That’s because the statement “the mugger will deliver on any promise he makes” carries with it an implied probability distribution over possible promises.
Agreed, but that’s not the whole picture. Let’s break this down a slightly different way: we know that p(mugger has magic) is very small number, and as you point out p(mugger will deliver on any promise) is a distribution, not a number. But we aren’t just dealing with p(mugger will deliver on any promise), we are dealing with the conditional probability of p(mugger will deliver on any promise|mugger has magic) times p(mugger has magic). Though this might be a distribution based on what exactly the mugger is promising, it is still different from p(mugger will deliver on any promise), and it might still allow for a Pascal’s Mugging.
This is why the card trick example doesn’t work: p(mugger performs card trick) is indeed very high, but what we are really dealing with is p(mugger performs card trick|mugger has magic) times p(mugger has magic), so our probability that he does a card trick using actual magic would be extremely low.
Agreed, but that’s not the whole picture. Let’s break this down a slightly different way: we know that p(mugger has magic) is very small number, and as you point out p(mugger will deliver on any promise) is a distribution, not a number. But we aren’t just dealing with p(mugger will deliver on any promise), we are dealing with the conditional probability of p(mugger will deliver on any promise|mugger has magic) times p(mugger has magic). Though this might be a distribution based on what exactly the mugger is promising, it is still different from p(mugger will deliver on any promise), and it might still allow for a Pascal’s Mugging.
This is why the card trick example doesn’t work: p(mugger performs card trick) is indeed very high, but what we are really dealing with is p(mugger performs card trick|mugger has magic) times p(mugger has magic), so our probability that he does a card trick using actual magic would be extremely low.