Until today, I always thought the right solution was given by Fitch in “A Goedelized Formulation of the Prediction Paradox”. Let’s define a sentence in Peano arithmetic called Surprise, which will refer to five other sentences Mon, Tue, Wed, Thu, Fri. Surprise will be defined recursively using the diagonal lemma, as a conjunction of these sentences:
1) Exactly one of Mon, Tue, Wed, Thu, Fri is true.
2) If Mon is true then “Surprise implies Mon” isn’t provable.
3) If Tue is true then “Surprise implies {Mon or Tue}” isn’t provable.
4) If Wed is true then “Surprise implies {Mon or Tue or Wed}” isn’t provable.
5) If Thu is true then “Surprise implies {Mon or Tue or Wed or Thu}” isn’t provable.
6) If Fri is true then “Surprise implies {Mon or Tue or Wed or Thu or Fri}” isn’t provable.
All self-references are legal because they occur inside “provable” quotes. Now it’s easy to prove that Surprise is false, no matter what Mon, Tue, Wed, Thu and Fri say. So the teacher is lying and that seems to be the end of the paradox.
But it just occurred to me that there’s a much simpler solution that doesn’t require Gödel encoding. Let’s define a sentence Surprise in naive logic with self-reference, as a conjunction of these:
1) Exactly one of Mon, Tue, Wed, Thu, Fri is true.
2) If Mon is true then “Surprise implies Mon” is false.
3) If Tue is true then “Surprise implies {Mon or Tue}” is false.
4) If Wed is true then “Surprise implies {Mon or Tue or Wed}” is false.
5) If Thu is true then “Surprise implies {Mon or Tue or Wed or Thu}” is false.
6) If Fri is true then “Surprise implies {Mon or Tue or Wed or Thu or Fri}” is false.
Now it’s even easier to prove that Surprise can’t be true (it’s either false or indeterminate). The proof is similar to Fitch’s reasoning, but without the complicated machinery. It seems to me that it resolves the paradox just as effectively, no?
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.
I’ve been thinking about the unexpected hanging paradox again.
Until today, I always thought the right solution was given by Fitch in “A Goedelized Formulation of the Prediction Paradox”. Let’s define a sentence in Peano arithmetic called Surprise, which will refer to five other sentences Mon, Tue, Wed, Thu, Fri. Surprise will be defined recursively using the diagonal lemma, as a conjunction of these sentences:
1) Exactly one of Mon, Tue, Wed, Thu, Fri is true.
2) If Mon is true then “Surprise implies Mon” isn’t provable.
3) If Tue is true then “Surprise implies {Mon or Tue}” isn’t provable.
4) If Wed is true then “Surprise implies {Mon or Tue or Wed}” isn’t provable.
5) If Thu is true then “Surprise implies {Mon or Tue or Wed or Thu}” isn’t provable.
6) If Fri is true then “Surprise implies {Mon or Tue or Wed or Thu or Fri}” isn’t provable.
All self-references are legal because they occur inside “provable” quotes. Now it’s easy to prove that Surprise is false, no matter what Mon, Tue, Wed, Thu and Fri say. So the teacher is lying and that seems to be the end of the paradox.
But it just occurred to me that there’s a much simpler solution that doesn’t require Gödel encoding. Let’s define a sentence Surprise in naive logic with self-reference, as a conjunction of these:
1) Exactly one of Mon, Tue, Wed, Thu, Fri is true.
2) If Mon is true then “Surprise implies Mon” is false.
3) If Tue is true then “Surprise implies {Mon or Tue}” is false.
4) If Wed is true then “Surprise implies {Mon or Tue or Wed}” is false.
5) If Thu is true then “Surprise implies {Mon or Tue or Wed or Thu}” is false.
6) If Fri is true then “Surprise implies {Mon or Tue or Wed or Thu or Fri}” is false.
Now it’s even easier to prove that Surprise can’t be true (it’s either false or indeterminate). The proof is similar to Fitch’s reasoning, but without the complicated machinery. It seems to me that it resolves the paradox just as effectively, no?
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.
It’s a finite probability (greater than zero) for Monday, that it’s a hanging day. So, the hangman will not be unexpected on Monday.
It’s a finite probability (greater than zero) for Tuesday, that it’s a hanging day. So, the hangman will not be unexpected on Tuesday.
....
It’s a high probability for Friday, that it’s a hanging day. So, the hangman will not be unexpected on Friday.
The “unexpected day” was a judge’s small lie. Much like when the judge says “you will be hanged alive until you die next week twice”.
He is just joking.
No real paradox here.
That is my position, although I consider it being reasonable rather than stubbornly insisting that nothing can be deep and mysterious.
I think liar type paradoxes are exactly as mysterious as this Python function:
If your intuition is like a Turing complete language, you can’t insist that all functions must terminate.
This function just never returns. That’s not what the Liar’s Paradox is about.