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Part 2, How are the alternatives, IRV section #3

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trevorhobson opened this issue Dec 15, 2016 · 17 comments
Open

Part 2, How are the alternatives, IRV section #3

trevorhobson opened this issue Dec 15, 2016 · 17 comments

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@trevorhobson
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it's possible for a winning candidate to lose, by becoming more popular. What a glitch!

It is possible for a sitting candidate to lose a subsequent election if they become more popular, but only if another candidate has become even more popular. In the example the triangle has become more popular but the hexagon has become even more popular. This is not a glitch, it is how the system is supposed to work.

When you move the voter base you are not changing an election, you are comparing the results of different elections.

  • In the first election, starting position, the two candidate preferred vote is between square and triangle so the hexagon votes are distributed to their second preference candidate, as the hexagon is closer to the triangle more hexagon votes go to the triangle so the triangle has won.
  • In the second election, voters move closer to triangle and hexagon, the two candidate preferred vote is between the hexagon and the triangle, so the square votes are distributed to their second preference candidate, as the square is closer to the hexagon more square votes go to the hexagon so the hexagon has won. Notice that the difference in round 1 votes between triangle and hexagon stay about the same as you move the voters up, they are both getting more popular.

You must remember that a candidate is only winning if they have greater than 50% of the vote in the current election. At no point in the second election is the triangle ever winning.

There may be problems with IRV but I don’t think that the example you have used demonstrates them.

Maybe the way you have chosen to represent the voters has influenced the way you view the results?

@ncase
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ncase commented Dec 15, 2016

Thank you for catching my mistake, @trevorhobson !

I was trying to simulate this example of IRV's non-monotonicity from Exploratorium, but I messed up. Wikipedia also has another concrete example, where Left Candidate gets more popular compared to Right Candidate… without changing any votes to Center Candidate... and that causes the election to swing to Centre Candidate.

...Hm. It may be hard to simulate that, given my voting simulation uses 2D geometry as its central metaphor. There isn't a way to increase votes for "left" and less for "right" without affecting "center" in some way.

I'll figure out a better simulation for that section. Thanks again!

Cheers,
~ Nicky

@trevorhobson
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I don't find the examples on Wikipedia very compelling. There seems to be no understanding that by changing the votes they are talking about different elections.

The Wikipedia example seems to gloss over the fact that in two people changing their votes, in the subsequent election, the Right Candidate is becoming less popular compared to the Center Candidate and that most Right Candidate electors have the Center Candidate as their second choice so of course the Center Candidate should win the second election.

In IRV no candidate is winning the current election until they have greater than 50% of the vote. Until that point it is impossible to identify who is winning, it is only possible to identify who is losing. Once a candidate is winning it is impossible for them to lose.

There is some discussion on this topic on the talk page of the Wikipedia article.

@endolith
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endolith commented Jun 27, 2017

Here's a 1D example, if that helps:

Population moving away from Red causes Red to win. Sounds like a glitch to me.

You could also adapt any of the non-monotonic examples from http://zesty.ca/voting/sim/, but the 3-candidate mostly-1D version is probably still the simplest:

You have to get the candidates really close together to show this effect, however:

2017-06-26 21_57_59-to build a better ballot - waterfox

@trevorhobson
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I am afraid that I still cannot see the issue even with your most recent examples.

Provided that green does not get 50%+1 on their own, blue voters put red second and are eliminated in the first round it does not matter how much the swing is away from red they are still going to win.

It is simply mathematically impossible for a winning candidate, that is they have achieved 50%+1 of the total votes cast, and still loose by getting more votes in further counting. Three is no need to bother continuing counting if a candidate reaches this point.

Fundamentally you cannot pick a winning candidate until one reaches 50%+1. It is a mistake to look at the ongoing count that way because the count will be influenced by a number of things including the random order that the counters pick up the ballots. Once you have counted the first round you have identified the least popular candidate and unless a candidate has reached 50%+1 you eliminate the least popular and count their second votes.

You are trying to get a run-off between the two most preferred candidates overall. Sure someone could attempt to place a strategic vote to try and get their preferred candidate elected, but this is not a very safe option as they don't have any real idea how others will vote.

@endolith
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endolith commented Dec 7, 2017

Provided that green does not get 50%+1 on their own, blue voters put red second and are eliminated in the first round it does not matter how much the swing is away from red they are still going to win.

I don't understand. Are you saying the images are wrong and IRV would not behave this way?

Fundamentally you cannot pick a winning candidate until one reaches 50%+1.

The way IRV does the rounds produces a false majority, though. You could equally well eliminate the top-rated candidate in each round until there are only two losers left, and then declare that one has majority support among the population because they're preferred by 50%+1 over the other loser.

Once you have counted the first round you have identified the least popular candidate

IRV only looks at first preference votes, not the entire ballot, so it is incapable of identifying the most/least popular candidate. In the 1D example above, Blue is the most popular candidate, but is eliminated first by IRV.

@trevorhobson
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I don't understand. Are you saying the images are wrong and IRV would not behave this way?

Sorry for the confusion. I am saying it is supposed to behave this way.

The way IRV does the rounds produces a false majority, though. You could equally well eliminate the top-rated candidate in each round until there are only two losers left, and then declare that one has majority support among the population because they're preferred by 50%+1 over the other loser.

First round identifies the candidate that most people don't want as their first choice and so on. You are trying to get a run-off between the two candidates with peoples vote closest to their first choice. I must be misunderstanding what you mean by false majority.

I am not sure why you would count the vote the way mention because if a candidate got more than 50%+1 in the first round they would be eliminated which really does not make sense.

IRV only looks at first preference votes, not the entire ballot, so it is incapable of identifying the most/least popular candidate. In the 1D example above, Blue is the most popular candidate, but is eliminated first by IRV.

I think this a fundamental difference between how we view most popular. I don't see blue as being the most popular. There is no way to tell how much less an individual wants their second candidate compared to their first.

@endolith
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endolith commented Dec 8, 2017

First round identifies the candidate that most people don't want as their first choice and so on. You are trying to get a run-off between the two candidates with peoples vote closest to their first choice. I must be misunderstanding what you mean by false majority.

I guess what I'm describing would be called an "artificial majority" in this essay.

I am not sure why you would count the vote the way mention because if a candidate got more than 50%+1 in the first round they would be eliminated which really does not make sense.

You wouldn't count in the way I described; it would produce a terrible result (the best of two losers). But by eliminating everyone else, the best of two losers could claim that they have majority support, in the same way that the winner of IRV does. It's only a majority because everyone better was eliminated first.

A better example would be Coombs' method: Same process as IRV, but with a different elimination rule:

  • IRV: Eliminate the candidate ranked first by the smallest number of voters
  • Coombs: Eliminate the candidate ranked last by the largest number of voters

Both rules eliminate the "worst" candidate in each round, and produce a "majority" winner, yet the winners are different. (Coombs chooses Nashville, IRV chooses Knoxville.)

I think this a fundamental difference between how we view most popular. I don't see blue as being the most popular.

Well here we know the preferences of each voter, so we know that blue is the most popular. Blue is preferred over red by a majority, and preferred over green by a majority. Blue is the best match for the average voter, the most representative choice that could be made.

There is no way to tell how much less an individual wants their second candidate compared to their first.

That's true for ranked-choice ballots, which is why ranked-choice ballots are a bad idea. :D They eliminate any information about strength of preference. Cardinal ballots allow you to express big preferences and small preferences.

@endolith
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endolith commented Dec 8, 2017

Anyway, back to your original point, referencing this example:

It is possible for a sitting candidate to lose a subsequent election if they become more popular

In this case, the population is shifting left from one election to the next, Blue is becoming more popular, and losing the subsequent election.

but only if another candidate has become even more popular.

In this case, Green is becoming more popular. But Red is the one who wins. Red is the only candidate that has become less popular from one election to the next. You think that's logical?

@trevorhobson
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While red is becoming less popular as a first choice they are still the preferred second choice of the blue voters, the percentage of blue > red > green is not changing and remains higher than the blue > green > red percentage. The apparent 'glitch' comes at the point that blue becomes less popular than green.

In this case, Green is becoming more popular. But Red is the one who wins. Red is the only candidate that has become less popular from one election to the next. You think that's logical?

To what lengths do you take this?
Two elections, 100 votes each
Election one:

  • Red 98 votes
  • Blue 1 vote
  • Green 1 vote

Election two

  • Red 51 votes
  • Blue 24 votes
  • Green 25 votes

Red has become dramatically less popular, 47 votes, so who should win?

Well here we know the preferences of each voter, so we know that blue is the most popular. Blue is preferred over red by a majority, and preferred over green by a majority. Blue is the best match for the average voter, the most representative choice that could be made.

Are you considering blue the best choice because both green and red put them second? I see this as the most begrudging compromise choice for the majority of the people and one that can not be considered as equally begrudging for both red and green voters.

Let us say we have three political parties, and my apologies for invoking the Nazis:

  1. Nazi party
  2. Nazi, but won't kill one particular group that the Nazi's will party
  3. Peace and rainbows party

In IRV the party 3 voters are forced begrudgingly to put party 2 as their second choice for their vote to count but it is a greater compromise than party 1 voters putting party 2 as their second choice so voters apparently prefer party 2 over party 1 and party 3 by a majority!

The second and subsequent choices are begrudging choices. The vote is saying that if my candidate is not in the run-off then I guess I would have to choose this candidate.

Coombs: Eliminate the candidate ranked last by the largest number of voters

In this situation:
A > B > C: 17% -1 vote
A > C > B: 8%
B > C > A: 8% +1 vote
B > A > C: 17%
C > D > A: 25%
C > A > D: 25%
The last total for C is 34% -1 vote, they are eliminated in the first round because the next closest is A with 33% +1 last total vote. So A and B run off against each other resulting in 75% of voters being upset by their first choice not being elected and 50% of voters being upset for not getting their candidate in the run-off. With IRV 50% of voters get their first choice being elected and 75% of voters get their first choice in the run-off.
It looks to me that Coombs could be worse than most first votes wins in terms of having the most upset voters.

@wolftune
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wolftune commented Jan 4, 2018

@trevorhobson you're missing the key points:

Nobody is saying that IRV is breaking it's own rules. Internally to the rules of IRV, everything works exactly as described. The question is whether those rules are good and fair.

Consider the collection of voters whose 1st choice does not win. Within that group, some of them have their 2nd choice counted and others have their 2nd choice completely ignored, never counted at all. All the problems with IRV fall out of this fact.

Because of that fact, the winner in IRV can depend on which voters get their 2nd choice counted and which do not. Thus, the only things to discuss about the fairness of the system are:

  • Is it okay to count of some and not all of the 2nd choices from the voters whose 1st choice does not win?
  • If that is okay (it's not though), does IRV reliably select which 2nd choices are counted in a fair way? (answer: definitely not)

The problem with the definition of "majority" here is that you are suggesting that the winner in the end has the majority of: (some voters' 1st choice votes) + (some of the other voters' 2nd choice votes)

Other candidates who don't win in IRV also have the same type of majority, only it uses a different mix of 1st and 2nd choice votes. There's no reason that isn't internal to the rules of IRV that one of these calculations of majority is the "right" one.

The reductio ad absurdem says that if everyone ranks 3 candidates, then every candidate has 100% vote, i.e. all candidates have majority support.

The only fully true version of majority support is a candidate preferred by the majority to all other candidates (i.e. the Condorcet principle). Whenever that isn't present, IRV's supposed "majority" is actually just a deceptive way to claim that a plurality (or even less-than-plurality) is a majority.

Are you considering blue the best choice because both green and red put them second? I see this as the most begrudging compromise choice for the majority of the people and one that can not be considered as equally begrudging for both red and green voters.

We have no way from the pure rankings to know the difference between a race where everyone totally loves both their 1st and 2nd choices and hates their 3rd choice and a race where everyone loves only their 1st choice and hates both their 2nd and 3rd choices (or even a race where voters love all the candidates or hate all the candidates). While it's fine to have a view that the way we select the winner should account for this (i.e. should or should not favor compromise candidates), there's nothing in IRV that accounts for these differences or identifies when something is a compromise.

Political issues are far more than one-dimensional too. It's fully possible that the Condorcet winner who is only ranked number one by only a plurality or less of the voters can be a compromise candidate in some cases and a candidate supported for reasons totally unrelated to the reasons people like or dislike the other candidates.

@endolith
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endolith commented Jan 4, 2018

@trevorhobson

Are you considering blue the best choice because both green and red put them second?

No, Blue is the best choice because Blue would beat both Red and Green in head-to-head elections. Blue is being eliminated by IRV prematurely, arbitrarily, just by the existence of other candidates in the race, which can be made even worse by adding more candidates:

image

More generally:

  • The population as a whole would be happier if Blue won than if Red or Green won
  • Blue's political position is the best match for the population as a whole, and they are being chosen to represent the entire population, not just the stronger half.

@trevorhobson
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trevorhobson commented Jan 4, 2018

@wolftune

Consider the collection of voters whose 1st choice does not win. Within that group, some of them have their 2nd choice counted and others have their 2nd choice completely ignored, never counted at all. All the problems with IRV fall out of this fact.

I think you are looking at it from the wrong perspective. One group gets their first choice counted and the others are forced to have their second choice counted. Getting your second choice counted is a begrudging compromise.

Are you saying that in this situation B should win?
A > B > C: 49%
B > A > C: 2%
B > C > A: 1%
C > B > A: 48%

We have no way from the pure rankings to know the difference between a race where everyone totally loves both their 1st and 2nd choices and hates their 3rd choice and a race where everyone loves only their 1st choice and hates both their 2nd and 3rd choices (or even a race where voters love all the candidates or hate all the candidates). While it's fine to have a view that the way we select the winner should account for this (i.e. should or should not favor compromise candidates), there's nothing in IRV that accounts for these differences or identifies when something is a compromise.

Nothing in IRV accounts for the level of compromise, but you can identify that there is a compromise by the fact that the voter did not put a candidate first.

@wolftune
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wolftune commented Jan 5, 2018

Getting your second choice counted is a begrudging compromise.

You have no basis to make that assertion. It only applies some uncertain fraction of the time. What if I like two candidates effectively equally and just have to choose one as the top? In today's system, I pick the one who seems most viable, but that's risky. The false promise of IRV is that it's okay to put them in either order, since (wrongly) the 2nd will count if the 1st doesn't win.

forced to have their second choice counted

Nobody is forced. If you don't want your second choice counted, you can leave the ballot blank besides your first choice.

Are you saying that in this situation B should win?

That is a change of topic. I have an opinion here, but I refuse to let the technical errors of IRV get sidetracked by a discussion of whether consensus candidates should or shouldn't win.

In your numeric example, after B is eliminated and A wins, all the C voters could say "what the hell?? What happened to our 2nd choice votes? They don't get counted? We HATE A. We care far more about stopping A than about electing C, screw C! C spoiled the election! Next time, we're going to put B first even though we actually like C better. Then the fact that we prefer B over A will actually get counted and we'll stop A. God, what a stupid system, we should never have changed to this dumb ranked choice thing!"

@trevorhobson
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trevorhobson commented Jan 5, 2018

@wolftune

You have no basis to make that assertion.

Living in a country with a IPV system I most certainly do from my own personal experience. Sometimes there are 2 or more candidates that I really like, but usually there is 1 and a whole stack of ones that I really don't.

It only applies some uncertain fraction of the time. What if I like two candidates effectively equally and just have to choose one as the top? In today's system, I pick the one who seems most viable, but that's risky.

One would assume that pretty much everyone will have some basis to make a decision on how to rank candidates in order of their preference if they like 2 or more the same.

The false promise of IRV is that it's okay to put them in either order, since (wrongly) the 2nd will count if the 1st doesn't win.

That is not the promise of IRV, IRV is Instant run-off. The promise of IRV is that if your 1st doesn't reach the last run-off then your 2nd, or 3rd etc will

Nobody is forced. If you don't want your second choice counted, you can leave the ballot blank besides your first choice.

That would be an informal vote where I live and my vote would not be counted, so yes I am forced to number all the boxes.

That is a change of topic.

My deepest apologies that was certainly not my intent, it would seem have have misunderstood your point.

What happened to our 2nd choice votes? They don't get counted?

Of course their 2nd choice doesn't get counted, their 1st choice did. You only get 1 vote.

@trevorhobson
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@endolith

No, Blue is the best choice because Blue would beat both Red and Green in head-to-head elections. Blue is being eliminated by IRV prematurely, arbitrarily, just by the existence of other candidates in the race

No there are other far more popular first choice candidates in this election and blue is the least popular candidate to reach the run-off. If green or red does not run next election and people don't change how they vote other than leaving off their first choice who is not running then blue would win.

The fact that blue is eliminated from the run-off, to me, makes IRV better than any other system where blue can reach the final run-off.

The population as a whole would be happier if Blue won than if Red or Green won

There is no way of knowing that as there is no ranking attached to the votes. If both red and green voters consider blue to be as equally bad as their last choice then blue is defiantly the worst choice.

Blue's political position is the best match for the population as a whole, and they are being chosen to represent the entire population, not just the stronger half.

This assumes that blues political position is the reason people put them second and not just to ensure that their vote is valid. You have no way of knowing this from the vote.

@wolftune
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wolftune commented Jan 5, 2018

@trevorhobson you are making a lot of assumptions that fall completely outside of the ballot system. They are all unrelated to evaluating the qualities of a ballot system. You talk about your own experience with how much you like or dislike candidates in the past, assumptions that voters have ways to come up with a preference order that accurately represents their wishes… these are all topics outside of evaluating how a ballot system works. They serve to muddy any discussion of the ballot system.

That is not the promise of IRV, IRV is Instant run-off.

You are right, my wording before was imprecise. Many IRV proponents make the false promise that if your 1st choice doesn't make it, you get to move to your 2nd choice. IRV itself doesn't make that promise. It is a separate, additional (and perhaps bigger) problem to sell IRV on false premises. Unfortunately, this false promise is close to universal in the way IRV proponents describe the system.

That would be an informal vote where I live and my vote would not be counted, so yes I am forced to number all the boxes.

Okay, but that's not inherent to IRV, and I think that's a bad approach. Out of curiosity, where is it that you live that uses IRV and requires complete rankings?

What happened to our 2nd choice votes? They don't get counted?
Of course their 2nd choice doesn't get counted, their 1st choice did. You only get 1 vote.

It is hard to read this as being argued in good faith (but I'll try). In IRV's process of determining the winner, some voters do get both their 1st and 2nd choices considered (counted). Other voters (particularly but not only those whose 1st choice loses in the final round) have only their 1st choice counted. "You only get 1 vote" — right if you voted 1st for one of the candidates in the final round. But if your 1st choice doesn't make the final round, you can get more than 1 vote.

Oh, you mean only 1 vote per round? And where you are, you can't leave any ranking blank? Okay. Well, so if my vote is A>B>C>D and C loses to D in the final round, then sure, my vote for A counted in round 1 where B gets eliminated. My vote for A then counts in round 2 where A is eliminated. What happened to my vote for B>C or for B>D, oops, that doesn't get counted. But I still get my 1 vote for C in the 3rd round. You could go ahead and say that we then eliminate C and there's a final round where only one candidate is left, and then my vote for D is counted. So, D wins with 100% of the vote!! (it's no change to the system to move every ballot to the final pile in the end versus stopping once a majority are in one pile).

Of course their 2nd choice doesn't get counted, their 1st choice did. You only get 1 vote.

The reason this looks like bad faith arguing is because it ignores the point. The point is unequal counting of ballots. Some voters get their 2nd choice counted and some don't, and that latter group includes people who do not get their 1st choice elected. "You only get 1 vote" completely ignores the point. Why should some voters get their 2nd choice counted (which could make their 2nd choice win) while other voters don't? There are cases in IRV where if you simply changed which voters got to have their 2nd choice count, it would change the outcome!

Thus, to strategically deal with this objectively bad balloting system, people who get neither their 1st or 2nd choice, can force their 2nd choice to get counted (and then to win in some cases!) by betraying their favorite and falsely down-ranking them. IRV creates a situation where favorite-betrayal gets you a better result. So, it incentivizes dishonest voting. The dishonesty is people trying to correct for a fundamentally unfair voting system.

In your response to @endolith you misunderstand the whole idea of the outcome representing the voters. The whole point is that you could simply look at the percentage of voters who prefer Blue over Red and notice that the majority prefer Blue over Red. We don't know how much happier they would be or any other details. What we know is that the majority of voters would prefer Blue over Red. The claim that IRV is respecting the majority here is plainly false. The majority is denied their wishes specifically because IRV ignores a portion of that majority, it simply throws away a chunk of votes that express the Blue>Red preference without ever counting them. It does that by eliminating Blue before some of the Blue>Red votes get counted.

This means that green is a spoiler candidate because if they dropped out of the race, Blue would win. Thus, just like with FPTP, voters could assure the stopping of their least favorite by betraying their favorite and instead voting for the compromise. Some Green voters switch to Blue>Green>Red so that Green is eliminated instead of Blue. That elects Blue and stops Red. In a Condorcet system, as just one alternative, this sort of thing cannot happen because the counting actually considers everyone's rankings equally. IRV does not treat all voters equally, so it is an objectively unfair ballot counting method.

@endolith
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endolith commented Jan 6, 2018

@trevorhobson

There is no way of knowing that

Yes, we know it because we know what all the simulated voters want. That's the point of the model. We know what's going on in their simulated heads and how it influences their votes. This model is 1-dimensional, but IRV performs just as badly in 2D or in reality (which is N-dimensional preferences).

This assumes that blues political position is the reason people put them second and not just to ensure that their vote is valid. You have no way of knowing this from the vote.

Same as above. We know this because it's what we're modeling. True, you can't derive real people's preferences from their votes, but you can derive simulated people's votes from their simulated preferences.

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