Re: [EM] Simplest paper count to produce a winner in the smith set.
Clinton Mead wrote: What would be the simplest paper count that will produce a winner in the smith set? I'm not completely sure how your method works, but how about this? Count the number of ballots on which each candidate is ranked (in any position). Call each candidate's count his approval score. Then while there is more than one candidate left, eliminate the pairwise loser of the two remaining candidates with the least approval score. Eliminating losers won't turn a ranked candidate into a non-ranked candidate (or vice versa), so unlike IRV, you don't have to do a recount for every round. It should also pick a winner from the Smith set. Say all but one of the Smith set candidates have been eliminated. Then the remaining Smith set candidate will, by definition of the Smith set, beat everybody else pairwise, and so nobody will be able to eliminate him. Therefore, that remaining Smith set member will be elected. QED. The counting burden isn't that hard, either: you need one pass to calculate the approval scores, and then (n-1) (for n candidates) pairwise counts. I don't think you can do better than (n-1) pairwise counts and still be sure to elect someone in the Smith set. You could replace the approval count with a Plurality count to get something simpler, but that could also be unfair to Smith set candidates that have few first place votes, and it wouldn't be consistent: eliminating candidates from a Plurality count would mean other candidates could be exposed, and so you'd have to recount as in IRV. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Simplest paper count to produce a winner in the smith set.
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
Re: [EM] Simplest paper count to produce a winner in the smith set.
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 Election-Methods mailing list - see http://electorama.com/em for list info
