Talk:Instant-runoff voting

Prior Republican/Libertarian example

There was an example that I decided to replace:

Suppose my true preference is for the Libertarian first and the Republican second. Suppose further that the Libertarians are the strongest "minor" party. At some round of the IRV counting process, all the candidates will be eliminated except the Republican, the Democrat, and the Libertarian. If the Libertarian then has the fewest first-choice votes, he or she will be eliminated and my vote will transfer to the Republican, just as I wanted. But what if the Republican is eliminated before the Libertarian? Unless all the Republican votes transfer to the Libertarian, which is extremely unlikely, the Democrat might then beat the Libertarian. If so, I will have helped the Democrat win by not strategically ranking the Republican first.

However, this thinking is flawed. The problem in this case is that the Republicans are not the major party, but the third party. They come in third, that makes them a third party. And they fail to vote second for their Libertarian second choice, so their second choice does not win. This is how it is supposed to work. When third-party voters don't vote for their second choice, the second choice might not win. The proposed solution is that the major-party voters should vote third-party first.

Try this reasoning when Republicans are the major party. They think that Libertarians will not vote Republican second, and there are enough Libertarians to keep the Republicans from winning. So -- even when they have 45% of the vote and Libertarians have 6%, Republicans might agree to vote Libertarian because they would prefer Libertarians to Democrats. Does that sound in any way plausible? So why would major-party Libertarians vote third-party on the assumption that the third-party voters won't vote second for their candidate? Because they have no self-respect.

I was going to try to keep using the Democratic/Republican/Libertarian example given, and explain it better, but it seemed that the party labels were making it hard to express the idea. I thought Andy Jennings' example that I linked to in the article (referred to in Aaron Hamlin's "The Limits of Ranked Choice Voting" essay from February 2019) was a good explanation of favorite betrayal. -- RobLa (talk) 05:58, 16 December 2019 (UTC)

Does anyone call it IRV any more?

At present, what groups or nations call this voting method "Instant Runoff"? In the US, advocates of this method have seized the generic term Ranked Choice Voting (RCV) as their preferred term. Should IRV still be considered the preferred term? - Frankie1969 (talk) 22:24, 27 September 2021 (UTC)

IRV is still the term which everybody uses. RCV is used by one organization in one country. Also, "Rank Choice" refers to the ballot not the system. Nobody in the voting theory community would use RCV in an academic paper. --Dr. Edmonds (talk) 02:19, 28 September 2021 (UTC)

My point is that AFAICT, none of the English-speaking nations that actually use IRV call it IRV. (update) Australia, NZ, Papua call it Preferential Voting. US & Canada call it RCV. UK calls it Alternative Vote. Ireland & India call it STV. So it can't possibly be correct to say it's "the term which everybody uses". - Frankie1969 (talk) 13:21, 28 September 2021 (UTC)

It appears to be called "instant-runoff voting" on English Wikipedia (see w:Instant-runoff voting). I'm not inclined to go against the consensus over there. -- RobLa (talk) 20:56, 28 September 2021 (UTC)

Fixing the shortcomings of IRV

I would like to fix the shortcomings of IRV. For discussion, I offer an eleven step instant runoff process to do that. Some of the examples on this website are reproduced here to illustrate how an IRV process can do better.

Compromise: regarding the Memphis example.

IRV elects Knoxville, some compromises are suggested to avoid electing Knoxville. Nashville will be elected when the 42 Memphis>Nashville>Chattanooga>Knoxville voters flip their votes to Nashville>Memphis>Chattanooga>Knoxville. Alternatively it is suggested that the 26 Nashville>Chattanooga>Knoxville>Memphis voters flip their votes to Chattanooga>Nashville>Knoxville>Memphis to elect Chattanooga, again avoiding the election of Knoxville.

I have Nashville elected in step 8. No need for anyone to compromise.

IRV can fail to pick a good compromise.

38 A>C>B

38 B>C>A

11 C>A>B

13 C>B>A

IRV elects B. I would elect C in step 8.

IRV fails to count the ballots in a way most favorable to the voters.

26 A>B

25 C>B

49 D

IRV elects D. I see a three way tie, and I will elect one of A, B or D in step 11.

IRV is vulnerable to center-squeeze (“the three candidate” example from center-squeeze).

1031 A>B>C

415 B>A>C

446 B>C>A

1108 C>B>A

IRV elects C. I would elect B in step 8. IRV needs to be less vulnerable to center-squeeze, the center candidate, B in this example, should win.

IRV is vulnerable to favorite betrayal: from the favorite betrayal criterion.

10 A>B>C>D

6 B>A>C>D

5 C>B>A>D

20 D>A>C>B

IRV elects B. I would elect A in step 8. IRV needs to be less vulnerable to favorite betrayal, voters should not need to change their votes.

IRV gives supporters of A an opportunity to win if they betray their favorite.

2 C>A>B>D

8 A>B>C>D

6 B>A>C>D

5 C>B>A>D

20 D>A>C>B

IRV elects A. I would still elect A, this time in step 5. Supporters of A should not have to be this sneaky.

Please consider using an eleven step instant runoff process to make IRV more fair. I have used Google Sheets to create a demonstrator. I'm old and not an expert in anything. But I'm pretty sure it works. I will be very embarrassed if I've made typos in the formulae and the results don't make sense. Point it out and I'll fix it. Here's a link to my spreadsheet:

https://docs.google.com/spreadsheets/d/1D1Aeoy3Y17gcnCyVx6AlGEIYneDidRpqn5a-lClC66I/edit#gid=664199959

--RalphInOttawa (talk) 05:00, 2 December 2023 (UTC)

A simpler way to improve IRV is already explained at Ranked Choice Including Pairwise Elimination. --VoteFair (talk) 23:46, 4 December 2023 (UTC)

I'm happy that Nashville wins. When given the Condorcet paradox, please confirm that RCIPE elects A.

4 A>B

3 B>C

2 C>A

I suspect RCIPE eliminates C as having the smallest top-choice count. B becomes the runner-up, and ideally would move to a two candidate runoff A vs B. Then assuming honest opinions have been cast and remain unchanged, A wins.

This lends itself to strategic campaigning where the leading candidate ensures the least preferred candidate finishes second. That’s IRV’s problem and RCIPE campaigns will do the same thing. Then it’s favorite betrayal time, where the supporters of B>C will have to decide if they need to change to C>B. That’s unfair to the B and C voters and very convenient for candidate A. RalphInOttawa (talk) 01:24, 5 December 2023 (UTC)

Every vote-counting method has specific cases where the method elects the "wrong" winner. What's more important is how often these specific "unfair" results occur.

I don't know what your method does because it's not available on a non-Google URL, but 11 steps is way too complex for voters to understand. And it certainly fails the IIA criterion sometimes because all method fail it.

Here's a plot that shows some calculations for how often RCIPE and other methods fail the IIA (independence of irrelevant alternatives) criterion, along with clone independence failures which are a special kind of IIA failure:

http://www.votefair.org/clone_iia_success_rates.png

--VoteFair (talk) 22:54, 5 December 2023 (UTC)

Regarding IIA, I think IRV can fail less.

Here’s a very strong example of IIA failing in IRV. I got this from somewhere on Wikipedia.

6 C > B > D > A

5 D > A > C > B

5 A > C > B > D

4 B > D > A > C

One-choice elects C. IRV elects C, Borda (4-3-2-1) elects C, Copeland elects C. I would elect A in step 8 or step 9.

Taking D out:

 6 C > B > A

10 A > C > B

 4 B > A > C

One-choice elects A, IRV elects A, Borda (3-2-1) elects A, Copeland elects A. I would elect A in step 5 or step 6.

In looking for the above example, I came across this one from the University of Nebraska where One-choice, Borda and IRV fail.

https://mathbooks.unl.edu/Contemporary/sec-5-5-arrow.html

6  A>B>D>C

5  D>B>C>A

4  C>D>B>A

2  B>C>D>A

IRV elects C. I will elect D in step 8

Taking A out:

6  B>D>C

5  D>B>C

4  C>D>B

2  B>C>D

IRV elects D. I will elect D in step 9.

I know my eleven steps can fail to satisfy IIA too. I spent a lot of time on the following example which shows the Condorcet winner being pushed out of the elevator on the third floor of a six storey building. I can’t stop that. I can’t fix that. If only the Condorcet winner had stayed on the elevator a little longer, or taken the stairs, it would have found a way to reach the top floor.

6 A > F > D

3 B >E > D

1 B > F

4 C > D > E

3 D > E

2 E > F > B

1 F > B

One-choice elects A. IRV elects B. The Condorcet winner is D. My eleven steps elect E in step 10. Take out C and I would elect D in step 5.

RalphInOttawa (talk) 01:44, 8 December 2023 (UTC)

Thanks for sending the PDFs via email. I looked at them and see you are attempting to do the same thing as RCIPE, which is to look deeper into the preferences. Also you are attempting to avoid IIA failures by identifying which candidates are irrelevant. The RCIPE method achieves both of those goals using the same "deeper" elimination process (by eliminating pairwise losing candidates). In contrast, your 11 step process is too complex. It seems to do three different calculation methods and chooses the "best" winner of the three. That same approach can be done by using any three methods. I'm not saying your method is bad. I'm saying there are simpler ways to reach the same goals. I commend you for spending time digging deep into the math behind fairer vote-counting methods, which are long overdue for adoption into real elections. --VoteFair (talk) 01:48, 9 December 2023 (UTC)

Variants: ways of dealing with equal rankings (see STV page)

For discussion, I offer this alternative. The following example (amending the example given) clearly shows how my improvement of IRV deals with equal rankings using three orders of Random Voter Hierarchy (RVH). The result changes as C reaches the border of a majority, going from a three way tie to a win by C.

15 A>C>B

30 A=C>B

35 B>A>C

20 C>B>A

With C claiming the better rank in the 1st order, C wins. With A claiming the better rank in the 1st order, it is a three way tie. When one or more votes honestly change from A>C to A=C, C wins regardless of rank within the 1st order. Alternatively, when one or more votes honestly change from A=C to A>C, it’s a three way tie regardless of rank within the 1st order (with the winner decided using a 3rd order of RVH).

In the other example given on “Ways to deal with equal ranking”

34 A=B=C

33 D

33 E

The winner is whichever of A, B or C claims the best ranking in the 1st order of RVH.

RalphInOttawa (talk) 12:27, 29 December 2023 (UTC)