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

[Clinton Mead]

On Wed, Nov 30, 2011 at 2:47 AM, Kristofer Munsterhjelm <
km_elmet at lavabit.com> wrote:

> Clinton Mead wrote:
>
>> What would be the simplest paper count that will produce a winner in the
>> smith set?
>
>
> Then while there is more than one candidate left, eliminate the pairwise
> loser of the two remaining candidates with the least approval score.
>
>
I was thinking the a similar thing, but then I looked at typical election
first preference counts, which are often like the following like the
following (at least in Australia).

Major Party 1: 41%
Major Party 2: 39%
Minor Party 1: 10%
Minor Party 2: 3%
Minor Party 3: 2%
Minor Party 4: 2%
Minor Party 5: 2%
Minor Party 6: 1%

Producing this entails one pass through the ballot (100% of papers need to
be examined for one preference). This is the base line for FPP.

For IRV, we can eliminate the 6 minor candidates in bulk, because they sum
to less than the vote of the second candidate. So then we reexamine the 6
minor party candidates for preferences. This means we have to look at 20%
more votes.

Our total work of IRV: 120% of FPP.

For the method you suggest (with the minor difference that I'll use first
preference to initially rank ballots) we need 100% work to do the first
count, then, with Minor Party 6 and Minor Party 5 on a total of 3%, we need
to look at the other 97% of ballots to do a pairwise comparison. Then we
eliminate the pairwise loser, then we have to look at 95% of the remaining
ballots to compare Minor 5/6 with Minor 4, etc.

We get a result which might involve 400-500% more looks at ballot papers
than FPP.

The method I suggest is slightly different in that it starts from the top.
We compare the top two candidates pairwise, this involves looking at 100%
of the votes to do the first preference count, and an additional 20% to do
the two candidate count. We're now at 120% of the work of FPP (one could
say you could do the first preference and two candidate in the same pass,
but this doesn't significantly lower the work, the time consuming work is
looking for preferences, not flicking through ballot papers).

Now, lets say Major Party 2 beats Major Party 1 pairwise. Then we
distribute Major Party 1's preferences. This takes looking at 41% of the
votes.

Now we're at 161%.

If Major Party 2 now has a majority, we have a winner. But if it doesn't,
we pairwise compare Major Party 2 with Minor Party 1.

This involves looking at all of the other minors vote, which is 10%. If
Major Party 2 beats Minor Party 1, it probably has a majority, if it
doesn't, it's very close, and will quickly get a majority as it defeats
other minor candidates pairwise and distributes their preferences.

If Minor Party 1 beats Major Party 2 however, then we need to distribute
Major Party 2's preferences,  which involves looking at another 39-50% of
the ballots.

At this point, it is likely Minor Party 1 has a majority (particularly
considering Major 1 and Major 2 are eliminated). If this is the case, we
then just need to check Minor Party 1 v Major Party 1 pairwise. This
involves looking at the 49% of votes which are not theirs. If Minor Party 1
pairwise beats Major Party 1, it is the winner, if Major Party 1 beats
Minor Party 1, we have a cycle (Major Party 1 < Major Party 2 < Minor Party
1 < Major Party 1) and we resolve this in favour of the first preference
winner.

The idea of this method is that it typically takes only 160% or so of the
work of a FPP, and in the worse (typical) case perhaps 250% or at most 300%
of the work of a FPP count. In particular, it's not particularly burdensome
compared to IRV, which in the best case is around 120% and in the worse
case around 200%.
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