If a>b, b>c, and c>a there's no Condorcet winner, and if a>b, b>c, and a>c, then a is the winner, but what is it considered when a=b, b>c, a>c (e.g. D: a>b>c, E: a>c>b, F: b>a>c, G: b>c>a), is that it then still considered that there is no Condorcet winner, or would we call it a tie between a and b in that case (or alternatively both a and b are winners)?
I'm a proponent of Condorcet's method, and it took even me a while to interpret what looks like unfairness. Evidence that the Condorcet win criterion is not very intuitive. The example of M winning only with Condorcet's method actually shows how the other methods are flawed. It is the interference of the less popular candidates, L and O, that creates the illusion that N has the most public support, when the truth is that M is preferred over N. Creating a hybrid, consisting of Condorcet's and other methods, can clear up confusion. For example: 1. Eliminate one candidate having the fewest 1st ranks. 2. Eliminate one candidate having the fewest 1st and 2nd ranks combined. 3. Condorcet comparison of remaining candidates. Or as a practical election method (Remember that most elections have far more than 11 voters): 1. Eliminate up to half the field by 1st ranks, but keep at least 3. 2. Condorcet comparison 3. Paradox or ties resolved by eliminating one candidate having the fewest 1st+2nd ranks.
the CWC is intuitively desirable but causes unworkable problems. Maybe this can be explained by Temkin's ideas on nontransitivty. Temkin says that a surviving option of many 1v1 contests shouldn't necessarily be the same winner as a single contest of many options. For instance, when you go to the dealership to buy a car you might be tempted to buy a fancier model than you originally intended. Then you may be tempted to get the slightly fancier model. Etc. Soon enough you'll be looking at a very expensive car that you would never have considered when you came in the door. Relatedly, you might not be entirely irrational to prefer candidate x to y, but then rank y over x when they are part of a larger set of options (e.g., you might make the ranking of "z>y>x>w" while you also prefer x over y when they are against each other one-on-one). Perhaps the CWC's concern with the winner being able to win each and every 1 on 1 contest is extraneous. Temkin talks about some of his ideas on nontransitivity here: philosophybites.com/2015/07/larry-temkin-on-transitivity.html
From my understanding Temkin is questioning the transitivity of preferences, he's arguing that one might rationally prefer a to b, b to c and c to a, if offered in 1v1 contests. If someone were to hold such preferences, then they could not even vote in this system since they would be unable to rationally rank their choices in an order with all the options included. If you think this is an issue, the problem is not with the CWC, but rather with the voting system to begin with. If people can have nontransitive preferences, then to express those preferences they need to vote only in 1 v 1 matches (which is not what Condorcet allows, rather it creates artificial 1v1 matches based on a full ranked list of preferences). Nontransitivity poses a problem for this whole system, not the CWC.
Could you give me a realistic example of preferring x>y in isolation but z>y>x>w with multiple options? Especially in the context of elections (because, let's be honest, we're probably mainly debating these systems and criteria for use in elections) Even the a>b, b>c, c>a seems quite unlikely for a single individual to hold, especially in an election contest. I think it's really only an issue when as a collective, a situation of a>b, b>c, c>a arises. I could for instance imagine it with the most recent French elections, where we had Macron's Renaissance (RE) party, LePen's Rassemblement National (RN) party, and the left wing Nouveau Front Populaire (NFP) alliance under Mélenchon as the three main parties. RE is establishment centre-right RN is anti-establishment far-right NFP is anti-establishment left to far-left RE voters, being centrists, could, depending on their leaning, go to either side as their second preference. Whilst RN and NFP voters might prefer RE for being ideological closer, or the other for being anti-establishment. Depending on how people would vote, that could feasibly lead to no (initial) winner under a Condorcet system (typically Condorcet systems have either come up with a way to still determine a winner from the Smith set, or use a secondary method to choose a winner from it). But I find it hard to believe that any individual voter would prefer RE over RN, NFP over RE, but then RN over NFP (right wing), or the other way around. Or take something like Brexit, let's say voters had been given the choice between a no-deal Brexit, May's deal Brexit, or no Brexit. I can imagine: -hard line Brexiteers who prefer no deal > May's deal > no Brexit -Europhiles who prefer no Brexit > May's deal > no deal -soft Brexiteers leaning to Brexit who prefer May's deal > no deal > no Brexit -soft Brexiteers leaning Europhilic who prefer May's deal > no Brexit> no deal -in-or-outers leaning to Brexit who prefer no deal > no Brexit > May's deal -in-or-outers leaning Europhilic who prefer no Brexit> no deal > May's deal I can also imagine thinking two options are equally good and the third is bad, or two options being equally bad and the third is good: -any Brexit: no deal=May's deal>no Brexit -any Euro ties: no Brexit=May's deal>no deal -in-or-outers without leaning: no Brexit=no deal>May's Deal -Hard Brexit only: no deal>May's deal=no Brexit -EU only: no Brexit>May's deal=no deal -May's deal only: May's deal>no deal=no Brexit But I can't imagine anyone who'd say no Brexit > May's deal, May's deal> no deal, no deal > no Brexit, or vice versa.
I'm having trouble finding this criterion trivially desirable. Imagine each member of the society gave the candidates a number from 0 to 10 showing how much they like them (higher better) and then rank the candidates according these numbers. Assume a society of 3: ABC. A and B give candidate D a 2 and candidate E a 5 while C gives D a 5 and E a 2. Assume E wins. if we then change the valuation to A and B: 0 and 0.0000001, C: 10 and 0 (for D, E respectively), if the system passes the Condorcet Criterion, E wins. but A and B clearly don't like their options too much, while C does. This criterion, in other words, cares about who you prefer, without worrying how much more you prefer him/her over your other option. I don't know if this is desirable. All least not trivially. Of course, maximizing happiness has its own problems, as it gives incentive to lie, and would not even be eligible as a system, as it does not give a list of candidates as input, but rather their valuations for each voter. But the question still stands, I think.
In the last video in the series we will look into each of the reasonability of each of these criteria. There are other voting systems that deal with how much you dislike a particular candidate, or take that into consideration. Clearly I need to do another more advanced series on voting systems after this one.
So C gets to be a dictator because the majority thinks both options are awful? That sounds like a terrible voting system Also, even A and B even have just a slight preference, even if minimal, why wouldn't they just tactically score and give 0 and 10, rather than 0 and 0.000001, to at least increase the option they marginally prefer?
Why would Ranked Vote eliminate M and O at the same time (2:36)? Wouldn't tie-breaking rules dictate considering O to be weaker in 1st 2 rounds (or most commonly chose as the last choice)?
You could have some loser-tie-breaker method specified in your Instant Runoff Voting method, but nonetheless, it is still possible for a non-condorcet winner to win sometimes. But the most simple iteration of IRV, as used here, would simultaneously eliminate all least first-(remaining-)preference votes if they are tied for that metric.
1:14 is there any evidence that there often is no Condorcet winner? Is this based on analysis of hypotheticals, or of real life scenarios? Considering that in most elections candidates can generally be (roughly) ranked on an axis voters on the extremes will tend to choose their own extreme, then the centre, then the other extreme, and the centre will split some leaning one way and some another but centre first, which in practice often means that the centre/median (within the ideological space) tends to be the condorcet winner. Only if the centre might be ruling long enough whilst being dissatisfying to voters, might people start to prefer anything non-establishment over the established centre, and thus choose extreme1>extreme2>centre, and then only if enough people have a preference like that, but also still enough stick with traditional preferences, would there reasonably arise a smith set rather than a single Condorcet winner. Intuitively, I'd therefore expect that in real life election scenarios, having no condorcet winner is quite unlikely.
Either I'm misunderstanding IRV or you've made a mistake. The first round is done correctly: L has one vote, M three, N four, and O three. This means L is indeed eliminated. His vote still counts, though, and therefore goes to M. Second round results are therefore: M four, N four, O three. This eliminates O, whose votes all go to M, leaving him with 7 votes and N with 3 votes. There are indeed scenarios where IRV does not pick the concordet winner, but this is not one of them.
K's order is L>N>M>O, so that vote goes to N after L is eliminated, not to M. In the second round that means 5N 3O 3M. It depends on the tiebreaker rule used how it proceeds from there. If a tie for last place removes all tied candidates, then N is the only one left and wins. If a tiebreaker is used that favours O, then M's votes go to N, causing 8N 3O and this N wins. If a tie breaker is used that favours M then O's votes go to M and it becomes 5N 6M and M wins. As he used the method of tied last place being both eliminated, N won.
@@nienke7713 nono, imagine if N and M were the only candidates in the election. That's what condorcet winner means, the candidate that would beat every other candidate if the election was head to head. If it was only N and M, D would pick N and N would win. Meaning N is the condorcet winner
If a>b, b>c, and c>a there's no Condorcet winner, and if a>b, b>c, and a>c, then a is the winner, but what is it considered when a=b, b>c, a>c (e.g. D: a>b>c, E: a>c>b, F: b>a>c, G: b>c>a), is that it then still considered that there is no Condorcet winner, or would we call it a tie between a and b in that case (or alternatively both a and b are winners)?
I'm a proponent of Condorcet's method, and it took even me a while to interpret what looks like unfairness. Evidence that the Condorcet win criterion is not very intuitive.
The example of M winning only with Condorcet's method actually shows how the other methods are flawed. It is the interference of the less popular candidates, L and O, that creates the illusion that N has the most public support, when the truth is that M is preferred over N.
Creating a hybrid, consisting of Condorcet's and other methods, can clear up confusion. For example:
1. Eliminate one candidate having the fewest 1st ranks.
2. Eliminate one candidate having the fewest 1st and 2nd ranks combined.
3. Condorcet comparison of remaining candidates.
Or as a practical election method (Remember that most elections have far more than 11 voters):
1. Eliminate up to half the field by 1st ranks, but keep at least 3.
2. Condorcet comparison
3. Paradox or ties resolved by eliminating one candidate having the fewest 1st+2nd ranks.
the CWC is intuitively desirable but causes unworkable problems. Maybe this can be explained by Temkin's ideas on nontransitivty. Temkin says that a surviving option of many 1v1 contests shouldn't necessarily be the same winner as a single contest of many options. For instance, when you go to the dealership to buy a car you might be tempted to buy a fancier model than you originally intended. Then you may be tempted to get the slightly fancier model. Etc. Soon enough you'll be looking at a very expensive car that you would never have considered when you came in the door. Relatedly, you might not be entirely irrational to prefer candidate x to y, but then rank y over x when they are part of a larger set of options (e.g., you might make the ranking of "z>y>x>w" while you also prefer x over y when they are against each other one-on-one). Perhaps the CWC's concern with the winner being able to win each and every 1 on 1 contest is extraneous.
Temkin talks about some of his ideas on nontransitivity here:
philosophybites.com/2015/07/larry-temkin-on-transitivity.html
From my understanding Temkin is questioning the transitivity of preferences, he's arguing that one might rationally prefer a to b, b to c and c to a, if offered in 1v1 contests. If someone were to hold such preferences, then they could not even vote in this system since they would be unable to rationally rank their choices in an order with all the options included. If you think this is an issue, the problem is not with the CWC, but rather with the voting system to begin with. If people can have nontransitive preferences, then to express those preferences they need to vote only in 1 v 1 matches (which is not what Condorcet allows, rather it creates artificial 1v1 matches based on a full ranked list of preferences). Nontransitivity poses a problem for this whole system, not the CWC.
Carneades.org you're right, it is more devastating than i implied!
Could you give me a realistic example of preferring x>y in isolation but z>y>x>w with multiple options? Especially in the context of elections (because, let's be honest, we're probably mainly debating these systems and criteria for use in elections)
Even the a>b, b>c, c>a seems quite unlikely for a single individual to hold, especially in an election contest.
I think it's really only an issue when as a collective, a situation of a>b, b>c, c>a arises.
I could for instance imagine it with the most recent French elections, where we had Macron's Renaissance (RE) party, LePen's Rassemblement National (RN) party, and the left wing Nouveau Front Populaire (NFP) alliance under Mélenchon as the three main parties.
RE is establishment centre-right
RN is anti-establishment far-right
NFP is anti-establishment left to far-left
RE voters, being centrists, could, depending on their leaning, go to either side as their second preference.
Whilst RN and NFP voters might prefer RE for being ideological closer, or the other for being anti-establishment.
Depending on how people would vote, that could feasibly lead to no (initial) winner under a Condorcet system (typically Condorcet systems have either come up with a way to still determine a winner from the Smith set, or use a secondary method to choose a winner from it).
But I find it hard to believe that any individual voter would prefer RE over RN, NFP over RE, but then RN over NFP (right wing), or the other way around.
Or take something like Brexit, let's say voters had been given the choice between a no-deal Brexit, May's deal Brexit, or no Brexit.
I can imagine:
-hard line Brexiteers who prefer no deal > May's deal > no Brexit
-Europhiles who prefer no Brexit > May's deal > no deal
-soft Brexiteers leaning to Brexit who prefer May's deal > no deal > no Brexit
-soft Brexiteers leaning Europhilic who prefer May's deal > no Brexit> no deal
-in-or-outers leaning to Brexit who prefer no deal > no Brexit > May's deal
-in-or-outers leaning Europhilic who prefer no Brexit> no deal > May's deal
I can also imagine thinking two options are equally good and the third is bad, or two options being equally bad and the third is good:
-any Brexit: no deal=May's deal>no Brexit
-any Euro ties: no Brexit=May's deal>no deal
-in-or-outers without leaning: no Brexit=no deal>May's Deal
-Hard Brexit only: no deal>May's deal=no Brexit
-EU only: no Brexit>May's deal=no deal
-May's deal only: May's deal>no deal=no Brexit
But I can't imagine anyone who'd say no Brexit > May's deal, May's deal> no deal, no deal > no Brexit, or vice versa.
I'm having trouble finding this criterion trivially desirable. Imagine each member of the society gave the candidates a number from 0 to 10 showing how much they like them (higher better) and then rank the candidates according these numbers. Assume a society of 3: ABC. A and B give candidate D a 2 and candidate E a 5 while C gives D a 5 and E a 2. Assume E wins. if we then change the valuation to A and B: 0 and 0.0000001, C: 10 and 0 (for D, E respectively), if the system passes the Condorcet Criterion, E wins. but A and B clearly don't like their options too much, while C does. This criterion, in other words, cares about who you prefer, without worrying how much more you prefer him/her over your other option. I don't know if this is desirable. All least not trivially. Of course, maximizing happiness has its own problems, as it gives incentive to lie, and would not even be eligible as a system, as it does not give a list of candidates as input, but rather their valuations for each voter. But the question still stands, I think.
In the last video in the series we will look into each of the reasonability of each of these criteria. There are other voting systems that deal with how much you dislike a particular candidate, or take that into consideration. Clearly I need to do another more advanced series on voting systems after this one.
@@CarneadesOfCyrene When you get around to it, perhaps include the STAR (score then automatic run-off) system.
So C gets to be a dictator because the majority thinks both options are awful? That sounds like a terrible voting system
Also, even A and B even have just a slight preference, even if minimal, why wouldn't they just tactically score and give 0 and 10, rather than 0 and 0.000001, to at least increase the option they marginally prefer?
Cooms' rule Would choose M. in the example given. So this is not a counter example for Cooms' rule.
At 3:23, the Borda Count for O is 14 not 13
You are correct! Thanks for the correction. If RUclips brings back annotations, I'll make a note in the video.
Why would Ranked Vote eliminate M and O at the same time (2:36)? Wouldn't tie-breaking rules dictate considering O to be weaker in 1st 2 rounds (or most commonly chose as the last choice)?
You could have some loser-tie-breaker method specified in your Instant Runoff Voting method, but nonetheless, it is still possible for a non-condorcet winner to win sometimes.
But the most simple iteration of IRV, as used here, would simultaneously eliminate all least first-(remaining-)preference votes if they are tied for that metric.
1:14 is there any evidence that there often is no Condorcet winner? Is this based on analysis of hypotheticals, or of real life scenarios?
Considering that in most elections candidates can generally be (roughly) ranked on an axis voters on the extremes will tend to choose their own extreme, then the centre, then the other extreme, and the centre will split some leaning one way and some another but centre first, which in practice often means that the centre/median (within the ideological space) tends to be the condorcet winner.
Only if the centre might be ruling long enough whilst being dissatisfying to voters, might people start to prefer anything non-establishment over the established centre, and thus choose extreme1>extreme2>centre, and then only if enough people have a preference like that, but also still enough stick with traditional preferences, would there reasonably arise a smith set rather than a single Condorcet winner.
Intuitively, I'd therefore expect that in real life election scenarios, having no condorcet winner is quite unlikely.
Either I'm misunderstanding IRV or you've made a mistake. The first round is done correctly: L has one vote, M three, N four, and O three. This means L is indeed eliminated. His vote still counts, though, and therefore goes to M. Second round results are therefore: M four, N four, O three. This eliminates O, whose votes all go to M, leaving him with 7 votes and N with 3 votes. There are indeed scenarios where IRV does not pick the concordet winner, but this is not one of them.
K's order is L>N>M>O, so that vote goes to N after L is eliminated, not to M.
In the second round that means 5N 3O 3M.
It depends on the tiebreaker rule used how it proceeds from there.
If a tie for last place removes all tied candidates, then N is the only one left and wins.
If a tiebreaker is used that favours O, then M's votes go to N, causing 8N 3O and this N wins.
If a tie breaker is used that favours M then O's votes go to M and it becomes 5N 6M and M wins.
As he used the method of tied last place being both eliminated, N won.
actually, dictatorship would fulfill the Condorcet win criterion since in a one on one race D picks N over M.
That's why it doesn't, M is the Condorcet winner, yet D picks N instead
@@nienke7713 nono, imagine if N and M were the only candidates in the election. That's what condorcet winner means, the candidate that would beat every other candidate if the election was head to head. If it was only N and M, D would pick N and N would win. Meaning N is the condorcet winner
@@shmendusel if the majority would vote for M over N but the dictator votes N over M, then M is the Condorcet winner but N is the dictatorship winner.