<html><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; ">Been a busy day on this thread. I will try for Condorcet, to sell that it can be good and usable by voters without requiring much new understanding by them.<div><br></div><div>Ranking: </div><div> IRV ranking, learned by many, is a start, with equal ranking a trivial addition.</div><div> Approval voting permissible and usable as valid Condorcet by using a single rank number.</div><div><br></div><div>How many rank numbers? Three, as in IRV, is probable reasonable minimum. More needs thought, but not necessarily many - usability of equal ranks minimizes true need for more. </div><div><br></div><div>Write-ins - need being doable and countable with reasonable effort.</div><div><br></div><div>Majority: Needed by plurality. Minimum truly needed can be different for other methods.</div><div><br></div><div>Counting: I promote an N*N matrix with a row and column for each candidate, and an N array with an element for each candidate. If a write-in is in one array/matrix and not in another, a merged array/matrix of the precincts will need to include such.</div><div> When reading a ballot each ranked candidate gets counted in the array. For each pair of ranked candidates this will later be counted in the matrix as if both won so, right now, adjust in the matrix for the losing candidate - and for equal ranks if that is needed.</div><div> In the final array/matrix, after reading all ballots, add each array element to each element in its row in the N*N matrix.</div><div> This both reduces labor while reading ballots and provides for proper counting of write-ins.</div><div><br></div><div>Cycles: Cycle members have to approach equality for the cycle to possibly complete, and each cycle member would be CW in absence of other cycle members. Thus it seems reasonable to look only among members to decide who should act as CW.</div><div><br></div><div>Hybrids such as Approval/Asset? I choke because their combined methods end up making complications for voters.</div><div><br></div><div>Dave Ketchum</div><div><br></div><div><div><div>On Jun 3, 2011, at 7:01 PM, Jameson Quinn wrote:</div><br class="Apple-interchange-newline"><blockquote type="cite"><div class="gmail_quote"><div>I thought of a simpler way to explain my "safety" fix. The full system description follows, with my new phrasing in bold.</div><div><br></div></div><blockquote class="webkit-indent-blockquote" style="margin: 0 0 0 40px; border: none; padding: 0px;"> <div class="gmail_quote"><div>N days before the election, all candidates (including declared write-in candidates) rank order all other candidates (including declared write-in candidates). These orderings are announced. In the election, all voters submit an approval ballot, with two spaces for write-ins. Total approvals and number of bullet votes are counted for each candidate and announced. (Bullet votes are votes for only one candidate, including all valid or invalid write-in votes.) Then each candidate may grant the number of bullet votes they received to N other candidates from the top of their preference list, where N can be any number including 0. All candidates decide what number N to use simultaneously, and then those decisions are announced publicly. <b>Take the two candidates with the highest approvals. Recount those two as if they hadn't approved each other (that is, without adding any bullet votes from one to the other).</b> The winner is the candidate <b>of those two</b> with the highest approval in this final count.</div> </div></blockquote><div class="gmail_quote"><div><br></div><div>The purpose of removing the mutual votes from the top two before deciding the pairwise winner of these two is, as I explained before, to make it so that one candidate will never lose because they approved another one. This frees candidates to be honest in their approvals.</div> <div><br></div><div>I believe that this system, as described, is pareto-dominant over plurality, asset, and approval.</div><div><br></div><div>It is also very Condorcet-compliant. That is, assuming that X% of all candidates' voters agree with their candidate's preference order, and that the other (100-X)% have preferences which cancel each other out (random noise); that this X is the same for all candidates; that all voters who do not agree with their candidate do not bullet-vote (voting for a random number of extra candidates), and all voters who do agree with their candidate do bullet vote; and that there is a true pairwise champion; then the pairwise champion will win in a (unique) strong Nash equilibrium. This is a very solid result, which relies on the perfect information of the candidates when choosing how to "delegate" their approvals; it is NOT true of systems such as Approval or even DYN (without the preference-ordering and top-two-pairwise-recount aspects). It is not even true of any Condorcet system I know of (because of strategy)! So this system (and some obvious variants) is in fact <b>the most Condorcet-compliant</b> system I know of.</div> <div><br></div><div>Since it is also relatively simple to understand - not as simple as approval, but not too far behind - I think it makes an excellent candidate for a practical proposal.</div><div><br></div><div>Jameson Quinn</div> <div> </div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;"><div class="gmail_quote"><div></div><div>I'd add my "safety" fix to the near-clone problem, <u><b>if</b></u> someone can think of an easy way to describe and motivate it. Basically, it looks at any candidate who mutually approve with the winner, and sees if they would beat the winner (pairwise) with those mutual approvals turned off. This helps when, for instance (honest preferences):</div> <div><br></div><div>35: X1>X2</div><div>25: X2>X1</div><div>21: Y>>X2</div><div>19: Y>>X1</div><div><br></div><div>If X1 and X2 approve each other, the right thing happens (X1 wins), no matter what Y voters do. If they do not, this fix does not attempt to read anyone's minds (or to ask people again in a runoff).</div></div></blockquote></div></blockquote></div></div></body></html>