bottom counts must be done fractionally to preserve clone independence.</div><div dir="auto"><br></div><div dir="auto">That's it ... so now let's do an example of Method 5 with incomplete rankings:</div><div dir="auto"><br></div><div dir="auto">48 C</div><div dir="auto">28 A>B</div><div dir="auto">24 B</div><div dir="auto"><br></div><div dir="auto">For the different TT tallies we need the following Too and Bottom (fractional) counts:</div><div dir="auto">Top(A)=28, Top(B)=24, Top(C)=48</div><div dir="auto">Bot(A)=(48+24)/2=36,</div><div dir="auto">Bot(B)=48/2=24,</div><div dir="auto">Bot(C)=(24/2)+28=40</div><div dir="auto"><br></div><div dir="auto">TT(A)=Bot(A)+Bot(B)+Top(C)=36+24+48</div><div dir="auto">=108</div><div dir="auto">TT(B)= Bot(B)+Bot(C)+Top(A)=24+40+28</div><div dir="auto">=92</div><div dir="auto">TT(C)=Bot(C)+Bot(A)+Top(B)=40+36+24=100</div><div dir="auto"><br></div><div dir="auto">It appears that argmin TT(Y) is B.</div><div dir="auto"><br></div><div dir="auto">Let's compare this with the total tallies TT'(Y) for method 4:</div><div dir="auto">TT'(A)=Top(A)+Top(B)+Bot(C)</div><div dir="auto">=28+24+40=92</div><div dir="auto">TT'(B)=Top(B)+Top(C)+Bot(A)</div><div dir="auto">=24+48+36=108</div><div dir="auto">TT'(C)=Top(C)+Too(A)+Bot(B)</div><div dir="auto">=48+28+24=100</div><div dir="auto">Argmax TT'(Y)=B, the same as the method 5 winner.</div><div dir="auto"><br></div><div dir="auto">The method 3 totals are the respective differences TT'-TT of the methods 4 and 5 ... 92-108=-16, 108-92=16, and 100-100=0. So B wins by method 3.</div><div dir="auto"><br></div><div dir="auto">Somebody should check my work on these methods and also calculate the Method 1 and 2 winners.</div><div dir="auto"><br></div><div dir="auto">Methods 3, 4, and 5 do not appear to be Chicken Strategy resistant.</div><div dir="auto"><br></div><div dir="auto">I hope that these sample calculations provide enough clarity for those who want to explore further.</div><div dir="auto"><br></div><div dir="auto">-Forest</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sun, Feb 12, 2023, 9:42 PM Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" rel="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">For simplicity we'll stick with complete rankings for now ... with three methods that are equivalent in that context, though not in general ... like MinMaxPaiwiseOpposition and MaxMinPairwiseSupport are equivalent in this context.<div dir="auto"><br></div><div dir="auto">We say candidate X is Friendly to Y if X doesn't defeat Y pairwise ... otherwise X Vexes Y.</div><div dir="auto"><br></div><div dir="auto">Also X has Top or Bottom Status on ballot B according to whether no candidate outranks it or it outranks no candidate on ballot B.</div><div dir="auto"><br></div><div dir="auto">Method 1:</div><div dir="auto"><br></div><div dir="auto">Elect the candidate argmax S(Y) defined as Sum{Bot(X)-Top(X)| X vexes Y} where Bot(X) and Top(X), respectively, are the number of ballots on which X has Top or Bottom status.</div><div dir="auto"><br></div><div dir="auto">Example 1.</div><div dir="auto"><br></div><div dir="auto">fpA: A>B>C</div><div dir="auto">foB: B>C>A</div><div dir="auto">fpC: C>A>B</div><div dir="auto"><br></div><div dir="auto">C is the only candidate that vexes A, so S(A) is merely Bot(C)-Too(C), which equals fpA-fpC.</div><div dir="auto"><br></div><div dir="auto">Method 2:</div><div dir="auto"><br></div><div dir="auto">Elect argmax S'(Y) defined as </div><div dir="auto">Sum{Top(X)-Bot(X)| X is friendly to Y }.</div><div dir="auto"><br></div><div dir="auto">Method 2 applied to the same ballot profile from example 1 goes as follows:</div><div dir="auto"><br></div><div dir="auto">S'(A)=(Top(A)-Bot(A))+(Too(B)-Bot(B)), which equals fpA-fpB+fpB-fpC, which simplifies to fpA-fpC, the same exact expression given by method 1.</div><div dir="auto"><br></div><div dir="auto">Method 3. Elect argmax (S(X)+S'(X)).</div><div dir="auto"><br></div><div dir="auto">Note that S(A)+S'(A) can (by rearrangement of terms) be written as</div><div dir="auto">Bot(C)+Top(A)+Top(B) minus</div><div dir="auto">[Top(C) +Bot(A)+Bot(B)].</div><div dir="auto"><br></div><div dir="auto">The three positive terms cinsist in the Top counts of friendly candidates and the Botoom count of the unfriendly candidate.</div><div dir="auto"><br></div><div dir="auto">The subtrahend consists of the bracketed terms which are the Bottom counts of the friendly candidates and the Top count of the unfriendly candidate, as it should be: we want our friends to have high top counts and low bottom counts ... and vice-versa for those who vex us.</div><div dir="auto"><br></div><div dir="auto">Does this heuristic make sense as a foundation for a Friendly Voting Method?</div><div dir="auto"><br></div><div dir="auto">-Forest</div></div>
</blockquote></div>
</blockquote></div></div></div>
</blockquote></div>