[EM] 3 or more choices - Condorcet
Robert Bristow-Johnson wrote (1 Oct 2012): my spin is similar. Ranked Pairs simply says that some elections (or runoffs) speak more loudly than others. those with higher margins are more definitive in expressing the will of the electorate than elections with small margins. of course, a margin of zero is a tie and this says *nothing* regarding the will of the electorate, since it can go either way. the reason i like margins over winning votes is that the margin, in vote count, is the product of the margin as a percent (that would be a measure of the decisiveness of the electorate) times the total number of votes (which is a measure of how important the election is). so the margin in votes is the product of salience of the race times how decisive the decision is. Say there are 3 candidates and the voters have the option to fully rank them, but instead they all just choose to vote FPP-style thus: 49: A 48: B 03: C Of course the only possible winner is A. Now say the election is held again (with the same voters and candidates), and the B voters change to BC giving: 49: A 48: BC 03: C Now to my mind this change adds strength to no candidate other than C, so the winner should either stay the same or change to C. Does anyone disagree? So how do you (Robert or whoever the cap fits) justify to the A voters (and any fair-minded person not infatuated with the Margins pairwise algorithm) that the new Margins winner is B?? The pairwise comparisons: BC 48-3, CA 51-49, AB 49-48. Ranked Pairs(Margins) gives the order BCA. I am happy with either A or C winning, but a win for C might look odd to people accustomed to FPP and/or IRV. *If* we insist on a Condorcet method that uses only information contained in the pairwise matrix (and so ignoring all positional or approval information) then *maybe* Losing Votes is the best way to weigh the pairwise results. (So the strongest pairwise results are those where the loser has the fewest votes and, put the other way, the weakest results are those where the loser gets the most votes). In the example Losing Votes elects A. Winning Votes elects C which I'm fine with, but I don't like Winning Votes for other reasons. Chris Benham Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] 3 or more choices - Condorcet
Hi Chris, You discuss Winning Votes vs. Margins below. What do you think about using the Cardinal-Weighted Pairwise array in conjunction with the traditional Condorcet array? In other words, either WV or Margins is used to decide whether there is a defeat, but the CWP array is used to determine the defeat strength, in either Ranked Pairs or Schulze. To recap for those not familiar with the technique (due to James Green-Armytage in 2004), a ratings ballot is used: give a score of a_i to candidate i. Ranks are inferred: candidate i receives one Condorcet vote over candidate j if a_i a_j. Whenever that Condorcet vote is recorded into the standard A_ij array, you also tally the difference (a_i - a_j) into the corresponding CWP_ij location. Ted On 08 Nov 2012 08:55:24 -0800, Chris Benham wrote: Robert Bristow-Johnson wrote (1 Oct 2012): my spin is similar. Ranked Pairs simply says that some elections (or runoffs) speak more loudly than others. those with higher margins are more definitive in expressing the will of the electorate than elections with small margins. of course, a margin of zero is a tie and this says *nothing* regarding the will of the electorate, since it can go either way. the reason i like margins over winning votes is that the margin, in vote count, is the product of the margin as a percent (that would be a measure of the decisiveness of the electorate) times the total number of votes (which is a measure of how important the election is). so the margin in votes is the product of salience of the race times how decisive the decision is. Say there are 3 candidates and the voters have the option to fully rank them, but instead they all just choose to vote FPP-style thus: 49: A 48: B 03: C Of course the only possible winner is A. Now say the election is held again (with the same voters and candidates), and the B voters change to BC giving: 49: A 48: BC 03: C Now to my mind this change adds strength to no candidate other than C, so the winner should either stay the same or change to C. Does anyone disagree? So how do you (Robert or whoever the cap fits) justify to the A voters (and any fair-minded person not infatuated with the Margins pairwise algorithm) that the new Margins winner is B?? The pairwise comparisons: BC 48-3, CA 51-49, AB 49-48. Ranked Pairs(Margins) gives the order BCA. I am happy with either A or C winning, but a win for C might look odd to people accustomed to FPP and/or IRV. *If* we insist on a Condorcet method that uses only information contained in the pairwise matrix (and so ignoring all positional or approval information) then *maybe* Losing Votes is the best way to weigh the pairwise results. (So the strongest pairwise results are those where the loser has the fewest votes and, put the other way, the weakest results are those where the loser gets the most votes). In the example Losing Votes elects A. Winning Votes elects C which I'm fine with, but I don't like Winning Votes for other reasons. Chris Benham Election-Methods mailing list - see http://electorama.com/em for list info -- PO Box 3707 MC 0R-JF (Google Voice) 206-552-9611 Seattle, WA 98124-2207 (Fax) 425-717-3652 http://directorysearch.web.boeing.com/bps/details.asp?bemsid=1660261 --- Frango ut patefaciam -- I break so that I may reveal --- -- araucaria dot araucana at gmail dot com Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] 3 or more choices - Condorcet
On 11/8/12 11:55 AM, Chris Benham wrote: Robert Bristow-Johnson wrote (1 Oct 2012): my spin is similar. Ranked Pairs simply says that some elections (or runoffs) speak more loudly than others. those with higher margins are more definitive in expressing the will of the electorate than elections with small margins. of course, a margin of zero is a tie and this says *nothing* regarding the will of the electorate, since it can go either way. the reason i like margins over winning votes is that the margin, in vote count, is the product of the margin as a percent (that would be a measure of the decisiveness of the electorate) times the total number of votes (which is a measure of how important the election is). so the margin in votes is the product of salience of the race times how decisive the decision is. Say there are 3 candidates and the voters have the option to fully rank them, but instead they all just choose to vote FPP-style thus: 49: A 48: B 03: C Of course the only possible winner is A. Now say the election is held again (with the same voters and candidates), and the B voters change to BC giving: 49: A 48: BC 03: C Now to my mind this change adds strength to no candidate other than C, so the winner should either stay the same or change to C. Does anyone disagree? So how do you (Robert or whoever the cap fits) justify to the A voters (and any fair-minded person not infatuated with the Margins pairwise algorithm) that the new Margins winner is B?? The pairwise comparisons: BC 48-3, CA 51-49, AB 49-48. Ranked Pairs(Margins) gives the order BCA. I am happy with either A or C winning, but a win for C might look odd to people accustomed to FPP and/or IRV. *If* we insist on a Condorcet method that uses only information contained in the pairwise matrix (and so ignoring all positional or approval information) then *maybe* Losing Votes is the best way to weigh the pairwise results. (So the strongest pairwise results are those where the loser has the fewest votes and, put the other way, the weakest results are those where the loser gets the most votes). In the example Losing Votes elects A. Winning Votes elects C which I'm fine with, but I don't like Winning Votes for other reasons. well, i'm not the guy with upper-case letters. i didn't comment on this response to what i said, but looking it over right now, whether people vote on a Ranked-Choice ballot as if it were FPP or not, any candidates *not* ranked are tied for last place on that ballot. so 49: A 48: B 03: C is really 49: AB AC 48: BA BC 03: CA CB the difference with this: 49: A 48: BC 03: C is that C gets a helluva lot more support from B voters than the other scenario. 49: AB AC 48: BCA 03: CA CB it's the same old complaint that Rob Ritchie (FairVote) and others make against Condorcet (justifying putting all their support behind IRV): that (from their POV) Condorcet can elect wishy-washy candidates with little primary support. i (and Condorcet) would say that in the second case, C is the best candidate even if he/she got only 3 first-choice votes. might be a nice centrist, no-drama candidate in a polarized environment. in Burlington VT 2009, the 3rd-place finisher from the POV or FPP or IRV was the Condorcet winner and nearly everyone i talked with would have been much happier with this candidate than with whom actually won the IRV or with whom would have won FPP (who suffered a decisive defeat in a repeat run in March 2012). but the margins weren't so wide as with this example, candidate C got a lot more than 3% primary support votes. again, i will repeat that probably, technically, Schulze is superior to Ranked-Pairs. but it doesn't matter with a Smith Set of 3 candidates or less. Condorcet cycles will be rare. cycles with more than 3 in the Smith set will be rare of the rare. it's best to get Condorcet of *some* method enacted into law. the most realistic path to accomplishing that is *not* to advocate a method that cannot be explained to citizen-legislators. and i still think that margins is better than either winning votes (or the logical complement regarding the most losing votes). margins encompasses *both* winning votes and losing votes (the latter with a negative sign, of course). -- r b-j r...@audioimagination.com Imagination is more important than knowledge. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] 3 or more choices - Condorcet
this is the 2nd resend. it's been hours since i first posted this and it hasn't shown up on the list. On 11/8/12 11:55 AM, Chris Benham wrote: Robert Bristow-Johnson wrote (1 Oct 2012): my spin is similar. Ranked Pairs simply says that some elections (or runoffs) speak more loudly than others. those with higher margins are more definitive in expressing the will of the electorate than elections with small margins. of course, a margin of zero is a tie and this says *nothing* regarding the will of the electorate, since it can go either way. the reason i like margins over winning votes is that the margin, in vote count, is the product of the margin as a percent (that would be a measure of the decisiveness of the electorate) times the total number of votes (which is a measure of how important the election is). so the margin in votes is the product of salience of the race times how decisive the decision is. Say there are 3 candidates and the voters have the option to fully rank them, but instead they all just choose to vote FPP-style thus: 49: A 48: B 03: C Of course the only possible winner is A. Now say the election is held again (with the same voters and candidates), and the B voters change to BC giving: 49: A 48: BC 03: C Now to my mind this change adds strength to no candidate other than C, so the winner should either stay the same or change to C. Does anyone disagree? So how do you (Robert or whoever the cap fits) justify to the A voters (and any fair-minded person not infatuated with the Margins pairwise algorithm) that the new Margins winner is B?? The pairwise comparisons: BC 48-3, CA 51-49, AB 49-48. Ranked Pairs(Margins) gives the order BCA. I am happy with either A or C winning, but a win for C might look odd to people accustomed to FPP and/or IRV. *If* we insist on a Condorcet method that uses only information contained in the pairwise matrix (and so ignoring all positional or approval information) then *maybe* Losing Votes is the best way to weigh the pairwise results. (So the strongest pairwise results are those where the loser has the fewest votes and, put the other way, the weakest results are those where the loser gets the most votes). In the example Losing Votes elects A. Winning Votes elects C which I'm fine with, but I don't like Winning Votes for other reasons. well, i'm not the guy with upper-case letters. i didn't comment on this response to what i said, but looking it over right now, whether people vote on a Ranked-Choice ballot as if it were FPP or not, any candidates *not* ranked are tied for last place on that ballot. so 49: A 48: B 03: C is really 49: AB AC 48: BA BC 03: CA CB the difference with this: 49: A 48: BC 03: C is that C gets a helluva lot more support from B voters than the other scenario. 49: AB AC 48: BCA 03: CA CB it's the same old complaint that Rob Ritchie (FairVote) and others make against Condorcet (justifying putting all their support behind IRV): that (from their POV) Condorcet can elect wishy-washy candidates with little primary support. i (and Condorcet) would say that in the second case, C is the best candidate even if he/she got only 3 first-choice votes. might be a nice centrist, no-drama candidate in a polarized environment. in Burlington VT 2009, the 3rd-place finisher from the POV or FPP or IRV was the Condorcet winner and nearly everyone i talked with would have been much happier with this candidate than with whom actually won the IRV or with whom would have won FPP (who suffered a decisive defeat in a repeat run in March 2012). but the margins weren't so wide as with this example, candidate C got a lot more than 3% primary support votes. again, i will repeat that probably, technically, Schulze is superior to Ranked-Pairs. but it doesn't matter with a Smith Set of 3 candidates or less. Condorcet cycles will be rare. cycles with more than 3 in the Smith set will be rare of the rare. it's best to get Condorcet of *some* method enacted into law. the most realistic path to accomplishing that is *not* to advocate a method that cannot be explained to citizen-legislators. and i still think that margins is better than either winning votes (or the logical complement regarding the most losing votes). margins encompasses *both* winning votes and losing votes (the latter with a negative sign, of course). -- r b-j r...@audioimagination.com Imagination is more important than knowledge. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] testing 1,2,3.... hellooooo?
none of my messages seem to be getting posted to the EM list, today. -- r b-j r...@audioimagination.com Imagination is more important than knowledge. Election-Methods mailing list - see http://electorama.com/em for list info
