…& of course multiply that 18090 by the number of ballots, to get an idea of what’s involved in the 18090 order-determinations to be done on each ballot, & recorded, & then summed, to obtain each of the 18090 pairwise vote-totals. …each of which then must be carried or transmitted to where the central count is done. On Sun, May 26, 2024 at 22:03 Michael Ossipoff <email9648742@gmail.com> wrote: > > > On Sun, May 26, 2024 at 12:56 Chris Benham <cbenhamau@yahoo.com.au> wrote: > >> >> In 2003 there was California gubernatorial election with 135 candidates. > > > That’s 18090 pairwise vote-totals to determine at each precinct from the > rankings, by examining each ranking to determine which member of each > possible candidate-pair is ranked over the other on that ballot. > > …& 18080 pairwise vote-totals for the precincts to sum, store, & transmit > or carry to the central count location. > > …& verify in an audit. > > > >> >> >> https://en.wikipedia.org/wiki/2003_California_gubernatorial_recall_election#Results >> >> Chris B. >> >> On 26/05/2024 9:29 pm, Kristofer Munsterhjelm wrote: >> > On 2024-05-26 07:28, Michael Ossipoff wrote: >> >> Someone keeps repeating that the voters shouldn’t have to vote >> >> strategically. He wants the method to do it all for us, after we >> >> merely state our sincere-rankings. >> >> >> >> That’s of course a common attitude: >> >> >> >> …wanting a high-tech, computation-intensive,computer-dependent system >> >> to do it all for us, taking all the actual choosing responsibility >> >> off of us.…sheltering & isolating us from the choice. >> > >> > I would prefer that you do not attribute opinions that the proponents >> > have not expressed. Nowhere have I said that Condorcet "[isolates] us >> > from the choice" we make. >> > >> > What you call "sheltering" and "isolating", I see as the method taking >> > proper responsibility - proper responsibility to turn the voters' >> > unambiguous honest opinions into an outcome without dumping the >> > algorithmic calculation upon te voter themselves. See my "kick the can >> > down the road" post for more info. >> > >> > Now, I could make a caricature of Approval itself. Perhaps something >> > about a calculator that just says "IDK, do the base conversion >> > yourself, I only accept input numbers in factoradic". But caricatures >> > only make people angry. Let's not stoop to them, shall we? >> > >> >> I’ll ask this for the 3^rd time: >> >> >> >> … >> >> >> >> How would like you to handount-audit a Condorcet count for a >> >> many-candidate national presidential election? >> > >> > I think your question assumes something that won't hold. If you have a >> > 25-candidate presidential election, you've already lost, because >> > nobody is going to rank 25 candidates, irrespective of whether the >> > method is Condorcet, IRV, or Borda. >> > >> > I'm not familiar with minor parties in the US. Has there ever been a >> > 25-candidate presidential election? >> > >> > I don't think I can comment beyond that: I don't know enough about >> > poll workers. I'll leave that to someone with experience. (Although >> > roughly calculating: suppose 6 candidates like in Burlington. That's >> > 30 pairs. Five times the work if counting a particular preference is >> > as before. So you'd either need 5x the workers, or five times the >> > time, or some combination of the two.) >> > >> > On an aside, though, I would say that I generally wouldn't want to get >> > computers anywhere near election counting. However, if you absolutely >> > have to have them, there are ways of making sure they don't cheat: >> > formal verification. You could also create special-purpose tools that >> > say, only turn ranks into matrices and nothing else: there's no reason >> > (apart from programmer convenience) why an election tool should be a >> > general purpose computer that you could hide all sorts of shenanigans >> in. >> > >> >> I’ve discussed that at length in previous posts, & it probably isn’t >> >> necessary to again post about ways of choosing how to vote in Approval. >> >> >> >> But, just summarize:It’s easy.Whichever of the various ways you >> >> prefer to use, for choosing whom to approve, it’s easy.…& no, it >> >> doesn’t require knowing your objectively-optimal vote. >> > >> >> >> >> I’ve many times pointed out that Approval’s Myerson-Weber equilibrium >> >> is the voter-median. >> >> >> >> i.e. Approval soon homes in on where the Condorcet-Winner is. >> > >> > Not necessarily. See the following paper: >> > https://arxiv.org/abs/2001.04216v2 >> > >> > According to the authors, under their model, if every voter follows a >> > particular thresholding rule, then iterative approval generally >> > arrives at the Condorcet winner (but not always). However, if voters >> > are left to choose any thresholding rule they want (as you propose), >> > then anything is possible. The arrangement might even elect a >> > Condorcet loser, and the outcome may be slow to converge or who wins >> > may depend on how many polls you hold. >> > >> > I'm reminded of a quote about distributed algorithms that I read >> > somewhere: "It's really easy to design distributed algorithms that >> > suffer from deadlock, network floods or widely unpredictable and >> > bizarre oscillations". That's in the context of computer science - >> > deadlocks might not be applicable to election methods. But it does >> > justify a starting position of skepticisim when considering schemes >> > that offload more of the work to the voters by turning a one-shot >> > method into a dynamical system. >> > >> >> It seems to me that, in every one of EM’s polls, including the recent >> >> one, Approval chose the CW. >> >> >> >> Have we forgotten that? >> > >> > I haven't, nor have I forgotten that Approval wasn't actually the poll >> > winner. >> > >> > But let's take this reasoning at face value. I don't have the other >> > polls' ballot data available at the moment, so let's consider the >> > latest one and pick... say, Borda. >> > >> > As one can see by going to >> > https://munsterhjelm.no/km/rbvote/calc.html, pasting in the data, and >> > clicking Borda, Ranked Pairs is also the Borda winner. >> > >> > But I don't think I'm going to start advocating for Borda. >> > >> > >> > To not be accused of tu quoque, let me clarify the point. As >> > Burlington shows, an election method needs to handle the hard cases, >> > not just the easy ones, or there may be a considerable uproar when the >> > method is faced with a hard case and then stumbles. (In addition, >> > people trying to make sure a stumble doesn't happen may start to do >> > mass compromising, further entrenching two-party rule.) >> > >> > So it's quite possible that our polls are easy cases. But like >> > FairVote claiming that IRV gets the Condorcet winner more than 90% of >> > the time, that doesn't by itself tell us much, because the failures >> > have such a strong impact. >> > >> > I illustrate the example above by picking a method we know to be bad >> > (Borda being so extremely easy to fool with cloning and burial), and >> > showing that the poll result comes out right. If a bad method can get >> > a good result, then "getting a good result" is less useful than it >> > might appear at first glance. >> > >> > -km >> > ---- >> > Election-Methods mailing list - see https://electorama.com/em for list >> > info >> >

Suppose that there are 300 million voters. Round 18090 off to 20000x 3E8 X 2E4 = 6E12 6 trillion. That’s roughly the number of miles in a light-year. Whatever the number of voters, nationally, is, suppose that every one of them participated in the handcount. With the work divided among them all, each has “only” an amount of work equal to that of determining which member of each one of 18090/2 pairs of candidates is ranked over the other on a ballot. 135 candidates in each ranking. On the average, s/he’d only have to look at half of the ranking’s candidates to find each of the 2 members of each pair. …but s/he’d have to do it for both. So the order-determination for each candidate-pair, on the average will require looking at 135 candidates. Suppose that s/he can skim over 10 of them in a second, when doing those searches. Then it would take (135)(18090/2)/10 seconds. That’s about 1.4 days. But say she only does it for 8 hours per day (with no breaks). Now it’s more like 4.2 days. But supervision is the whole point. If s/he’s working alone, she can say that the pairwise vote-totals are whatever s/he wants them to be. So in reality, it would be counting *teams* each wit representatives of several parties. Say (optimistically) there are 10 parties. Now it will take 42 days. But the parties don’t really have equal numbers of members. It’s going to take longer. …&, realistically, they aren’t going to scan 10 ranked candidates every second for 8 hours with no breaks. It would obviously take months. Might it not, in fact, be measured in years…with every one of the nation’s voters participating in that handcount? On Sun, May 26, 2024 at 22:11 Michael Ossipoff <email9648742@gmail.com> wrote: > …& of course multiply that 18090 by the number of ballots, to get an idea > of what’s involved in the 18090 order-determinations to be done on each > ballot, & recorded, & then summed, to obtain each of the 18090 pairwise > vote-totals. > > …each of which then must be carried or transmitted to where the central > count is done. > > On Sun, May 26, 2024 at 22:03 Michael Ossipoff <email9648742@gmail.com> > wrote: > >> >> >> On Sun, May 26, 2024 at 12:56 Chris Benham <cbenhamau@yahoo.com.au> >> wrote: >> >>> >>> In 2003 there was California gubernatorial election with 135 >>> candidates. >> >> >> That’s 18090 pairwise vote-totals to determine at each precinct from the >> rankings, by examining each ranking to determine which member of each >> possible candidate-pair is ranked over the other on that ballot. >> >> …& 18080 pairwise vote-totals for the precincts to sum, store, & transmit >> or carry to the central count location. >> >> …& verify in an audit. >> >> >> >>> >>> >>> https://en.wikipedia.org/wiki/2003_California_gubernatorial_recall_election#Results >>> >>> Chris B. >>> >>> On 26/05/2024 9:29 pm, Kristofer Munsterhjelm wrote: >>> > On 2024-05-26 07:28, Michael Ossipoff wrote: >>> >> Someone keeps repeating that the voters shouldn’t have to vote >>> >> strategically. He wants the method to do it all for us, after we >>> >> merely state our sincere-rankings. >>> >> >>> >> That’s of course a common attitude: >>> >> >>> >> …wanting a high-tech, computation-intensive,computer-dependent system >>> >> to do it all for us, taking all the actual choosing responsibility >>> >> off of us.…sheltering & isolating us from the choice. >>> > >>> > I would prefer that you do not attribute opinions that the proponents >>> > have not expressed. Nowhere have I said that Condorcet "[isolates] us >>> > from the choice" we make. >>> > >>> > What you call "sheltering" and "isolating", I see as the method taking >>> > proper responsibility - proper responsibility to turn the voters' >>> > unambiguous honest opinions into an outcome without dumping the >>> > algorithmic calculation upon te voter themselves. See my "kick the can >>> > down the road" post for more info. >>> > >>> > Now, I could make a caricature of Approval itself. Perhaps something >>> > about a calculator that just says "IDK, do the base conversion >>> > yourself, I only accept input numbers in factoradic". But caricatures >>> > only make people angry. Let's not stoop to them, shall we? >>> > >>> >> I’ll ask this for the 3^rd time: >>> >> >>> >> … >>> >> >>> >> How would like you to handount-audit a Condorcet count for a >>> >> many-candidate national presidential election? >>> > >>> > I think your question assumes something that won't hold. If you have a >>> > 25-candidate presidential election, you've already lost, because >>> > nobody is going to rank 25 candidates, irrespective of whether the >>> > method is Condorcet, IRV, or Borda. >>> > >>> > I'm not familiar with minor parties in the US. Has there ever been a >>> > 25-candidate presidential election? >>> > >>> > I don't think I can comment beyond that: I don't know enough about >>> > poll workers. I'll leave that to someone with experience. (Although >>> > roughly calculating: suppose 6 candidates like in Burlington. That's >>> > 30 pairs. Five times the work if counting a particular preference is >>> > as before. So you'd either need 5x the workers, or five times the >>> > time, or some combination of the two.) >>> > >>> > On an aside, though, I would say that I generally wouldn't want to get >>> > computers anywhere near election counting. However, if you absolutely >>> > have to have them, there are ways of making sure they don't cheat: >>> > formal verification. You could also create special-purpose tools that >>> > say, only turn ranks into matrices and nothing else: there's no reason >>> > (apart from programmer convenience) why an election tool should be a >>> > general purpose computer that you could hide all sorts of shenanigans >>> in. >>> > >>> >> I’ve discussed that at length in previous posts, & it probably isn’t >>> >> necessary to again post about ways of choosing how to vote in >>> Approval. >>> >> >>> >> But, just summarize:It’s easy.Whichever of the various ways you >>> >> prefer to use, for choosing whom to approve, it’s easy.…& no, it >>> >> doesn’t require knowing your objectively-optimal vote. >>> > >>> >> >>> >> I’ve many times pointed out that Approval’s Myerson-Weber equilibrium >>> >> is the voter-median. >>> >> >>> >> i.e. Approval soon homes in on where the Condorcet-Winner is. >>> > >>> > Not necessarily. See the following paper: >>> > https://arxiv.org/abs/2001.04216v2 >>> > >>> > According to the authors, under their model, if every voter follows a >>> > particular thresholding rule, then iterative approval generally >>> > arrives at the Condorcet winner (but not always). However, if voters >>> > are left to choose any thresholding rule they want (as you propose), >>> > then anything is possible. The arrangement might even elect a >>> > Condorcet loser, and the outcome may be slow to converge or who wins >>> > may depend on how many polls you hold. >>> > >>> > I'm reminded of a quote about distributed algorithms that I read >>> > somewhere: "It's really easy to design distributed algorithms that >>> > suffer from deadlock, network floods or widely unpredictable and >>> > bizarre oscillations". That's in the context of computer science - >>> > deadlocks might not be applicable to election methods. But it does >>> > justify a starting position of skepticisim when considering schemes >>> > that offload more of the work to the voters by turning a one-shot >>> > method into a dynamical system. >>> > >>> >> It seems to me that, in every one of EM’s polls, including the recent >>> >> one, Approval chose the CW. >>> >> >>> >> Have we forgotten that? >>> > >>> > I haven't, nor have I forgotten that Approval wasn't actually the poll >>> > winner. >>> > >>> > But let's take this reasoning at face value. I don't have the other >>> > polls' ballot data available at the moment, so let's consider the >>> > latest one and pick... say, Borda. >>> > >>> > As one can see by going to >>> > https://munsterhjelm.no/km/rbvote/calc.html, pasting in the data, and >>> > clicking Borda, Ranked Pairs is also the Borda winner. >>> > >>> > But I don't think I'm going to start advocating for Borda. >>> > >>> > >>> > To not be accused of tu quoque, let me clarify the point. As >>> > Burlington shows, an election method needs to handle the hard cases, >>> > not just the easy ones, or there may be a considerable uproar when the >>> > method is faced with a hard case and then stumbles. (In addition, >>> > people trying to make sure a stumble doesn't happen may start to do >>> > mass compromising, further entrenching two-party rule.) >>> > >>> > So it's quite possible that our polls are easy cases. But like >>> > FairVote claiming that IRV gets the Condorcet winner more than 90% of >>> > the time, that doesn't by itself tell us much, because the failures >>> > have such a strong impact. >>> > >>> > I illustrate the example above by picking a method we know to be bad >>> > (Borda being so extremely easy to fool with cloning and burial), and >>> > showing that the poll result comes out right. If a bad method can get >>> > a good result, then "getting a good result" is less useful than it >>> > might appear at first glance. >>> > >>> > -km >>> > ---- >>> > Election-Methods mailing list - see https://electorama.com/em for >>> list >>> > info >>> >>

…or of course you could, for the voting, just ask each voter to write on hir (big) ballot, which member of each of the 18090/2 candidate-pairs s/he prefers to the other. S/he might be in the voting/booth for a while… On Sun, May 26, 2024 at 23:51 Michael Ossipoff <email9648742@gmail.com> wrote: > Suppose that there are 300 million voters. Round 18090 off to 20000x > > 3E8 X 2E4 = 6E12 > > 6 trillion. > > That’s roughly the number of miles in a light-year. > > Whatever the number of voters, nationally, is, suppose that every one of > them participated in the handcount. > > With the work divided among them all, each has “only” an amount of work > equal to that of determining which member of each one of 18090/2 pairs of > candidates is ranked over the other on a ballot. > > 135 candidates in each ranking. On the average, s/he’d only have to look > at half of the ranking’s candidates to find each of the 2 members of each > pair. > > …but s/he’d have to do it for both. So the order-determination for each > candidate-pair, on the average will require looking at 135 candidates. > > Suppose that s/he can skim over 10 of them in a second, when doing those > searches. > > Then it would take (135)(18090/2)/10 seconds. That’s about 1.4 days. But > say she only does it for 8 hours per day (with no breaks). Now it’s more > like 4.2 days. > > But supervision is the whole point. If s/he’s working alone, she can say > that the pairwise vote-totals are whatever s/he wants them to be. > > So in reality, it would be counting *teams* each wit representatives of > several parties. Say (optimistically) there are 10 parties. > > Now it will take 42 days. But the parties don’t really have equal numbers > of members. It’s going to take longer. > > …&, realistically, they aren’t going to scan 10 ranked candidates every > second for 8 hours with no breaks. > > It would obviously take months. Might it not, in fact, be measured in > years…with every one of the nation’s voters participating in that handcount? > > > > On Sun, May 26, 2024 at 22:11 Michael Ossipoff <email9648742@gmail.com> > wrote: > >> …& of course multiply that 18090 by the number of ballots, to get an idea >> of what’s involved in the 18090 order-determinations to be done on each >> ballot, & recorded, & then summed, to obtain each of the 18090 pairwise >> vote-totals. >> >> …each of which then must be carried or transmitted to where the central >> count is done. >> >> On Sun, May 26, 2024 at 22:03 Michael Ossipoff <email9648742@gmail.com> >> wrote: >> >>> >>> >>> On Sun, May 26, 2024 at 12:56 Chris Benham <cbenhamau@yahoo.com.au> >>> wrote: >>> >>>> >>>> In 2003 there was California gubernatorial election with 135 >>>> candidates. >>> >>> >>> That’s 18090 pairwise vote-totals to determine at each precinct from the >>> rankings, by examining each ranking to determine which member of each >>> possible candidate-pair is ranked over the other on that ballot. >>> >>> …& 18080 pairwise vote-totals for the precincts to sum, store, & >>> transmit or carry to the central count location. >>> >>> …& verify in an audit. >>> >>> >>> >>>> >>>> >>>> https://en.wikipedia.org/wiki/2003_California_gubernatorial_recall_election#Results >>>> >>>> Chris B. >>>> >>>> On 26/05/2024 9:29 pm, Kristofer Munsterhjelm wrote: >>>> > On 2024-05-26 07:28, Michael Ossipoff wrote: >>>> >> Someone keeps repeating that the voters shouldn’t have to vote >>>> >> strategically. He wants the method to do it all for us, after we >>>> >> merely state our sincere-rankings. >>>> >> >>>> >> That’s of course a common attitude: >>>> >> >>>> >> …wanting a high-tech, computation-intensive,computer-dependent >>>> system >>>> >> to do it all for us, taking all the actual choosing responsibility >>>> >> off of us.…sheltering & isolating us from the choice. >>>> > >>>> > I would prefer that you do not attribute opinions that the proponents >>>> > have not expressed. Nowhere have I said that Condorcet "[isolates] us >>>> > from the choice" we make. >>>> > >>>> > What you call "sheltering" and "isolating", I see as the method >>>> taking >>>> > proper responsibility - proper responsibility to turn the voters' >>>> > unambiguous honest opinions into an outcome without dumping the >>>> > algorithmic calculation upon te voter themselves. See my "kick the >>>> can >>>> > down the road" post for more info. >>>> > >>>> > Now, I could make a caricature of Approval itself. Perhaps something >>>> > about a calculator that just says "IDK, do the base conversion >>>> > yourself, I only accept input numbers in factoradic". But caricatures >>>> > only make people angry. Let's not stoop to them, shall we? >>>> > >>>> >> I’ll ask this for the 3^rd time: >>>> >> >>>> >> … >>>> >> >>>> >> How would like you to handount-audit a Condorcet count for a >>>> >> many-candidate national presidential election? >>>> > >>>> > I think your question assumes something that won't hold. If you have >>>> a >>>> > 25-candidate presidential election, you've already lost, because >>>> > nobody is going to rank 25 candidates, irrespective of whether the >>>> > method is Condorcet, IRV, or Borda. >>>> > >>>> > I'm not familiar with minor parties in the US. Has there ever been a >>>> > 25-candidate presidential election? >>>> > >>>> > I don't think I can comment beyond that: I don't know enough about >>>> > poll workers. I'll leave that to someone with experience. (Although >>>> > roughly calculating: suppose 6 candidates like in Burlington. That's >>>> > 30 pairs. Five times the work if counting a particular preference is >>>> > as before. So you'd either need 5x the workers, or five times the >>>> > time, or some combination of the two.) >>>> > >>>> > On an aside, though, I would say that I generally wouldn't want to >>>> get >>>> > computers anywhere near election counting. However, if you absolutely >>>> > have to have them, there are ways of making sure they don't cheat: >>>> > formal verification. You could also create special-purpose tools that >>>> > say, only turn ranks into matrices and nothing else: there's no >>>> reason >>>> > (apart from programmer convenience) why an election tool should be a >>>> > general purpose computer that you could hide all sorts of shenanigans >>>> in. >>>> > >>>> >> I’ve discussed that at length in previous posts, & it probably isn’t >>>> >> necessary to again post about ways of choosing how to vote in >>>> Approval. >>>> >> >>>> >> But, just summarize:It’s easy.Whichever of the various ways you >>>> >> prefer to use, for choosing whom to approve, it’s easy.…& no, it >>>> >> doesn’t require knowing your objectively-optimal vote. >>>> > >>>> >> >>>> >> I’ve many times pointed out that Approval’s Myerson-Weber >>>> equilibrium >>>> >> is the voter-median. >>>> >> >>>> >> i.e. Approval soon homes in on where the Condorcet-Winner is. >>>> > >>>> > Not necessarily. See the following paper: >>>> > https://arxiv.org/abs/2001.04216v2 >>>> > >>>> > According to the authors, under their model, if every voter follows a >>>> > particular thresholding rule, then iterative approval generally >>>> > arrives at the Condorcet winner (but not always). However, if voters >>>> > are left to choose any thresholding rule they want (as you propose), >>>> > then anything is possible. The arrangement might even elect a >>>> > Condorcet loser, and the outcome may be slow to converge or who wins >>>> > may depend on how many polls you hold. >>>> > >>>> > I'm reminded of a quote about distributed algorithms that I read >>>> > somewhere: "It's really easy to design distributed algorithms that >>>> > suffer from deadlock, network floods or widely unpredictable and >>>> > bizarre oscillations". That's in the context of computer science - >>>> > deadlocks might not be applicable to election methods. But it does >>>> > justify a starting position of skepticisim when considering schemes >>>> > that offload more of the work to the voters by turning a one-shot >>>> > method into a dynamical system. >>>> > >>>> >> It seems to me that, in every one of EM’s polls, including the >>>> recent >>>> >> one, Approval chose the CW. >>>> >> >>>> >> Have we forgotten that? >>>> > >>>> > I haven't, nor have I forgotten that Approval wasn't actually the >>>> poll >>>> > winner. >>>> > >>>> > But let's take this reasoning at face value. I don't have the other >>>> > polls' ballot data available at the moment, so let's consider the >>>> > latest one and pick... say, Borda. >>>> > >>>> > As one can see by going to >>>> > https://munsterhjelm.no/km/rbvote/calc.html, pasting in the data, >>>> and >>>> > clicking Borda, Ranked Pairs is also the Borda winner. >>>> > >>>> > But I don't think I'm going to start advocating for Borda. >>>> > >>>> > >>>> > To not be accused of tu quoque, let me clarify the point. As >>>> > Burlington shows, an election method needs to handle the hard cases, >>>> > not just the easy ones, or there may be a considerable uproar when >>>> the >>>> > method is faced with a hard case and then stumbles. (In addition, >>>> > people trying to make sure a stumble doesn't happen may start to do >>>> > mass compromising, further entrenching two-party rule.) >>>> > >>>> > So it's quite possible that our polls are easy cases. But like >>>> > FairVote claiming that IRV gets the Condorcet winner more than 90% of >>>> > the time, that doesn't by itself tell us much, because the failures >>>> > have such a strong impact. >>>> > >>>> > I illustrate the example above by picking a method we know to be bad >>>> > (Borda being so extremely easy to fool with cloning and burial), and >>>> > showing that the poll result comes out right. If a bad method can get >>>> > a good result, then "getting a good result" is less useful than it >>>> > might appear at first glance. >>>> > >>>> > -km >>>> > ---- >>>> > Election-Methods mailing list - see https://electorama.com/em for >>>> list >>>> > info >>>> >>>

Please try to fit it all in one email instead of splitting it into so many chunks! If someone could relay the message to him as he has me blocked I’d be much obliged. On Mon, May 27, 2024 at 9:04 AM Michael Ossipoff <email9648742@gmail.com> wrote: > …or of course you could, for the voting, just ask each voter to write on > hir (big) ballot, which member of each of the 18090/2 candidate-pairs s/he > prefers to the other. S/he might be in the voting/booth for a while… > > On Sun, May 26, 2024 at 23:51 Michael Ossipoff <email9648742@gmail.com> > wrote: > >> Suppose that there are 300 million voters. Round 18090 off to 20000x >> >> 3E8 X 2E4 = 6E12 >> >> 6 trillion. >> >> That’s roughly the number of miles in a light-year. >> >> Whatever the number of voters, nationally, is, suppose that every one of >> them participated in the handcount. >> >> With the work divided among them all, each has “only” an amount of work >> equal to that of determining which member of each one of 18090/2 pairs of >> candidates is ranked over the other on a ballot. >> >> 135 candidates in each ranking. On the average, s/he’d only have to look >> at half of the ranking’s candidates to find each of the 2 members of each >> pair. >> >> …but s/he’d have to do it for both. So the order-determination for each >> candidate-pair, on the average will require looking at 135 candidates. >> >> Suppose that s/he can skim over 10 of them in a second, when doing those >> searches. >> >> Then it would take (135)(18090/2)/10 seconds. That’s about 1.4 days. But >> say she only does it for 8 hours per day (with no breaks). Now it’s more >> like 4.2 days. >> >> But supervision is the whole point. If s/he’s working alone, she can say >> that the pairwise vote-totals are whatever s/he wants them to be. >> >> So in reality, it would be counting *teams* each wit representatives of >> several parties. Say (optimistically) there are 10 parties. >> >> Now it will take 42 days. But the parties don’t really have equal numbers >> of members. It’s going to take longer. >> >> …&, realistically, they aren’t going to scan 10 ranked candidates every >> second for 8 hours with no breaks. >> >> It would obviously take months. Might it not, in fact, be measured in >> years…with every one of the nation’s voters participating in that handcount? >> >> >> >> On Sun, May 26, 2024 at 22:11 Michael Ossipoff <email9648742@gmail.com> >> wrote: >> >>> …& of course multiply that 18090 by the number of ballots, to get an >>> idea of what’s involved in the 18090 order-determinations to be done on >>> each ballot, & recorded, & then summed, to obtain each of the 18090 >>> pairwise vote-totals. >>> >>> …each of which then must be carried or transmitted to where the central >>> count is done. >>> >>> On Sun, May 26, 2024 at 22:03 Michael Ossipoff <email9648742@gmail.com> >>> wrote: >>> >>>> >>>> >>>> On Sun, May 26, 2024 at 12:56 Chris Benham <cbenhamau@yahoo.com.au> >>>> wrote: >>>> >>>>> >>>>> In 2003 there was California gubernatorial election with 135 >>>>> candidates. >>>> >>>> >>>> That’s 18090 pairwise vote-totals to determine at each precinct from >>>> the rankings, by examining each ranking to determine which member of each >>>> possible candidate-pair is ranked over the other on that ballot. >>>> >>>> …& 18080 pairwise vote-totals for the precincts to sum, store, & >>>> transmit or carry to the central count location. >>>> >>>> …& verify in an audit. >>>> >>>> >>>> >>>>> >>>>> >>>>> https://en.wikipedia.org/wiki/2003_California_gubernatorial_recall_election#Results >>>>> >>>>> Chris B. >>>>> >>>>> On 26/05/2024 9:29 pm, Kristofer Munsterhjelm wrote: >>>>> > On 2024-05-26 07:28, Michael Ossipoff wrote: >>>>> >> Someone keeps repeating that the voters shouldn’t have to vote >>>>> >> strategically. He wants the method to do it all for us, after we >>>>> >> merely state our sincere-rankings. >>>>> >> >>>>> >> That’s of course a common attitude: >>>>> >> >>>>> >> …wanting a high-tech, computation-intensive,computer-dependent >>>>> system >>>>> >> to do it all for us, taking all the actual choosing responsibility >>>>> >> off of us.…sheltering & isolating us from the choice. >>>>> > >>>>> > I would prefer that you do not attribute opinions that the >>>>> proponents >>>>> > have not expressed. Nowhere have I said that Condorcet "[isolates] >>>>> us >>>>> > from the choice" we make. >>>>> > >>>>> > What you call "sheltering" and "isolating", I see as the method >>>>> taking >>>>> > proper responsibility - proper responsibility to turn the voters' >>>>> > unambiguous honest opinions into an outcome without dumping the >>>>> > algorithmic calculation upon te voter themselves. See my "kick the >>>>> can >>>>> > down the road" post for more info. >>>>> > >>>>> > Now, I could make a caricature of Approval itself. Perhaps something >>>>> > about a calculator that just says "IDK, do the base conversion >>>>> > yourself, I only accept input numbers in factoradic". But >>>>> caricatures >>>>> > only make people angry. Let's not stoop to them, shall we? >>>>> > >>>>> >> I’ll ask this for the 3^rd time: >>>>> >> >>>>> >> … >>>>> >> >>>>> >> How would like you to handount-audit a Condorcet count for a >>>>> >> many-candidate national presidential election? >>>>> > >>>>> > I think your question assumes something that won't hold. If you have >>>>> a >>>>> > 25-candidate presidential election, you've already lost, because >>>>> > nobody is going to rank 25 candidates, irrespective of whether the >>>>> > method is Condorcet, IRV, or Borda. >>>>> > >>>>> > I'm not familiar with minor parties in the US. Has there ever been a >>>>> > 25-candidate presidential election? >>>>> > >>>>> > I don't think I can comment beyond that: I don't know enough about >>>>> > poll workers. I'll leave that to someone with experience. (Although >>>>> > roughly calculating: suppose 6 candidates like in Burlington. That's >>>>> > 30 pairs. Five times the work if counting a particular preference is >>>>> > as before. So you'd either need 5x the workers, or five times the >>>>> > time, or some combination of the two.) >>>>> > >>>>> > On an aside, though, I would say that I generally wouldn't want to >>>>> get >>>>> > computers anywhere near election counting. However, if you >>>>> absolutely >>>>> > have to have them, there are ways of making sure they don't cheat: >>>>> > formal verification. You could also create special-purpose tools >>>>> that >>>>> > say, only turn ranks into matrices and nothing else: there's no >>>>> reason >>>>> > (apart from programmer convenience) why an election tool should be a >>>>> > general purpose computer that you could hide all sorts of >>>>> shenanigans in. >>>>> > >>>>> >> I’ve discussed that at length in previous posts, & it probably >>>>> isn’t >>>>> >> necessary to again post about ways of choosing how to vote in >>>>> Approval. >>>>> >> >>>>> >> But, just summarize:It’s easy.Whichever of the various ways you >>>>> >> prefer to use, for choosing whom to approve, it’s easy.…& no, it >>>>> >> doesn’t require knowing your objectively-optimal vote. >>>>> > >>>>> >> >>>>> >> I’ve many times pointed out that Approval’s Myerson-Weber >>>>> equilibrium >>>>> >> is the voter-median. >>>>> >> >>>>> >> i.e. Approval soon homes in on where the Condorcet-Winner is. >>>>> > >>>>> > Not necessarily. See the following paper: >>>>> > https://arxiv.org/abs/2001.04216v2 >>>>> > >>>>> > According to the authors, under their model, if every voter follows >>>>> a >>>>> > particular thresholding rule, then iterative approval generally >>>>> > arrives at the Condorcet winner (but not always). However, if voters >>>>> > are left to choose any thresholding rule they want (as you propose), >>>>> > then anything is possible. The arrangement might even elect a >>>>> > Condorcet loser, and the outcome may be slow to converge or who wins >>>>> > may depend on how many polls you hold. >>>>> > >>>>> > I'm reminded of a quote about distributed algorithms that I read >>>>> > somewhere: "It's really easy to design distributed algorithms that >>>>> > suffer from deadlock, network floods or widely unpredictable and >>>>> > bizarre oscillations". That's in the context of computer science - >>>>> > deadlocks might not be applicable to election methods. But it does >>>>> > justify a starting position of skepticisim when considering schemes >>>>> > that offload more of the work to the voters by turning a one-shot >>>>> > method into a dynamical system. >>>>> > >>>>> >> It seems to me that, in every one of EM’s polls, including the >>>>> recent >>>>> >> one, Approval chose the CW. >>>>> >> >>>>> >> Have we forgotten that? >>>>> > >>>>> > I haven't, nor have I forgotten that Approval wasn't actually the >>>>> poll >>>>> > winner. >>>>> > >>>>> > But let's take this reasoning at face value. I don't have the other >>>>> > polls' ballot data available at the moment, so let's consider the >>>>> > latest one and pick... say, Borda. >>>>> > >>>>> > As one can see by going to >>>>> > https://munsterhjelm.no/km/rbvote/calc.html, pasting in the data, >>>>> and >>>>> > clicking Borda, Ranked Pairs is also the Borda winner. >>>>> > >>>>> > But I don't think I'm going to start advocating for Borda. >>>>> > >>>>> > >>>>> > To not be accused of tu quoque, let me clarify the point. As >>>>> > Burlington shows, an election method needs to handle the hard cases, >>>>> > not just the easy ones, or there may be a considerable uproar when >>>>> the >>>>> > method is faced with a hard case and then stumbles. (In addition, >>>>> > people trying to make sure a stumble doesn't happen may start to do >>>>> > mass compromising, further entrenching two-party rule.) >>>>> > >>>>> > So it's quite possible that our polls are easy cases. But like >>>>> > FairVote claiming that IRV gets the Condorcet winner more than 90% >>>>> of >>>>> > the time, that doesn't by itself tell us much, because the failures >>>>> > have such a strong impact. >>>>> > >>>>> > I illustrate the example above by picking a method we know to be bad >>>>> > (Borda being so extremely easy to fool with cloning and burial), and >>>>> > showing that the poll result comes out right. If a bad method can >>>>> get >>>>> > a good result, then "getting a good result" is less useful than it >>>>> > might appear at first glance. >>>>> > >>>>> > -km >>>>> > ---- >>>>> > Election-Methods mailing list - see https://electorama.com/em for >>>>> list >>>>> > info >>>>> >>>> ---- > Election-Methods mailing list - see https://electorama.com/em for list > info >

 Suppose that there are 300 million voters. Round 18090 off to 20000x

 3E8 X 2E4 = 6E12

 6 trillion.

 That’s roughly the number of miles in a light-year.

 Whatever the number of voters, nationally, is, suppose that every
 one of them participated in the handcount.

 With the work divided among them all, each has “only” an amount of
 work equal to that of determining which member of each one of
 18090/2 pairs of candidates is ranked over the other on a ballot.

 135 candidates in each ranking. On the average, s/he’d only have
 to look at half of the ranking’s candidates to find each of the 2
 members of each pair.

 …but s/he’d have to do it for both. So the order-determination for
 each candidate-pair, on the average will require looking at 135
 candidates.

 Suppose that s/he can skim over 10 of them in a second, when doing
 those searches.

 Then it would take (135)(18090/2)/10 seconds. That’s about 1.4
 days. But say she only does it for 8 hours per day (with no
 breaks). Now it’s more like 4.2 days.

 But supervision is the whole point. If s/he’s working alone, she
 can say that the pairwise vote-totals are whatever s/he wants them
 to be.

 So in reality, it would be counting *teams* each wit
 representatives of several parties. Say (optimistically) there are
 10 parties.

 Now it will take 42 days. But the parties don’t really have equal
 numbers of members. It’s going to take longer.

 …&, realistically, they aren’t going to scan 10 ranked candidates
 every second for 8 hours with no breaks.

 It would obviously take months. Might it not, in fact, be measured
 in years…with every one of the nation’s voters participating in
 that handcount?


 On Sun, May 26, 2024 at 22:11 Michael Ossipoff
 <email9648742@gmail.com> wrote:

     …& of course multiply that 18090 by the number of ballots, to
     get an idea of what’s involved in the 18090
     order-determinations to be done on each ballot, & recorded, &
      then summed, to obtain each of the 18090 pairwise vote-totals.

     …each of which then must be carried or transmitted to where
     the central count is done.

     On Sun, May 26, 2024 at 22:03 Michael Ossipoff
     <email9648742@gmail.com> wrote:



         On Sun, May 26, 2024 at 12:56 Chris Benham
         <cbenhamau@yahoo.com.au> wrote:


             In 2003  there was California gubernatorial election
             with 135 candidates.


         That’s 18090 pairwise vote-totals to determine at each
         precinct from the rankings, by examining each ranking to
         determine which member of each possible candidate-pair is
         ranked over the other on that ballot.

         …& 18080 pairwise vote-totals for the precincts to sum,
         store, & transmit or carry to the central count location.

         …& verify in an audit.




             https://en.wikipedia.org/wiki/2003_California_gubernatorial_recall_election#Results

             Chris B.

             On 26/05/2024 9:29 pm, Kristofer Munsterhjelm wrote:

             have to vote

             us, after we

             computation-intensive,computer-dependent system

             responsibility

             that the proponents

             Condorcet "[isolates] us

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             turn the voters'

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             See my "kick the can

             Perhaps something

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             shall we?

             count for a

             hold. If you have a

             lost, because

             of whether the

             there ever been a

             know enough about

             experience. (Although

             Burlington. That's

             particular preference is

             five times the

             wouldn't want to get

             if you absolutely

             they don't cheat:

             special-purpose tools that

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             tool should be a

             sorts of shenanigans in.

             it probably isn’t

             to vote in Approval.

             various ways you

             easy.…& no, it

             Myerson-Weber equilibrium

             Condorcet-Winner is.

             every voter follows a

             approval generally

             However, if voters

             (as you propose),

             even elect a

             converge or who wins

             that I read

             algorithms that

             unpredictable and

             computer science -

             methods. But it does

             considering schemes

             turning a one-shot

             including the recent

             actually the poll

             have the other

             consider the

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             point. As

             the hard cases,

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             may start to do

             cases. But like

             more than 90% of

             because the failures

             we know to be bad

             and burial), and

             bad method can get

             useful than it

             https://electorama.com/em for list

Are your computers really necessarily so untrustworthy? I think counting Hare would be somewhat easier. Probably very weak candidates could be eliminated in batches (while ensuring the correct one-at-a-time elimination result). One possible kludgey solution could be to apply the Condorcet method to the Hare last X candidates.  X could be say 7. Something I've been meaning to point out: the Approval Sorted Margins method I promoted in the poll but the other voters here either ignored or claimed they couldn't understand  (and then failed to give it even majority approval) has the big practical advantage over other Condorcet methods that it normally doesn't use (and therefore doesn't have to know) the full pairwise matrix.  The more candidates there are I should say the greater would this advantage be. This is because (as you all recall) it only needs to check if adjacent pairs of candidates in the Approval-score order are in "pairwise harmony" with that order.  Probably most of the time most of them will be. Chris B. On 27/05/2024 4:28 pm, Michael Ossipoff wrote: > …or of course you could, for the voting, just ask each voter to write > on hir (big) ballot, which member of each of the 18090/2 > candidate-pairs s/he prefers to the other. S/he might be in the > voting/booth for a while… > > On Sun, May 26, 2024 at 23:51 Michael Ossipoff > <email9648742@gmail.com> wrote: > > Suppose that there are 300 million voters. Round 18090 off to 20000x > > 3E8 X 2E4 = 6E12 > > 6 trillion. > > That’s roughly the number of miles in a light-year. > > Whatever the number of voters, nationally, is, suppose that every > one of them participated in the handcount. > > With the work divided among them all, each has “only” an amount of > work equal to that of determining which member of each one of > 18090/2 pairs of candidates is ranked over the other on a ballot. > > 135 candidates in each ranking. On the average, s/he’d only have > to look at half of the ranking’s candidates to find each of the 2 > members of each pair. > > …but s/he’d have to do it for both. So the order-determination for > each candidate-pair, on the average will require looking at 135 > candidates. > > Suppose that s/he can skim over 10 of them in a second, when doing > those searches. > > Then it would take (135)(18090/2)/10 seconds. That’s about 1.4 > days. But say she only does it for 8 hours per day (with no > breaks). Now it’s more like 4.2 days. > > But supervision is the whole point. If s/he’s working alone, she > can say that the pairwise vote-totals are whatever s/he wants them > to be. > > So in reality, it would be counting *teams* each wit > representatives of several parties. Say (optimistically) there are > 10 parties. > > Now it will take 42 days. But the parties don’t really have equal > numbers of members. It’s going to take longer. > > …&, realistically, they aren’t going to scan 10 ranked candidates > every second for 8 hours with no breaks. > > It would obviously take months. Might it not, in fact, be measured > in years…with every one of the nation’s voters participating in > that handcount? > > > On Sun, May 26, 2024 at 22:11 Michael Ossipoff > <email9648742@gmail.com> wrote: > > …& of course multiply that 18090 by the number of ballots, to > get an idea of what’s involved in the 18090 > order-determinations to be done on each ballot, & recorded, & >  then summed, to obtain each of the 18090 pairwise vote-totals. > > …each of which then must be carried or transmitted to where > the central count is done. > > On Sun, May 26, 2024 at 22:03 Michael Ossipoff > <email9648742@gmail.com> wrote: > > > > On Sun, May 26, 2024 at 12:56 Chris Benham > <cbenhamau@yahoo.com.au> wrote: > > > In 2003  there was California gubernatorial election > with 135 candidates. > > > That’s 18090 pairwise vote-totals to determine at each > precinct from the rankings, by examining each ranking to > determine which member of each possible candidate-pair is > ranked over the other on that ballot. > > …& 18080 pairwise vote-totals for the precincts to sum, > store, & transmit or carry to the central count location. > > …& verify in an audit. > > > > > https://en.wikipedia.org/wiki/2003_California_gubernatorial_recall_election#Results > > Chris B. > > On 26/05/2024 9:29 pm, Kristofer Munsterhjelm wrote: > > On 2024-05-26 07:28, Michael Ossipoff wrote: > >> Someone keeps repeating that the voters shouldn’t > have to vote > >> strategically. He wants the method to do it all for > us, after we > >> merely state our sincere-rankings. > >> > >> That’s of course a common attitude: > >> > >> …wanting a high-tech, > computation-intensive,computer-dependent system > >> to do it all for us, taking all the actual choosing > responsibility > >> off of us.…sheltering & isolating us from the choice. > > > > I would prefer that you do not attribute opinions > that the proponents > > have not expressed. Nowhere have I said that > Condorcet "[isolates] us > > from the choice" we make. > > > > What you call "sheltering" and "isolating", I see as > the method taking > > proper responsibility - proper responsibility to > turn the voters' > > unambiguous honest opinions into an outcome without > dumping the > > algorithmic calculation upon te voter themselves. > See my "kick the can > > down the road" post for more info. > > > > Now, I could make a caricature of Approval itself. > Perhaps something > > about a calculator that just says "IDK, do the base > conversion > > yourself, I only accept input numbers in > factoradic". But caricatures > > only make people angry. Let's not stoop to them, > shall we? > > > >> I’ll ask this for the 3^rd time: > >> > >> … > >> > >> How would like you to handount-audit a Condorcet > count for a > >> many-candidate national presidential election? > > > > I think your question assumes something that won't > hold. If you have a > > 25-candidate presidential election, you've already > lost, because > > nobody is going to rank 25 candidates, irrespective > of whether the > > method is Condorcet, IRV, or Borda. > > > > I'm not familiar with minor parties in the US. Has > there ever been a > > 25-candidate presidential election? > > > > I don't think I can comment beyond that: I don't > know enough about > > poll workers. I'll leave that to someone with > experience. (Although > > roughly calculating: suppose 6 candidates like in > Burlington. That's > > 30 pairs. Five times the work if counting a > particular preference is > > as before. So you'd either need 5x the workers, or > five times the > > time, or some combination of the two.) > > > > On an aside, though, I would say that I generally > wouldn't want to get > > computers anywhere near election counting. However, > if you absolutely > > have to have them, there are ways of making sure > they don't cheat: > > formal verification. You could also create > special-purpose tools that > > say, only turn ranks into matrices and nothing else: > there's no reason > > (apart from programmer convenience) why an election > tool should be a > > general purpose computer that you could hide all > sorts of shenanigans in. > > > >> I’ve discussed that at length in previous posts, & > it probably isn’t > >> necessary to again post about ways of choosing how > to vote in Approval. > >> > >> But, just summarize:It’s easy.Whichever of the > various ways you > >> prefer to use, for choosing whom to approve, it’s > easy.…& no, it > >> doesn’t require knowing your objectively-optimal vote. > > > >> > >> I’ve many times pointed out that Approval’s > Myerson-Weber equilibrium > >> is the voter-median. > >> > >> i.e. Approval soon homes in on where the > Condorcet-Winner is. > > > > Not necessarily. See the following paper: > > https://arxiv.org/abs/2001.04216v2 > > > > According to the authors, under their model, if > every voter follows a > > particular thresholding rule, then iterative > approval generally > > arrives at the Condorcet winner (but not always). > However, if voters > > are left to choose any thresholding rule they want > (as you propose), > > then anything is possible. The arrangement might > even elect a > > Condorcet loser, and the outcome may be slow to > converge or who wins > > may depend on how many polls you hold. > > > > I'm reminded of a quote about distributed algorithms > that I read > > somewhere: "It's really easy to design distributed > algorithms that > > suffer from deadlock, network floods or widely > unpredictable and > > bizarre oscillations". That's in the context of > computer science - > > deadlocks might not be applicable to election > methods. But it does > > justify a starting position of skepticisim when > considering schemes > > that offload more of the work to the voters by > turning a one-shot > > method into a dynamical system. > > > >> It seems to me that, in every one of EM’s polls, > including the recent > >> one, Approval chose the CW. > >> > >> Have we forgotten that? > > > > I haven't, nor have I forgotten that Approval wasn't > actually the poll > > winner. > > > > But let's take this reasoning at face value. I don't > have the other > > polls' ballot data available at the moment, so let's > consider the > > latest one and pick... say, Borda. > > > > As one can see by going to > > https://munsterhjelm.no/km/rbvote/calc.html, pasting > in the data, and > > clicking Borda, Ranked Pairs is also the Borda winner. > > > > But I don't think I'm going to start advocating for > Borda. > > > > > > To not be accused of tu quoque, let me clarify the > point. As > > Burlington shows, an election method needs to handle > the hard cases, > > not just the easy ones, or there may be a > considerable uproar when the > > method is faced with a hard case and then stumbles. > (In addition, > > people trying to make sure a stumble doesn't happen > may start to do > > mass compromising, further entrenching two-party rule.) > > > > So it's quite possible that our polls are easy > cases. But like > > FairVote claiming that IRV gets the Condorcet winner > more than 90% of > > the time, that doesn't by itself tell us much, > because the failures > > have such a strong impact. > > > > I illustrate the example above by picking a method > we know to be bad > > (Borda being so extremely easy to fool with cloning > and burial), and > > showing that the poll result comes out right. If a > bad method can get > > a good result, then "getting a good result" is less > useful than it > > might appear at first glance. > > > > -km > > ---- > > Election-Methods mailing list - see > https://electorama.com/em for list > > info >

No, but some humans are. Count-fraud has been demonstrated to have occurred, in 2000 & 2004. On Mon, May 27, 2024 at 01:39 Chris Benham <cbenhamau@yahoo.com.au> wrote: > Are your computers really necessarily so untrustworthy? > > I think counting Hare would be somewhat easier. Probably very weak > candidates could be eliminated in batches (while ensuring the correct > one-at-a-time elimination result). One possible kludgey solution could be > to apply the Condorcet method to the Hare last X candidates. X could be > say 7. > > Something I've been meaning to point out: the Approval Sorted Margins > method I promoted in the poll but the other voters here either ignored or > claimed they couldn't understand (and then failed to give it even majority > approval) has the big practical advantage over > other Condorcet methods that it normally doesn't use (and therefore > doesn't have to know) the full pairwise matrix. The more candidates there > are I should say the greater would this advantage be. > > This is because (as you all recall) it only needs to check if adjacent > pairs of candidates in the Approval-score order are in "pairwise harmony" > with that order. Probably most of the time most of them will be. > > Chris B. > > On 27/05/2024 4:28 pm, Michael Ossipoff wrote: > > …or of course you could, for the voting, just ask each voter to write on > hir (big) ballot, which member of each of the 18090/2 candidate-pairs s/he > prefers to the other. S/he might be in the voting/booth for a while… > > On Sun, May 26, 2024 at 23:51 Michael Ossipoff <email9648742@gmail.com> > wrote: > >> Suppose that there are 300 million voters. Round 18090 off to 20000x >> >> 3E8 X 2E4 = 6E12 >> >> 6 trillion. >> >> That’s roughly the number of miles in a light-year. >> >> Whatever the number of voters, nationally, is, suppose that every one of >> them participated in the handcount. >> >> With the work divided among them all, each has “only” an amount of work >> equal to that of determining which member of each one of 18090/2 pairs of >> candidates is ranked over the other on a ballot. >> >> 135 candidates in each ranking. On the average, s/he’d only have to look >> at half of the ranking’s candidates to find each of the 2 members of each >> pair. >> >> …but s/he’d have to do it for both. So the order-determination for each >> candidate-pair, on the average will require looking at 135 candidates. >> >> Suppose that s/he can skim over 10 of them in a second, when doing those >> searches. >> >> Then it would take (135)(18090/2)/10 seconds. That’s about 1.4 days. But >> say she only does it for 8 hours per day (with no breaks). Now it’s more >> like 4.2 days. >> >> But supervision is the whole point. If s/he’s working alone, she can say >> that the pairwise vote-totals are whatever s/he wants them to be. >> >> So in reality, it would be counting *teams* each wit representatives of >> several parties. Say (optimistically) there are 10 parties. >> >> Now it will take 42 days. But the parties don’t really have equal numbers >> of members. It’s going to take longer. >> >> …&, realistically, they aren’t going to scan 10 ranked candidates every >> second for 8 hours with no breaks. >> >> It would obviously take months. Might it not, in fact, be measured in >> years…with every one of the nation’s voters participating in that handcount? >> >> >> >> On Sun, May 26, 2024 at 22:11 Michael Ossipoff <email9648742@gmail.com> >> wrote: >> >>> …& of course multiply that 18090 by the number of ballots, to get an >>> idea of what’s involved in the 18090 order-determinations to be done on >>> each ballot, & recorded, & then summed, to obtain each of the 18090 >>> pairwise vote-totals. >>> >>> …each of which then must be carried or transmitted to where the central >>> count is done. >>> >>> On Sun, May 26, 2024 at 22:03 Michael Ossipoff <email9648742@gmail.com> >>> wrote: >>> >>>> >>>> >>>> On Sun, May 26, 2024 at 12:56 Chris Benham <cbenhamau@yahoo.com.au> >>>> wrote: >>>> >>>>> >>>>> In 2003 there was California gubernatorial election with 135 >>>>> candidates. >>>> >>>> >>>> That’s 18090 pairwise vote-totals to determine at each precinct from >>>> the rankings, by examining each ranking to determine which member of each >>>> possible candidate-pair is ranked over the other on that ballot. >>>> >>>> …& 18080 pairwise vote-totals for the precincts to sum, store, & >>>> transmit or carry to the central count location. >>>> >>>> …& verify in an audit. >>>> >>>> >>>> >>>>> >>>>> >>>>> https://en.wikipedia.org/wiki/2003_California_gubernatorial_recall_election#Results >>>>> >>>>> Chris B. >>>>> >>>>> On 26/05/2024 9:29 pm, Kristofer Munsterhjelm wrote: >>>>> > On 2024-05-26 07:28, Michael Ossipoff wrote: >>>>> >> Someone keeps repeating that the voters shouldn’t have to vote >>>>> >> strategically. He wants the method to do it all for us, after we >>>>> >> merely state our sincere-rankings. >>>>> >> >>>>> >> That’s of course a common attitude: >>>>> >> >>>>> >> …wanting a high-tech, computation-intensive,computer-dependent >>>>> system >>>>> >> to do it all for us, taking all the actual choosing responsibility >>>>> >> off of us.…sheltering & isolating us from the choice. >>>>> > >>>>> > I would prefer that you do not attribute opinions that the >>>>> proponents >>>>> > have not expressed. Nowhere have I said that Condorcet "[isolates] >>>>> us >>>>> > from the choice" we make. >>>>> > >>>>> > What you call "sheltering" and "isolating", I see as the method >>>>> taking >>>>> > proper responsibility - proper responsibility to turn the voters' >>>>> > unambiguous honest opinions into an outcome without dumping the >>>>> > algorithmic calculation upon te voter themselves. See my "kick the >>>>> can >>>>> > down the road" post for more info. >>>>> > >>>>> > Now, I could make a caricature of Approval itself. Perhaps something >>>>> > about a calculator that just says "IDK, do the base conversion >>>>> > yourself, I only accept input numbers in factoradic". But >>>>> caricatures >>>>> > only make people angry. Let's not stoop to them, shall we? >>>>> > >>>>> >> I’ll ask this for the 3^rd time: >>>>> >> >>>>> >> … >>>>> >> >>>>> >> How would like you to handount-audit a Condorcet count for a >>>>> >> many-candidate national presidential election? >>>>> > >>>>> > I think your question assumes something that won't hold. If you have >>>>> a >>>>> > 25-candidate presidential election, you've already lost, because >>>>> > nobody is going to rank 25 candidates, irrespective of whether the >>>>> > method is Condorcet, IRV, or Borda. >>>>> > >>>>> > I'm not familiar with minor parties in the US. Has there ever been a >>>>> > 25-candidate presidential election? >>>>> > >>>>> > I don't think I can comment beyond that: I don't know enough about >>>>> > poll workers. I'll leave that to someone with experience. (Although >>>>> > roughly calculating: suppose 6 candidates like in Burlington. That's >>>>> > 30 pairs. Five times the work if counting a particular preference is >>>>> > as before. So you'd either need 5x the workers, or five times the >>>>> > time, or some combination of the two.) >>>>> > >>>>> > On an aside, though, I would say that I generally wouldn't want to >>>>> get >>>>> > computers anywhere near election counting. However, if you >>>>> absolutely >>>>> > have to have them, there are ways of making sure they don't cheat: >>>>> > formal verification. You could also create special-purpose tools >>>>> that >>>>> > say, only turn ranks into matrices and nothing else: there's no >>>>> reason >>>>> > (apart from programmer convenience) why an election tool should be a >>>>> > general purpose computer that you could hide all sorts of >>>>> shenanigans in. >>>>> > >>>>> >> I’ve discussed that at length in previous posts, & it probably >>>>> isn’t >>>>> >> necessary to again post about ways of choosing how to vote in >>>>> Approval. >>>>> >> >>>>> >> But, just summarize:It’s easy.Whichever of the various ways you >>>>> >> prefer to use, for choosing whom to approve, it’s easy.…& no, it >>>>> >> doesn’t require knowing your objectively-optimal vote. >>>>> > >>>>> >> >>>>> >> I’ve many times pointed out that Approval’s Myerson-Weber >>>>> equilibrium >>>>> >> is the voter-median. >>>>> >> >>>>> >> i.e. Approval soon homes in on where the Condorcet-Winner is. >>>>> > >>>>> > Not necessarily. See the following paper: >>>>> > https://arxiv.org/abs/2001.04216v2 >>>>> > >>>>> > According to the authors, under their model, if every voter follows >>>>> a >>>>> > particular thresholding rule, then iterative approval generally >>>>> > arrives at the Condorcet winner (but not always). However, if voters >>>>> > are left to choose any thresholding rule they want (as you propose), >>>>> > then anything is possible. The arrangement might even elect a >>>>> > Condorcet loser, and the outcome may be slow to converge or who wins >>>>> > may depend on how many polls you hold. >>>>> > >>>>> > I'm reminded of a quote about distributed algorithms that I read >>>>> > somewhere: "It's really easy to design distributed algorithms that >>>>> > suffer from deadlock, network floods or widely unpredictable and >>>>> > bizarre oscillations". That's in the context of computer science - >>>>> > deadlocks might not be applicable to election methods. But it does >>>>> > justify a starting position of skepticisim when considering schemes >>>>> > that offload more of the work to the voters by turning a one-shot >>>>> > method into a dynamical system. >>>>> > >>>>> >> It seems to me that, in every one of EM’s polls, including the >>>>> recent >>>>> >> one, Approval chose the CW. >>>>> >> >>>>> >> Have we forgotten that? >>>>> > >>>>> > I haven't, nor have I forgotten that Approval wasn't actually the >>>>> poll >>>>> > winner. >>>>> > >>>>> > But let's take this reasoning at face value. I don't have the other >>>>> > polls' ballot data available at the moment, so let's consider the >>>>> > latest one and pick... say, Borda. >>>>> > >>>>> > As one can see by going to >>>>> > https://munsterhjelm.no/km/rbvote/calc.html, pasting in the data, >>>>> and >>>>> > clicking Borda, Ranked Pairs is also the Borda winner. >>>>> > >>>>> > But I don't think I'm going to start advocating for Borda. >>>>> > >>>>> > >>>>> > To not be accused of tu quoque, let me clarify the point. As >>>>> > Burlington shows, an election method needs to handle the hard cases, >>>>> > not just the easy ones, or there may be a considerable uproar when >>>>> the >>>>> > method is faced with a hard case and then stumbles. (In addition, >>>>> > people trying to make sure a stumble doesn't happen may start to do >>>>> > mass compromising, further entrenching two-party rule.) >>>>> > >>>>> > So it's quite possible that our polls are easy cases. But like >>>>> > FairVote claiming that IRV gets the Condorcet winner more than 90% >>>>> of >>>>> > the time, that doesn't by itself tell us much, because the failures >>>>> > have such a strong impact. >>>>> > >>>>> > I illustrate the example above by picking a method we know to be bad >>>>> > (Borda being so extremely easy to fool with cloning and burial), and >>>>> > showing that the poll result comes out right. If a bad method can >>>>> get >>>>> > a good result, then "getting a good result" is less useful than it >>>>> > might appear at first glance. >>>>> > >>>>> > -km >>>>> > ---- >>>>> > Election-Methods mailing list - see https://electorama.com/em for >>>>> list >>>>> > info >>>>> >>>>

>> This forum, like most, in its rules of conduct, asks that we not keep repeating an already-answered claim without answering the objections to it. In that case, kindly stop repeating your unsubstantiated conspiracy theories about fraud from 2000/2004! On Mon, May 27, 2024 at 10:53 AM Michael Ossipoff <email9648742@gmail.com> wrote: > > No, but some humans are. Count-fraud has been demonstrated to have > occurred, in 2000 & 2004. > > On Mon, May 27, 2024 at 01:39 Chris Benham <cbenhamau@yahoo.com.au> wrote: > >> Are your computers really necessarily so untrustworthy? >> >> I think counting Hare would be somewhat easier. Probably very weak >> candidates could be eliminated in batches (while ensuring the correct >> one-at-a-time elimination result). One possible kludgey solution could be >> to apply the Condorcet method to the Hare last X candidates. X could be >> say 7. >> >> Something I've been meaning to point out: the Approval Sorted Margins >> method I promoted in the poll but the other voters here either ignored or >> claimed they couldn't understand (and then failed to give it even majority >> approval) has the big practical advantage over >> other Condorcet methods that it normally doesn't use (and therefore >> doesn't have to know) the full pairwise matrix. The more candidates there >> are I should say the greater would this advantage be. >> >> This is because (as you all recall) it only needs to check if adjacent >> pairs of candidates in the Approval-score order are in "pairwise harmony" >> with that order. Probably most of the time most of them will be. >> >> Chris B. >> >> On 27/05/2024 4:28 pm, Michael Ossipoff wrote: >> >> …or of course you could, for the voting, just ask each voter to write on >> hir (big) ballot, which member of each of the 18090/2 candidate-pairs s/he >> prefers to the other. S/he might be in the voting/booth for a while… >> >> On Sun, May 26, 2024 at 23:51 Michael Ossipoff <email9648742@gmail.com> >> wrote: >> >>> Suppose that there are 300 million voters. Round 18090 off to 20000x >>> >>> 3E8 X 2E4 = 6E12 >>> >>> 6 trillion. >>> >>> That’s roughly the number of miles in a light-year. >>> >>> Whatever the number of voters, nationally, is, suppose that every one of >>> them participated in the handcount. >>> >>> With the work divided among them all, each has “only” an amount of work >>> equal to that of determining which member of each one of 18090/2 pairs of >>> candidates is ranked over the other on a ballot. >>> >>> 135 candidates in each ranking. On the average, s/he’d only have to look >>> at half of the ranking’s candidates to find each of the 2 members of each >>> pair. >>> >>> …but s/he’d have to do it for both. So the order-determination for each >>> candidate-pair, on the average will require looking at 135 candidates. >>> >>> Suppose that s/he can skim over 10 of them in a second, when doing those >>> searches. >>> >>> Then it would take (135)(18090/2)/10 seconds. That’s about 1.4 days. But >>> say she only does it for 8 hours per day (with no breaks). Now it’s more >>> like 4.2 days. >>> >>> But supervision is the whole point. If s/he’s working alone, she can say >>> that the pairwise vote-totals are whatever s/he wants them to be. >>> >>> So in reality, it would be counting *teams* each wit representatives of >>> several parties. Say (optimistically) there are 10 parties. >>> >>> Now it will take 42 days. But the parties don’t really have equal >>> numbers of members. It’s going to take longer. >>> >>> …&, realistically, they aren’t going to scan 10 ranked candidates every >>> second for 8 hours with no breaks. >>> >>> It would obviously take months. Might it not, in fact, be measured in >>> years…with every one of the nation’s voters participating in that handcount? >>> >>> >>> >>> On Sun, May 26, 2024 at 22:11 Michael Ossipoff <email9648742@gmail.com> >>> wrote: >>> >>>> …& of course multiply that 18090 by the number of ballots, to get an >>>> idea of what’s involved in the 18090 order-determinations to be done on >>>> each ballot, & recorded, & then summed, to obtain each of the 18090 >>>> pairwise vote-totals. >>>> >>>> …each of which then must be carried or transmitted to where the central >>>> count is done. >>>> >>>> On Sun, May 26, 2024 at 22:03 Michael Ossipoff <email9648742@gmail.com> >>>> wrote: >>>> >>>>> >>>>> >>>>> On Sun, May 26, 2024 at 12:56 Chris Benham <cbenhamau@yahoo.com.au> >>>>> wrote: >>>>> >>>>>> >>>>>> In 2003 there was California gubernatorial election with 135 >>>>>> candidates. >>>>> >>>>> >>>>> That’s 18090 pairwise vote-totals to determine at each precinct from >>>>> the rankings, by examining each ranking to determine which member of each >>>>> possible candidate-pair is ranked over the other on that ballot. >>>>> >>>>> …& 18080 pairwise vote-totals for the precincts to sum, store, & >>>>> transmit or carry to the central count location. >>>>> >>>>> …& verify in an audit. >>>>> >>>>> >>>>> >>>>>> >>>>>> >>>>>> https://en.wikipedia.org/wiki/2003_California_gubernatorial_recall_election#Results >>>>>> >>>>>> Chris B. >>>>>> >>>>>> On 26/05/2024 9:29 pm, Kristofer Munsterhjelm wrote: >>>>>> > On 2024-05-26 07:28, Michael Ossipoff wrote: >>>>>> >> Someone keeps repeating that the voters shouldn’t have to vote >>>>>> >> strategically. He wants the method to do it all for us, after we >>>>>> >> merely state our sincere-rankings. >>>>>> >> >>>>>> >> That’s of course a common attitude: >>>>>> >> >>>>>> >> …wanting a high-tech, computation-intensive,computer-dependent >>>>>> system >>>>>> >> to do it all for us, taking all the actual choosing responsibility >>>>>> >> off of us.…sheltering & isolating us from the choice. >>>>>> > >>>>>> > I would prefer that you do not attribute opinions that the >>>>>> proponents >>>>>> > have not expressed. Nowhere have I said that Condorcet "[isolates] >>>>>> us >>>>>> > from the choice" we make. >>>>>> > >>>>>> > What you call "sheltering" and "isolating", I see as the method >>>>>> taking >>>>>> > proper responsibility - proper responsibility to turn the voters' >>>>>> > unambiguous honest opinions into an outcome without dumping the >>>>>> > algorithmic calculation upon te voter themselves. See my "kick the >>>>>> can >>>>>> > down the road" post for more info. >>>>>> > >>>>>> > Now, I could make a caricature of Approval itself. Perhaps >>>>>> something >>>>>> > about a calculator that just says "IDK, do the base conversion >>>>>> > yourself, I only accept input numbers in factoradic". But >>>>>> caricatures >>>>>> > only make people angry. Let's not stoop to them, shall we? >>>>>> > >>>>>> >> I’ll ask this for the 3^rd time: >>>>>> >> >>>>>> >> … >>>>>> >> >>>>>> >> How would like you to handount-audit a Condorcet count for a >>>>>> >> many-candidate national presidential election? >>>>>> > >>>>>> > I think your question assumes something that won't hold. If you >>>>>> have a >>>>>> > 25-candidate presidential election, you've already lost, because >>>>>> > nobody is going to rank 25 candidates, irrespective of whether the >>>>>> > method is Condorcet, IRV, or Borda. >>>>>> > >>>>>> > I'm not familiar with minor parties in the US. Has there ever been >>>>>> a >>>>>> > 25-candidate presidential election? >>>>>> > >>>>>> > I don't think I can comment beyond that: I don't know enough about >>>>>> > poll workers. I'll leave that to someone with experience. (Although >>>>>> > roughly calculating: suppose 6 candidates like in Burlington. >>>>>> That's >>>>>> > 30 pairs. Five times the work if counting a particular preference >>>>>> is >>>>>> > as before. So you'd either need 5x the workers, or five times the >>>>>> > time, or some combination of the two.) >>>>>> > >>>>>> > On an aside, though, I would say that I generally wouldn't want to >>>>>> get >>>>>> > computers anywhere near election counting. However, if you >>>>>> absolutely >>>>>> > have to have them, there are ways of making sure they don't cheat: >>>>>> > formal verification. You could also create special-purpose tools >>>>>> that >>>>>> > say, only turn ranks into matrices and nothing else: there's no >>>>>> reason >>>>>> > (apart from programmer convenience) why an election tool should be >>>>>> a >>>>>> > general purpose computer that you could hide all sorts of >>>>>> shenanigans in. >>>>>> > >>>>>> >> I’ve discussed that at length in previous posts, & it probably >>>>>> isn’t >>>>>> >> necessary to again post about ways of choosing how to vote in >>>>>> Approval. >>>>>> >> >>>>>> >> But, just summarize:It’s easy.Whichever of the various ways you >>>>>> >> prefer to use, for choosing whom to approve, it’s easy.…& no, it >>>>>> >> doesn’t require knowing your objectively-optimal vote. >>>>>> > >>>>>> >> >>>>>> >> I’ve many times pointed out that Approval’s Myerson-Weber >>>>>> equilibrium >>>>>> >> is the voter-median. >>>>>> >> >>>>>> >> i.e. Approval soon homes in on where the Condorcet-Winner is. >>>>>> > >>>>>> > Not necessarily. See the following paper: >>>>>> > https://arxiv.org/abs/2001.04216v2 >>>>>> > >>>>>> > According to the authors, under their model, if every voter follows >>>>>> a >>>>>> > particular thresholding rule, then iterative approval generally >>>>>> > arrives at the Condorcet winner (but not always). However, if >>>>>> voters >>>>>> > are left to choose any thresholding rule they want (as you >>>>>> propose), >>>>>> > then anything is possible. The arrangement might even elect a >>>>>> > Condorcet loser, and the outcome may be slow to converge or who >>>>>> wins >>>>>> > may depend on how many polls you hold. >>>>>> > >>>>>> > I'm reminded of a quote about distributed algorithms that I read >>>>>> > somewhere: "It's really easy to design distributed algorithms that >>>>>> > suffer from deadlock, network floods or widely unpredictable and >>>>>> > bizarre oscillations". That's in the context of computer science - >>>>>> > deadlocks might not be applicable to election methods. But it does >>>>>> > justify a starting position of skepticisim when considering schemes >>>>>> > that offload more of the work to the voters by turning a one-shot >>>>>> > method into a dynamical system. >>>>>> > >>>>>> >> It seems to me that, in every one of EM’s polls, including the >>>>>> recent >>>>>> >> one, Approval chose the CW. >>>>>> >> >>>>>> >> Have we forgotten that? >>>>>> > >>>>>> > I haven't, nor have I forgotten that Approval wasn't actually the >>>>>> poll >>>>>> > winner. >>>>>> > >>>>>> > But let's take this reasoning at face value. I don't have the other >>>>>> > polls' ballot data available at the moment, so let's consider the >>>>>> > latest one and pick... say, Borda. >>>>>> > >>>>>> > As one can see by going to >>>>>> > https://munsterhjelm.no/km/rbvote/calc.html, pasting in the data, >>>>>> and >>>>>> > clicking Borda, Ranked Pairs is also the Borda winner. >>>>>> > >>>>>> > But I don't think I'm going to start advocating for Borda. >>>>>> > >>>>>> > >>>>>> > To not be accused of tu quoque, let me clarify the point. As >>>>>> > Burlington shows, an election method needs to handle the hard >>>>>> cases, >>>>>> > not just the easy ones, or there may be a considerable uproar when >>>>>> the >>>>>> > method is faced with a hard case and then stumbles. (In addition, >>>>>> > people trying to make sure a stumble doesn't happen may start to do >>>>>> > mass compromising, further entrenching two-party rule.) >>>>>> > >>>>>> > So it's quite possible that our polls are easy cases. But like >>>>>> > FairVote claiming that IRV gets the Condorcet winner more than 90% >>>>>> of >>>>>> > the time, that doesn't by itself tell us much, because the failures >>>>>> > have such a strong impact. >>>>>> > >>>>>> > I illustrate the example above by picking a method we know to be >>>>>> bad >>>>>> > (Borda being so extremely easy to fool with cloning and burial), >>>>>> and >>>>>> > showing that the poll result comes out right. If a bad method can >>>>>> get >>>>>> > a good result, then "getting a good result" is less useful than it >>>>>> > might appear at first glance. >>>>>> > >>>>>> > -km >>>>>> > ---- >>>>>> > Election-Methods mailing list - see https://electorama.com/em for >>>>>> list >>>>>> > info >>>>>> >>>>> ---- > Election-Methods mailing list - see https://electorama.com/em for list > info >

If this Ossipoff character held himself to the same standards he holds everyone else, he would have blocked himself a long time ago. Toby On Monday 27 May 2024 at 10:09:01 BST, Michael Garman <michael.garman@rankthevote.us> wrote: >> This forum, like most, in its rules of conduct, asks that we not keep repeating an already-answered claim without answering the objections to it. In that case, kindly stop repeating your unsubstantiated conspiracy theories about fraud from 2000/2004!

Just one more astronomical reference: Next time you look at the Milky-Way, or a long-exposure photo of it or any galaxy: Realize that, for every one of its estimated half-trillion stars, there are 12 instances of having to find a candidate-name in a ranking of 135 candidate-names, when doing a handcount of Condorcet with 135 candidates (the number that a recent California governor-election had), & 300 million voters. BTW, when I posted my reply to one of KM’s last posts before he temporarily left this list, I worded it as a reply to him, even though I’d removed his name from the post’s “To:” field, as he’d requested. In that reply, I often said that I didn’t know what he meant. In philosophy it isn’t unusual for what one person says to be completely meaningless to someone else. That doesn’t mean that one of those people is wrong. It’s just that one person’s philosophical framework & topic isn’t the other person’s. KM’s philosophy of voting systems, & the philosophical framework in which he was discussing them can rightly, validly & reasonably influence KM’s preferences & choices among voting-systems. But it doesn’t apply to the objective matter-of-fact Approval-properties that I’d stated. On Sun, May 26, 2024 at 23:58 Michael Ossipoff <email9648742@gmail.com> wrote: > …or of course you could, for the voting, just ask each voter to write on > hir (big) ballot, which member of each of the 18090/2 candidate-pairs s/he > prefers to the other. S/he might be in the voting/booth for a while… > > On Sun, May 26, 2024 at 23:51 Michael Ossipoff <email9648742@gmail.com> > wrote: > >> Suppose that there are 300 million voters. Round 18090 off to 20000x >> >> 3E8 X 2E4 = 6E12 >> >> 6 trillion. >> >> That’s roughly the number of miles in a light-year. >> >> Whatever the number of voters, nationally, is, suppose that every one of >> them participated in the handcount. >> >> With the work divided among them all, each has “only” an amount of work >> equal to that of determining which member of each one of 18090/2 pairs of >> candidates is ranked over the other on a ballot. >> >> 135 candidates in each ranking. On the average, s/he’d only have to look >> at half of the ranking’s candidates to find each of the 2 members of each >> pair. >> >> …but s/he’d have to do it for both. So the order-determination for each >> candidate-pair, on the average will require looking at 135 candidates. >> >> Suppose that s/he can skim over 10 of them in a second, when doing those >> searches. >> >> Then it would take (135)(18090/2)/10 seconds. That’s about 1.4 days. But >> say she only does it for 8 hours per day (with no breaks). Now it’s more >> like 4.2 days. >> >> But supervision is the whole point. If s/he’s working alone, she can say >> that the pairwise vote-totals are whatever s/he wants them to be. >> >> So in reality, it would be counting *teams* each wit representatives of >> several parties. Say (optimistically) there are 10 parties. >> >> Now it will take 42 days. But the parties don’t really have equal numbers >> of members. It’s going to take longer. >> >> …&, realistically, they aren’t going to scan 10 ranked candidates every >> second for 8 hours with no breaks. >> >> It would obviously take months. Might it not, in fact, be measured in >> years…with every one of the nation’s voters participating in that handcount? >> >> >> >> On Sun, May 26, 2024 at 22:11 Michael Ossipoff <email9648742@gmail.com> >> wrote: >> >>> …& of course multiply that 18090 by the number of ballots, to get an >>> idea of what’s involved in the 18090 order-determinations to be done on >>> each ballot, & recorded, & then summed, to obtain each of the 18090 >>> pairwise vote-totals. >>> >>> …each of which then must be carried or transmitted to where the central >>> count is done. >>> >>> On Sun, May 26, 2024 at 22:03 Michael Ossipoff <email9648742@gmail.com> >>> wrote: >>> >>>> >>>> >>>> On Sun, May 26, 2024 at 12:56 Chris Benham <cbenhamau@yahoo.com.au> >>>> wrote: >>>> >>>>> >>>>> In 2003 there was California gubernatorial election with 135 >>>>> candidates. >>>> >>>> >>>> That’s 18090 pairwise vote-totals to determine at each precinct from >>>> the rankings, by examining each ranking to determine which member of each >>>> possible candidate-pair is ranked over the other on that ballot. >>>> >>>> …& 18080 pairwise vote-totals for the precincts to sum, store, & >>>> transmit or carry to the central count location. >>>> >>>> …& verify in an audit. >>>> >>>> >>>> >>>>> >>>>> >>>>> https://en.wikipedia.org/wiki/2003_California_gubernatorial_recall_election#Results >>>>> >>>>> Chris B. >>>>> >>>>> On 26/05/2024 9:29 pm, Kristofer Munsterhjelm wrote: >>>>> > On 2024-05-26 07:28, Michael Ossipoff wrote: >>>>> >> Someone keeps repeating that the voters shouldn’t have to vote >>>>> >> strategically. He wants the method to do it all for us, after we >>>>> >> merely state our sincere-rankings. >>>>> >> >>>>> >> That’s of course a common attitude: >>>>> >> >>>>> >> …wanting a high-tech, computation-intensive,computer-dependent >>>>> system >>>>> >> to do it all for us, taking all the actual choosing responsibility >>>>> >> off of us.…sheltering & isolating us from the choice. >>>>> > >>>>> > I would prefer that you do not attribute opinions that the >>>>> proponents >>>>> > have not expressed. Nowhere have I said that Condorcet "[isolates] >>>>> us >>>>> > from the choice" we make. >>>>> > >>>>> > What you call "sheltering" and "isolating", I see as the method >>>>> taking >>>>> > proper responsibility - proper responsibility to turn the voters' >>>>> > unambiguous honest opinions into an outcome without dumping the >>>>> > algorithmic calculation upon te voter themselves. See my "kick the >>>>> can >>>>> > down the road" post for more info. >>>>> > >>>>> > Now, I could make a caricature of Approval itself. Perhaps something >>>>> > about a calculator that just says "IDK, do the base conversion >>>>> > yourself, I only accept input numbers in factoradic". But >>>>> caricatures >>>>> > only make people angry. Let's not stoop to them, shall we? >>>>> > >>>>> >> I’ll ask this for the 3^rd time: >>>>> >> >>>>> >> … >>>>> >> >>>>> >> How would like you to handount-audit a Condorcet count for a >>>>> >> many-candidate national presidential election? >>>>> > >>>>> > I think your question assumes something that won't hold. If you have >>>>> a >>>>> > 25-candidate presidential election, you've already lost, because >>>>> > nobody is going to rank 25 candidates, irrespective of whether the >>>>> > method is Condorcet, IRV, or Borda. >>>>> > >>>>> > I'm not familiar with minor parties in the US. Has there ever been a >>>>> > 25-candidate presidential election? >>>>> > >>>>> > I don't think I can comment beyond that: I don't know enough about >>>>> > poll workers. I'll leave that to someone with experience. (Although >>>>> > roughly calculating: suppose 6 candidates like in Burlington. That's >>>>> > 30 pairs. Five times the work if counting a particular preference is >>>>> > as before. So you'd either need 5x the workers, or five times the >>>>> > time, or some combination of the two.) >>>>> > >>>>> > On an aside, though, I would say that I generally wouldn't want to >>>>> get >>>>> > computers anywhere near election counting. However, if you >>>>> absolutely >>>>> > have to have them, there are ways of making sure they don't cheat: >>>>> > formal verification. You could also create special-purpose tools >>>>> that >>>>> > say, only turn ranks into matrices and nothing else: there's no >>>>> reason >>>>> > (apart from programmer convenience) why an election tool should be a >>>>> > general purpose computer that you could hide all sorts of >>>>> shenanigans in. >>>>> > >>>>> >> I’ve discussed that at length in previous posts, & it probably >>>>> isn’t >>>>> >> necessary to again post about ways of choosing how to vote in >>>>> Approval. >>>>> >> >>>>> >> But, just summarize:It’s easy.Whichever of the various ways you >>>>> >> prefer to use, for choosing whom to approve, it’s easy.…& no, it >>>>> >> doesn’t require knowing your objectively-optimal vote. >>>>> > >>>>> >> >>>>> >> I’ve many times pointed out that Approval’s Myerson-Weber >>>>> equilibrium >>>>> >> is the voter-median. >>>>> >> >>>>> >> i.e. Approval soon homes in on where the Condorcet-Winner is. >>>>> > >>>>> > Not necessarily. See the following paper: >>>>> > https://arxiv.org/abs/2001.04216v2 >>>>> > >>>>> > According to the authors, under their model, if every voter follows >>>>> a >>>>> > particular thresholding rule, then iterative approval generally >>>>> > arrives at the Condorcet winner (but not always). However, if voters >>>>> > are left to choose any thresholding rule they want (as you propose), >>>>> > then anything is possible. The arrangement might even elect a >>>>> > Condorcet loser, and the outcome may be slow to converge or who wins >>>>> > may depend on how many polls you hold. >>>>> > >>>>> > I'm reminded of a quote about distributed algorithms that I read >>>>> > somewhere: "It's really easy to design distributed algorithms that >>>>> > suffer from deadlock, network floods or widely unpredictable and >>>>> > bizarre oscillations". That's in the context of computer science - >>>>> > deadlocks might not be applicable to election methods. But it does >>>>> > justify a starting position of skepticisim when considering schemes >>>>> > that offload more of the work to the voters by turning a one-shot >>>>> > method into a dynamical system. >>>>> > >>>>> >> It seems to me that, in every one of EM’s polls, including the >>>>> recent >>>>> >> one, Approval chose the CW. >>>>> >> >>>>> >> Have we forgotten that? >>>>> > >>>>> > I haven't, nor have I forgotten that Approval wasn't actually the >>>>> poll >>>>> > winner. >>>>> > >>>>> > But let's take this reasoning at face value. I don't have the other >>>>> > polls' ballot data available at the moment, so let's consider the >>>>> > latest one and pick... say, Borda. >>>>> > >>>>> > As one can see by going to >>>>> > https://munsterhjelm.no/km/rbvote/calc.html, pasting in the data, >>>>> and >>>>> > clicking Borda, Ranked Pairs is also the Borda winner. >>>>> > >>>>> > But I don't think I'm going to start advocating for Borda. >>>>> > >>>>> > >>>>> > To not be accused of tu quoque, let me clarify the point. As >>>>> > Burlington shows, an election method needs to handle the hard cases, >>>>> > not just the easy ones, or there may be a considerable uproar when >>>>> the >>>>> > method is faced with a hard case and then stumbles. (In addition, >>>>> > people trying to make sure a stumble doesn't happen may start to do >>>>> > mass compromising, further entrenching two-party rule.) >>>>> > >>>>> > So it's quite possible that our polls are easy cases. But like >>>>> > FairVote claiming that IRV gets the Condorcet winner more than 90% >>>>> of >>>>> > the time, that doesn't by itself tell us much, because the failures >>>>> > have such a strong impact. >>>>> > >>>>> > I illustrate the example above by picking a method we know to be bad >>>>> > (Borda being so extremely easy to fool with cloning and burial), and >>>>> > showing that the poll result comes out right. If a bad method can >>>>> get >>>>> > a good result, then "getting a good result" is less useful than it >>>>> > might appear at first glance. >>>>> > >>>>> > -km >>>>> > ---- >>>>> > Election-Methods mailing list - see https://electorama.com/em for >>>>> list >>>>> > info >>>>> >>>>