Do you intend “‘Surprise implies Mon’ is false” to mean “Not (Surprise and Mon)”? I’m just a little confused because I think in classical logic, if Surprise is false then all the implication statements are true. Therefore if at least one of Mon-Fri is true, Surprise cannot be false. Maybe I should have read the paper you reference.
Personally, I think the original presentation is already sufficient. If the judge really refuses to hang the prisoner if they’re not surprised, and the judge is in some sense transparent to the prisoner, so that the prisoner knows what logic they will follow, then the judge will not hang the prisoner.
The paradox is in the punchline that if the judge then hangs the prisoner on wednesday, the prisoner is surprised. But this is simply because it’s a counterfactual that won’t happen, given the toy model of how the judge works. If the prisoner merely has a bad model of the judge and therefore makes wrong predictions, this isn’t much of a paradox, though still a decent joke.
I think this problem is clearer if we imagine a game where the judge outputs a probability distribution over which day the prisoner will be executed, and the prisoner is postulated to be clever enough to figure out what probability distribution the judge will output.
In this scenario, it is very easy for the judge to surprise the prisoner most of the time—a uniform distribution will suffice. If the judge’s goal is to minimize the expected information the prisoner has about whether they’ll die, on the day of the execution, then maybe you get some slightly more complicated probability distribution. But within this game, it is impossible for the judge to output a probability distribution such that the prisoner is always surprised, never unsurprised. Every probability distribution supported on Mon-Fri has a final day, on which the prisoner is not surprised.
So if the judge says they’re going to output a probability distribution such that the prisoner will definitely be surprised when they die, they are telling a lie.
Do you intend “‘Surprise implies Mon’ is false” to mean “Not (Surprise and Mon)”? I’m just a little confused because I think in classical logic, if Surprise is false then all the implication statements are true. Therefore if at least one of Mon-Fri is true, Surprise cannot be false. Maybe I should have read the paper you reference.
Personally, I think the original presentation is already sufficient. If the judge really refuses to hang the prisoner if they’re not surprised, and the judge is in some sense transparent to the prisoner, so that the prisoner knows what logic they will follow, then the judge will not hang the prisoner.
The paradox is in the punchline that if the judge then hangs the prisoner on wednesday, the prisoner is surprised. But this is simply because it’s a counterfactual that won’t happen, given the toy model of how the judge works. If the prisoner merely has a bad model of the judge and therefore makes wrong predictions, this isn’t much of a paradox, though still a decent joke.
I think this problem is clearer if we imagine a game where the judge outputs a probability distribution over which day the prisoner will be executed, and the prisoner is postulated to be clever enough to figure out what probability distribution the judge will output.
In this scenario, it is very easy for the judge to surprise the prisoner most of the time—a uniform distribution will suffice. If the judge’s goal is to minimize the expected information the prisoner has about whether they’ll die, on the day of the execution, then maybe you get some slightly more complicated probability distribution. But within this game, it is impossible for the judge to output a probability distribution such that the prisoner is always surprised, never unsurprised. Every probability distribution supported on Mon-Fri has a final day, on which the prisoner is not surprised.
So if the judge says they’re going to output a probability distribution such that the prisoner will definitely be surprised when they die, they are telling a lie.