Ranked Choice Voting Could Determine Maine Winner
The rules for counting votes matter.
Whether Susan Collins is re-elected—and, potentially, whether Democrats regain control of the United States Senate—could be impacted by the state’s relatively novel method of counting votes.
NPR (“Ranked-Choice Voting Could Play A Deciding Role In Maine’s Senate Race“):
One of this November’s closest and highest-profile U.S. Senate races could turn on a unique way of voting in Maine.
There, Republican incumbent Susan Collins is defending her seat against Democratic state House Speaker Sara Gideon as well as two independents. Polls show a close contest between Collins, who’s seeking her fifth term, and the well-funded Gideon.
And Maine voters will use the ranked-choice voting system, which allows them to rank their choices among the four Senate candidates.
In this year’s U.S. Senate race, one candidate, independent Lisa Savage, has used ranked-choice voting in her messaging, telling voters to make her their first choice and the Democrat Gideon their second one.
University of Maine political science professor Mark Brewer says that’s problematic for Collins.
“In a ranked-choice voting scheme, assuming that Savage’s supporters list Gideon second, which Savage is asking them to do, this now turns into a disadvantage for Collins,” he says.
But it’s unclear how the votes for the two independent candidates will break, with recent polls showing each with roughly the same share of the vote.
In principle, I support the idea of ranked-choice or instant-runoff voting. It simply makes sense to elect officeholders who have the support of a majority of the constituency.
In practice, I’m less sure. In most ranked-choice systems, and I gather in Maine’s case, the way it works is like this: If any candidate gets a majority of 1st-choice votes, they’re the winner. If not, the bottom finisher is eliminated from the contest and 2nd choice of that candidate’s voters now get a vote. This continues until someone has a majority.
In a race where there are two major-party candidates and one significant third party, that makes sense. So, for example, if there is a Green Party candidate who takes 8 percent of the vote and almost all of those who voted for said candidate preferred the Democrat to the Republican, those voters are allowed to both express their true preference and still have their vote “count” in deciding between the two major-party candidates.
But what about this situation? Here, we have four candidates. Let’s say that the fourth-place candidate’s voters preferred Collins to Gideon and that said candidate’s votes were sufficient to put Collins over the top? But what if the third-place candidate, who by definition received more votes than the fourth-place candidate, was preferred by Gideon second-choicers? Yet, even though the point of ranked-choice voting is to eliminate the “wasted vote,” that’s exactly what happens here. Even though there may well have been enough of their votes to give Gideon the win, they don’t matter under this system. Obviously, this would be compounded further in a race with five, six, or seven candidates who drew significant votes. Wouldn’t it make more sense to eliminate all of the non-finalists (that is, everyone other than the top two finishers) and redistribute all of their second-choice votes? Of course, that assumes most people’s second choice would be one of the major-party candidates. If that’s not the case, then multiple elimination rounds could also have perverse effects.
UPDATE: To quote the late, great Roseanne Roseannadanna, “Never mind.” As just about every commenter figured out, it is mathematically impossible for this method to push a candidate over the 50 percent threshold and there be enough votes remaining for another candidate to get more votes. I would have figured this out myself had I drawn the table and done the numbers.