You can eliminate all candidates that have less than half the top approval score, immediately. Australia is a bad example because they require full ranking; but without that requirement, you could expect that each candidate's approval total will be something on the order of min(50%, 2xFirst choices). Using the numbers you gave, that would mean only 3 parties left after elimination.
Jameson 2011/11/29 Clinton Mead <clintonm...@gmail.com> > > > On Wed, Nov 30, 2011 at 2:47 AM, Kristofer Munsterhjelm < > km_el...@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%. > > ---- > Election-Methods mailing list - see http://electorama.com/em for list info > >
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