Yeah, that’s the question. Saying that |−P(X)⋅X|<ϵ means that P(X)<ϵX. So if X doubles, then it’s required that P is at least cut in half. I doubt there is a proof of this per se, but in a situation as strange as this it seems reasonable to me that if you claim you can do 10 times as much of something, then that is at least 10 times less likely.
I guess the main point I wanted to make is that in the usual phrasings of Pascal’s Mugging the choice of X is oftentimes taken after the choice of P. But P should be a function of X. So the mugger at least has to include this in his argument, and (some of) the burden of proof is on him.
But does the probability decrease fast enough?
Yeah, that’s the question. Saying that |−P(X)⋅X|<ϵ means that P(X)<ϵX. So if X doubles, then it’s required that P is at least cut in half. I doubt there is a proof of this per se, but in a situation as strange as this it seems reasonable to me that if you claim you can do 10 times as much of something, then that is at least 10 times less likely.
I guess the main point I wanted to make is that in the usual phrasings of Pascal’s Mugging the choice of X is oftentimes taken after the choice of P. But P should be a function of X. So the mugger at least has to include this in his argument, and (some of) the burden of proof is on him.