Electoral Reform Amendment/Statute

[Gabu]

Understood. This would bring us back where we started on this one, huh?

I suppose, though, let me check the logic of the situation that I gave; something about it doesn't seem quite right...

Well, tell me what you come up with.

Geez, the entire electoral system is going to be designed by three people... ;-P

Well, if anyone else wants to chime in, they're more than welcome to.

Okay, I've figured it out.  Suppose we have five votes for candidates A, B, and C, as follows:

Vote 1: A, B, C
Vote 2: A, B, C
Vote 3: A, C, B
Vote 4: B, C, A
Vote 5: C, A, B

This results in the following situation:

A: 3
B: 1
C: 1

with both B and C having 2 second- and third-preference votes, leading to an unbreakable tie.

Okay!  Good.  I just wanted to make sure that the situation I thought of actually can happen, and so it can.

Now we need to figure out what to do about it...

[WMS]

Understood. This would bring us back where we started on this one, huh?

I suppose, though, let me check the logic of the situation that I gave; something about it doesn't seem quite right...

Well, tell me what you come up with.

Geez, the entire electoral system is going to be designed by three people... ;-P

Well, if anyone else wants to chime in, they're more than welcome to.

Okay, I've figured it out.  Suppose we have five votes for candidates A, B, and C, as follows:

Vote 1: A, B, C
Vote 2: A, B, C
Vote 3: A, C, B
Vote 4: B, C, A
Vote 5: C, A, B

This results in the following situation:

A: 3
B: 1
C: 1

with both B and C having 2 second- and third-preference votes, leading to an unbreakable tie.

Okay!  Good.  I just wanted to make sure that the situation I thought of actually can happen, and so it can.

Now we need to figure out what to do about it...

I think they're letting us figure it out.

Hmm. I don't know enough about Condorcet to know if that would solve the last-place unbreakable tie issue...

[Gabu]

Actually, come to think of it, a new vote might work... if you look at the votes that I gave, 3 people preferenced B over C while only 2 people preferenced C over B, so even though counting every preference results in an unbreakable tie, it would appear to me that a new vote would make B win 3-2.

I need one more bit of information: is an unbreakable tie for last place possible with an even number of voters?  If it's not, then I think we have our solution.

[Gabu]

Offhand, I can't think of a scenario in which it could... if it did, I think it would definitely need more than 3 candidates.

I wonder if there's a mathematical way of figuring this out; there probably is, but I can't think of it at the moment.  Maybe I should sleep on it and see what I can come up with.

[WMS]

Actually, come to think of it, a new vote might work... if you look at the votes that I gave, 3 people preferenced B over C while only 2 people preferenced C over B, so even though counting every preference results in an unbreakable tie, it would appear to me that a new vote would make B win 3-2.

I need one more bit of information: is an unbreakable tie for last place possible with an even number of voters?  If it's not, then I think we have our solution.

Interesting! The SoFA had better be on top of his (or her, if we get a female SoFA) game to handle elections!

Not sure about that last part, although the number of voters in elections is so random it's hard to plan for.
Using your votes:
Vote 1: A, B, C
Vote 2: A, B, C
Vote 3: A, C, B
Vote 4: B, C, A
Vote 5: C, A, B
Vote 6: A, C, B

Err, is this an unbreakable tie?

Edited Part: If there's a mathematical way to determine this, I yield the floor to you.

[Gabu]

Using your votes:
Vote 1: A, B, C
Vote 2: A, B, C
Vote 3: A, C, B
Vote 4: B, C, A
Vote 5: C, A, B
Vote 6: A, C, B

Err, is this an unbreakable tie?

No, it isn't; C has 3 second-place votes while B only has 2, so B would be eliminated.

For it to be an unbreakable tie, every candidate in the tie must have equal numbers of every level of preferencing.

[Sam Spade]

Link to a sort of explanation of the Condorcet method.  I don't understand it that well either, frankly.

http://electionmethods.org/CondorcetEx.htm

[WMS]

Using your votes:
Vote 1: A, B, C
Vote 2: A, B, C
Vote 3: A, C, B
Vote 4: B, C, A
Vote 5: C, A, B
Vote 6: B, A, C
Vote 7: C, A, B

Err, is this an unbreakable tie?

No, it isn't; C has 3 second-place votes while B only has 2, so B would be eliminated.

For it to be an unbreakable tie, every candidate in the tie must have equal numbers of every level of preferencing.

Edited: Well, it can sure happen with an odd number of voters...

[Gabu]

Edited Part: If there's a mathematical way to determine this, I yield the floor to you.

There must be one, but I'm too tired at the moment to think mathematically.

Given that this isn't incredibly urgent, I think I'm gonna head to bed and see what I can come up with regarding this tomorrow.  If we can prove that it's impossible to have an unbreakable tie with a number of voters divisible by the number of candidates in the tie (for whatever reason, I feel like that assertion is true, but I don't know why), then we can go with a new vote to resolve an unbreakable tie.

This issue was a lot more complicated than I thought it would be when I started.

[WMS]

Link to a sort of explanation of the Condorcet method.  I don't understand it that well either, frankly.

http://electionmethods.org/CondorcetEx.htm

Thanks! I'll take a look at it.

[Gabu]

Using your votes:
Vote 1: A, B, C
Vote 2: A, B, C
Vote 3: A, C, B
Vote 4: B, C, A
Vote 5: C, A, B
Vote 6: B, A, C
Vote 7: C, A, B

Err, is this an unbreakable tie?

No, it isn't; C has 3 second-place votes while B only has 2, so B would be eliminated.

For it to be an unbreakable tie, every candidate in the tie must have equal numbers of every level of preferencing.

Edited: Well, it can sure happen with an odd number of voters...

Yes, but with an odd number of voters in a 2-candidate tie, it's impossible to have a tie between the two candidates in a new vote, so that wouldn't be a problem.

The only problem that would arise is when you have a number of voters divisible by the number of candidates in the unbreakable tie.  If it's not possible to have an unbreakable tie under those circumstances, as I said before, then a new vote will work.

[WMS]

Edited Part: If there's a mathematical way to determine this, I yield the floor to you.

There must be one, but I'm too tired at the moment to think mathematically.

Given that this isn't incredibly urgent, I think I'm gonna head to bed and see what I can come up with regarding this tomorrow.  If we can prove that it's impossible to have an unbreakable tie with a number of voters divisible by the number of candidates in the tie (for whatever reason, I feel like that assertion is true, but I don't know why), then we can go with a new vote to resolve an unbreakable tie.

This issue was a lot more complicated than I thought it would be when I started.

Sounds good. And yes, Lewis suuuuuuuuuure opened up Pandora's Box on this one.

[WMS]

Using your votes:
Vote 1: A, B, C
Vote 2: A, B, C
Vote 3: A, C, B
Vote 4: B, C, A
Vote 5: C, A, B
Vote 6: B, A, C
Vote 7: C, A, B

Err, is this an unbreakable tie?

No, it isn't; C has 3 second-place votes while B only has 2, so B would be eliminated.

For it to be an unbreakable tie, every candidate in the tie must have equal numbers of every level of preferencing.

Edited: Well, it can sure happen with an odd number of voters...

Yes, but with an odd number of voters in a 2-candidate vote, it's impossible to have a tie, so that wouldn't be a problem.

The only problem that would arise is when you have a number of voters divisible by the number of candidates in the unbreakable tie.

Ah! Recognition Dawns! That's what you meant! I feel smarter now. Good luck with this one...

[Gabu]

Using your votes:
Vote 1: A, B, C
Vote 2: A, B, C
Vote 3: A, C, B
Vote 4: B, C, A
Vote 5: C, A, B
Vote 6: B, A, C
Vote 7: C, A, B

Err, is this an unbreakable tie?

No, it isn't; C has 3 second-place votes while B only has 2, so B would be eliminated.

For it to be an unbreakable tie, every candidate in the tie must have equal numbers of every level of preferencing.

Edited: Well, it can sure happen with an odd number of voters...

Yes, but with an odd number of voters in a 2-candidate vote, it's impossible to have a tie, so that wouldn't be a problem.

The only problem that would arise is when you have a number of voters divisible by the number of candidates in the unbreakable tie.

Ah! Recognition Dawns! That's what you meant! I feel smarter now. Good luck with this one...

Thanks; I'll probably need it.

Good night, I'll post here tomorrow with what I've got at that time.

[WMS]

Using your votes:
Vote 1: A, B, C
Vote 2: A, B, C
Vote 3: A, C, B
Vote 4: B, C, A
Vote 5: C, A, B
Vote 6: B, A, C
Vote 7: C, A, B

Err, is this an unbreakable tie?

No, it isn't; C has 3 second-place votes while B only has 2, so B would be eliminated.

For it to be an unbreakable tie, every candidate in the tie must have equal numbers of every level of preferencing.

Edited: Well, it can sure happen with an odd number of voters...

Yes, but with an odd number of voters in a 2-candidate vote, it's impossible to have a tie, so that wouldn't be a problem.

The only problem that would arise is when you have a number of voters divisible by the number of candidates in the unbreakable tie.

Ah! Recognition Dawns! That's what you meant! I feel smarter now. Good luck with this one...

Thanks; I'll probably need it.

Good night, I'll post here tomorrow with what I've got at that time.

You're welcome, fellow Senator.

And I'm sleeping on this one too.

[minionofmidas]

Well, in your 9-3-3 or 3-1-1 scenarios, there is no need to break the tie since candidate A already has a majority of the vote.
Getting back to the 5-3-3 scenario that really happened, therefore (it was Bono btw, not Keystone - mea culpa) and amending it this way...
Hockeydude had a second preference.
One Bono voter did not exist.
Of the other four, two sec.pref Umengus, two Migrendel.
Leaves us with
ABC
ABC
ACB
ACB
BCA
BCA
BCA
CBA
CBA
CBA
and an unbreakable tie for last place, except by tossing a coin.
So it certainly still could happen.
Under Condorcet, you cast a ballot as under IRV. You're not required to list all candidates.
You then draw up a matrix in which all candidates are compared to each other.
in this case:
   A   B   C
A x   4   4
B 6   x   5
C 6   5   x
6 voters prefer B to A, 4 voters prefer A to B
6 voters prefer C to A, 4 voters prefer A to C
5 voters prefer B to C, 5 voters prefer C to B

There are three different possible results really:
1 (by far the most common, would have happened in our presidential elections): One candidate beats all other candidates in one-on-one matchups and is the winner.
2 : There is a circular chain, A beats B, B beats C, C beats A. In that case, the margin of these defeats is used as a chain-breaker.
3: Two (or more) candidates beat everybody else, and tie among themselves. That's what happens in the above example, and frankly I don't have a clue what's supposed to happen.
Here's what would have happened in the original example (B being Umengus):
   A   B   C
A x   5   5
B 5   x   3
C 6   3   x
A and B tie, B and C tie, C beats A. I honestly don't know what happens in this case actually. At first glance I guess it would also count as a Migrendel victory. (Hey, don't blame me. I'm writing this from memory.)

[Peter]

Once again, this draft amendment contains provisions that really ought to be statutory. The 25-post rule should not be in the Constitution as I can see us getting about a month down the line and then everybody suddenly thinking - Thats too much or thats not enough. This once again leads us on the labourious endeavour of amending the Constitution (which frankly I'm really tired of now).

The 25 post rule is fine as part of the statute as it simply defines what the word active in the constitution means and is subject to further revision by the Senate at a later date.

[minionofmidas]

The 25 post rule is fine as part of the statute as it simply defines what the word active in the constitution means and is subject to further revision by the Senate at a later date.

No, Pete. The constitution clearly defines as active "anybody who has participated in other threads and has not joined for the purposes of trolling." The 25 post rule would require a change of the constitution.
And I agree that 25 posts is much too high. Make that ten.

[Peter]

Ooops. Good to know I still make mistakes.

My point still stands that using the Constitution to define active is a bad idea. Allow the Senate to define what active is, don't use the Constitution.

[Gabu]

Once again, this draft amendment contains provisions that really ought to be statutory. The 25-post rule should not be in the Constitution as I can see us getting about a month down the line and then everybody suddenly thinking - Thats too much or thats not enough. This once again leads us on the labourious endeavour of amending the Constitution (which frankly I'm really tired of now).

The 25 post rule is fine as part of the statute as it simply defines what the word active in the constitution means and is subject to further revision by the Senate at a later date.

So, your suggestion is that we just remove the current Constitutional definition of "active" and then redefine it later in the statute?

Okay, I'll change that in the next version of this legislation.

[Gabu]

Well, in your 9-3-3 or 3-1-1 scenarios, there is no need to break the tie since candidate A already has a majority of the vote.

Oh.  Right.  Eh heh.

Getting back to the 5-3-3 scenario that really happened, therefore (it was Bono btw, not Keystone - mea culpa) and amending it this way...
Hockeydude had a second preference.
One Bono voter did not exist.
Of the other four, two sec.pref Umengus, two Migrendel.
Leaves us with
ABC
ABC
ACB
ACB
BCA
BCA
BCA
CBA
CBA
CBA
and an unbreakable tie for last place, except by tossing a coin.
So it certainly still could happen.
Under Condorcet, you cast a ballot as under IRV. You're not required to list all candidates.
You then draw up a matrix in which all candidates are compared to each other.
in this case:
   A   B   C
A x   4   4
B 6   x   5
C 6   5   x
6 voters prefer B to A, 4 voters prefer A to B
6 voters prefer C to A, 4 voters prefer A to C
5 voters prefer B to C, 5 voters prefer C to B

There are three different possible results really:
1 (by far the most common, would have happened in our presidential elections): One candidate beats all other candidates in one-on-one matchups and is the winner.
2 : There is a circular chain, A beats B, B beats C, C beats A. In that case, the margin of these defeats is used as a chain-breaker.
3: Two (or more) candidates beat everybody else, and tie among themselves. That's what happens in the above example, and frankly I don't have a clue what's supposed to happen.
Here's what would have happened in the original example (B being Umengus):
   A   B   C
A x   5   5
B 5   x   3
C 6   3   x
A and B tie, B and C tie, C beats A. I honestly don't know what happens in this case actually. At first glance I guess it would also count as a Migrendel victory. (Hey, don't blame me. I'm writing this from memory.)

I'm going to need to put aside more time than I've currently got to study what's going on in this Condorcet business, as I currently don't quite understand it.

[Peter]

So, your suggestion is that we just remove the current Constitutional definition of "active" and then redefine it later in the statute?

My suggestion would be to strike the present definition of active and then say that the Senate shall have power to define what an active voter is by appropriate legislation. Then get the Senate to do it by appropriate legislation.

It should be noted that as a scholar of this sort of stuff I have a bias towards a constitutional structure that leaves as much to the legislature as possible. I'm a great believer that the constitution provides the skeleton of the law and the legislature provides the flesh of it. There are certainly viewpoints out there that disagree with my methods, so don't necessarily defer to me all the time (but certainly don't let me stop you!)

[Gabu]

So, your suggestion is that we just remove the current Constitutional definition of "active" and then redefine it later in the statute?

My suggestion would be to strike the present definition of active and then say that the Senate shall have power to define what an active voter is by appropriate legislation. Then get the Senate to do it by appropriate legislation.

It should be noted that as a scholar of this sort of stuff I have a bias towards a constitutional structure that leaves as much to the legislature as possible. I'm a great believer that the constitution provides the skeleton of the law and the legislature provides the flesh of it. There are certainly viewpoints out there that disagree with my methods, so don't necessarily defer to me all the time (but certainly don't let me stop you!)

It seems logical to me; if we screwed something up that desperately needs to be changed, it would be a serious pain and time-waster if we needed to go through the process of amending the whole Constitution yet again.

[Gabu]

Well, I've examined the mechanisms behind Condorcet, and as far as I can tell, it unfortunately won't resolve unbreakable ties, as far as I can tell:

Getting back to the 5-3-3 scenario that really happened, therefore (it was Bono btw, not Keystone - mea culpa) and amending it this way...
Hockeydude had a second preference.
One Bono voter did not exist.
Of the other four, two sec.pref Umengus, two Migrendel.
Leaves us with
ABC
ABC
ACB
ACB
BCA
BCA
BCA
CBA
CBA
CBA
and an unbreakable tie for last place, except by tossing a coin.
So it certainly still could happen.
Under Condorcet, you cast a ballot as under IRV. You're not required to list all candidates.
You then draw up a matrix in which all candidates are compared to each other.
in this case:
   A   B   C
A x   4   4
B 6   x   5
C 6   5   x
6 voters prefer B to A, 4 voters prefer A to B
6 voters prefer C to A, 4 voters prefer A to C
5 voters prefer B to C, 5 voters prefer C to B

(snip)

3: Two (or more) candidates beat everybody else, and tie among themselves. That's what happens in the above example, and frankly I don't have a clue what's supposed to happen.

From what I can tell, an unbreakable tie in IRV will result in an unbreakable tie in Condorcet, as well.  If all candidates in the tie have exactly the same amounts of first-, second-, third-, etc. preferences, then each candidate is prefered over the other candidates the exact same amount.

Unless someone can prove me wrong on this, I think it's back to the examination of the possibility of a new vote to break an unbreakable tie.

PS: JFK, is there any way to have this get postponed for debating time?  I think we're gonna need longer than what we'll get if it takes its place in the queue.

[Gabu]

Well, damn.  It isn't true that it's not possible to have an unbreakable tie with an even number of voters.  Here's a scenario where that happens:

ABC
ABC
ACB
ACB
BCA
BCA
BCA
CBA
CBA
CBA

First-preference results:

A: 4
B: 3
C: 3

Second-preference results:

A: 0
B: 5
C: 5

Third-preference results:

A: 6
B: 2
C: 2

If put to a vote between B and C, the final results would be 5-5 because 5 voters placed B above C and 5 voters placed C above 5, and we're at yet another unbreakable tie.  This would still be an unbreakable tie using Condorcet, because B would beat C 5 times and C would beat B 5 times.

So, neither a new vote between the tying candidates nor Condorcet will always work to resolve an unbreakable tie.

Back to the drawing board...

EDIT: I should add that there are, in fact, an infinite number of examples of votes that will create this.  Let's number the vote types:

1: ABC
2: ACB
3: BAC
4: BCA
5: CAB
6: CBA

If we let n be the total number of voters and v_k be the number of votes of type k cast, then my mathematical analysis of this situation says that we will have an unbreakable tie if and only if

v₃ + v₄ = v₅ +  v₆
v₁ + v₆ = v₂ +  v₄
v₃ + v₄ < n/3
v₅ +  v₆ < n/3
v₁ + v₂ < n/2

If you find a set of votes that satisfies those conditions, they will provide an unbreakable tie.