STAR voting is known to fail the favorite betrayal criterion, but there is disagreement regarding which situations incentivize favorite betrayal. In A Farewell to Pass/Fail: Why We Ditched Later No Harm, Emily Dempsey says:
Specifically, STAR Voting only incentivizes Favorite Betrayal when there is a Condorcet Cycle. Not much is known about how likely Condorcet Cycles are in real-world elections, but they are certainly less likely (and, I would wager, much more difficult for a voter to predict) than other problematic phenomena such as Vote Splitting and Center Squeeze Scenarios, in which many other voting systems incentivize Favorite Betrayal and STAR Voting does not.
On the other hand, Wikipedia states the following:
In STAR voting, in order for favorite betrayal to be strategically advantageous, four separate things must be true: the favorite candidate X must be in the runoff under an honest vote, X must lose the runoff under an honest vote, the betrayal beneficiary Y must not be in the runoff under an honest vote, and the Y must win the runoff under a strategic vote.
It is fairly straightforward to show that all four of these criteria are indeed necessary:
On the other hand, it is possible to devise a counterexample to the claim that a Condorcet cycle is necessary:
Number of Voters | Candidate A | Candidate B | Candidate C |
---|---|---|---|
1 | 5 | 1 | 0 |
1 | 5 | 0 | 2 |
2 | 4 | 0 | 5 |
6 | 3 | 5 | 4 |
In this election, Candidate A receives 36 points, Candidate B receives 31 points, and Candidate C receives 36 points, so A and C are the finalists. A is favored over C on 2 ballots while C is favored over A on 8 ballots, so C is the winner. However, the voter in the first row can change this outcome by betraying their favorite candidate:
Number of Voters | Candidate A | Candidate B | Candidate C |
---|---|---|---|
1 | 1 | 5 | 0 |
1 | 5 | 0 | 2 |
2 | 4 | 0 | 5 |
6 | 3 | 5 | 4 |
Now Candidate A receives 32 points, Candidate B receives 35 points, and Candidate C receives 36 points, so B and C are the finalists. B is favored over C on 7 ballots while C is favored over B on 3 ballots, so B wins the election. Notice that this voter cannot achieve this result by simply rating A and B the same; A will still have more points than B, preventing B from reaching the runoff.
Importantly, this example lacks a Condorcet cycle. Instead, B is the Condorcet winner, and will always win in the runoff. Thus, this disproves the claim that favorite betrayal is only incentivized under STAR voting when a Condorcet cycle exists.