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Does ranked-choice voting guarantee a majority winner?

Does it matter if a candidate wins with a plurality of the vote or a majority? And if it does matter, why does it matter?

2013 Minneapolis sample ballot
myballotmn.sos.state.mn.us

Last week, in a post titled Ask Me Anything: Ranked Choice Voting, I invited readers to ask me questions about Ranked Choice Voting, the system of voting being used in Minneapolis and St. Paul this year. And if you haven’t guessed it yet, this post is where I start answering those questions.

Here we go:

With 35 candidates for mayor in Minneapolis, do you think that the winner will actually end up with a majority after three (the maximum) rounds?

This depends on what you mean by the word “majority.” If you mean a majority of all votes, then probably not, no. If you mean a majority of all the non-exhausted ballots, then the answer is certainly yes, because that’s the only way to win an RCV election. You have to break the 50% threshold.

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Before I get further, I should also point out that there is no maximum number of ballot counting rounds. They keep going until someone breaks 50%, and with 35 candidates it’s possible the number of ballot counting rounds could be in the thirties. When the StarTribune polled this race back in September I used its results to run a ranked choice voting simulation. The Strib only asked about eight candidates in the poll, with options for “other candidate” and “not sure,” as well, for a total of ten choices. The simulation took nine rounds for someone to get to >50%.

Embedded below is a spreadsheet of that simulation:

On the first round of counting everything is a muddle, with no one candidate breaking 16%. In fact, it takes until the fifth round for anyone to even hit 20% and the eighth round for anyone to hit 30%.

In a race with no clear front runner and eighty thousand candidates, this is the dynamic you can expect to play out. The candidates who get eliminated in the early rounds won’t have many votes to get redistributed, meaning the results won’t change much as they drop off. It’s not until candidates with a substantial number of first place votes start to get eliminated that the numbers really start to change. Meaning there will be a lot of rounds of vote counting in Minneapolis on election night. And the next day.

But now back to the point of this question. You can see on the above embedded spreadsheet that the winner, Don Samuels, did indeed cross the 50% threshold. What you will also see is that Don Samuels only has 324 ballots at the end, which is less than 50% of the 800 ballots the simulation started with. If you look at the bottom row you can see the total number of ballots counted in each round, and what you can see is that by the time we get to the ninth ballot, there are only 605 of the original 800 ballots still in play.

So what happened to the other 195 ballots? In ranked choice voting terms, they were exhausted. Which means that all of the candidates ranked on the ballot have been eliminated, and the ballot is no longer part of the count. You know that recycling bin on your computer where you put your less than satisfactory selfies? That’s essentially where these ballots go.

This means that the winner of a ranked choice voting election will receive >50% of the non-exhausted ballots, but not necessarily >50% of all ballots.

I would be surprised if the winner of the Minneapolis Mayoral race ends up with >50% of all ballots cast. The sheer number of candidates, combined with the fact that there are at least six serious candidates none of whom are likely to break 25% on the first round of ballot counting, but will nonetheless receive some minimal amount of support, pretty much guarantees that the winner will fall short of getting >50% of all ballots.

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To be fair, if you were to figure out Don Samuels winning percentage based on 324 ballots in the ninth round, and the original 800 ballots as the denominator, it comes out to 40%. 40% is less than 50%, but it’s a lot closer to 50% than the 16% that Samuels started the simulation out with. So there’s that.

If 100,000 people go to the poll and cast a ballot in the Mayor’s race, how many votes does the candidate need to get to win?

This is an interesting question. There is no absolute answer as it all depends on how the votes are distributed among the candidates. Clearly, if a candidate got 50,001 or more first place votes that would be it. But a candidate could also win with a lot fewer votes than that.

To try and find the least number of votes a candidate could conceivably win with, I will assume that all 35 candidates will receive a roughly equal number of votes, with the second and third place votes also distributed evenly among them. This is certainly not how the election will play out in practice, it’s merely an exercise in extremes.

With the above assumptions it would be possible for the winner of the election to end up with as little as around 10,500 votes or slightly more than 10% of the total votes cast. As I said though, this is not how things will actually play out as the vast majority of the 35 candidates will get very few first place votes to begin with.

A better way of answering this question would be to use the one source of possible vote distribution we have, the September StarTribune poll. As I discussed earlier, the person who won that simulation ended up with just over 40% of the overall ballots cast. So, for 100,000 ballots, that works out to about 40,000 votes.

RCV was sold, at least in part, on the theory that it would insure that the winning candidate would have a majority of the votes. Many candidates at all levels of politics in Minnesota and elsewhere have won elections with a plurality of the votes. Al Franken, Mark Dayton, Jesse Ventura and Tim Pawlenty come to mind.

If it doesn’t produce a majority winner, is RCV a failure for not producing a “legitimate” winner?

This is a philosophical question about which I don’t know if there is a correct answer.

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First, as I indicated before, this hinges on what is meant by majority winner. Since Ranked Choice Voting is not a classic election, is it fair to apply classical election metrics, like what constitutes a majority, to it? Within the framework of an RCV election, the word majority has a different meaning than it does within the framework of a classic election.

On the other hand, if the above seemed a bit tortured, it was. When you start talking about what a majority actually means you’re too deep in the weeds. That’s why I think it’s disingenuous for proponents of Ranked Choice Voting to wield “Upholds the principle of majority rule” as their number one talking point. It’s a dubious claim that is easily refuted with basic math. And when the number one talking point for a thing is factually incorrect, it makes me wonder about the rest of the talking points. And the thing itself.

Additionally, as pointed out in the question, Al Franken, Mark Dayton, Jesse Ventura and Tim Pawlenty all won with less than 50% of the vote under the classic system. That didn’t make them any less legitimate winners. Likewise, if the winner of an RCV election receives less then 50% of the overall votes, but >50% of the non-exhausted ballots, that person is no less a legitimate winner than someone who won on the first ballot.

It’s the rules of the election itself that defines legitimacy.

So, to answer the question, whether the possible, or in the case of the Minneapolis Mayors race likely, failure of RCV to produce a winner who received >50% of the overall votes counts as a failure of Ranked Choice Voting or not isn’t something that this post is going to deal with. That’s for you to decide on your own.

While you’re deciding though, you should also ask yourself this question; does it matter if a candidate wins with a plurality of the vote or a majority? And if it does matter, why does it matter?

This post was written by Tony Petrangelo and originally published on LeftMN. Follow Tony on Twitter: @TonyAngelo.

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