The existence of an irrelevant or non sequitur choice does not render the effective binary choice (A or B) false.
For instance, if the question is who will win the boxing match between Mike Tyson and Oleksandr Usyk?
Then you can indeed choose Taylor Swift as your response. But given that Taylor Swift is not even a rational choice then the choice is moot and anyone betting on her to even be present for the Tyson-Usyk match is stupid.
@MeganC Couple of things here that is wrong with your logic. First off, Taylor Swift isn't an option in a fight between two people, neither of which is Taylor Swift. Second, even this isn't a binary choice because of the way you worded it. Your question is who would win. Tyson and Usyk could end up in a draw, in which case neither wins. So there are three outcomes:
- Tyson wins
- Usyk wins
- Neither wins
Not binary. One can choose Tyson to win, Usyk to win, or neither to win. Taylor Swift is not an option because only two persons are allowed in the fight, and she is not one of them.
This scenario differs significantly from the US Presidential election where. There are more than two persons in the fight. And there is no possibility of a draw. Here are the possible outcomes:
- The Democrat Party nominee wins
- The Republican Party nominee wins
- The Green Party nominee wins
- The Libertarian Party nominee wins
- The Constitution Party nominee wins
- Independent candidate #1 wins
- Independent candidate #2 wins
- etc.
Again, not a binary choice. And by definition, the existence of a third choice does indeed render void a true-false binary choice. 'A or B' is a binary choice. 'A or B or C' is not, regardless of how unlikely the outcome of C as long as 'C' is possible.