[EM] Simplest paper count to produce a winner in the smith set.

[Clinton Mead]

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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