What would be the simplest paper count that will produce a winner in the smith set?
I was thinking of just collecting the first preference totals for all candidates, and comparing the top two candidates on a two candidate preferred basis (by examining other ballots). Transfers the votes of the loser to their next preferred candidates, and then do a two candidate count between the last winner and the third candidate. Then, the winner of this contest has a two candidate preferred count against the fourth candidate, and this continues til a candidate has collected a majority of the remaining vote. This candidate remaining with a majority is our notional winner. Eliminate permanently all candidates which were beaten by this candidate directly, and with the remaining candidates, do a two candidate preferred vote against this notional winner in order of their first preference vote. If they beat this notional winner, they are the winner, however if no-one beats this notional winner, the notional winner becomes the winner. 5: AC 4: B 2: CB 1: D 1: E B beats A pairwise 6 to 5. Transfer As votes and we get: 7 for C (5: C, 2: CB) 4 for B 1 for D 1 for E C beats B pairwise 7 to 4. C now has a majority of the remaining votes, so it's the notional winner. Eliminate all candidates C has directly defeated, namely B, D, E. Transfer all votes: A: 5 B: 2 A beats C. A is the winner. (if the 4: B voters voted 4: BC, then C would have won). This method I believe will always pick the member of the smith set with the highest first plurality vote. I don't think it's too much harder to count than IRV, except perhaps in very unusual circumstances. In most cases the notional winner will quickly reach a majority through transfers, and in the few circumstances that this isn't one of the top two candidates, it'll probably be the third. And if it is that just involves a quick pairwise check, of which most of the work has probably already been done, as A and C have already been counted, a number of Bs votes would have already been moved next to A and C's pile, it's only a matter of transferring votes which currently sit with the 4th and lower candidates. Is there a case where this method won't pick a member of the smith set, or is there a method which still picks a member of the smith set but is easier to do a paper count for? Clinton
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