What you present is the basic fallacy of Pascal’s Mugging: treating the probability of B and of C as independent the fact that a threat of given magnitude is made.
The prior probability of X is 2^-(K-complexity of X). There are more possible universes where they carry out smaller threats, so the K-complexity is lower. What I showed is that, even if there were only a single possible universe where the threat was carried out, it’s still simple enough that the K-complexity is small enough that it’s worth paying the threatener.
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You gave a vague argument. Rather than giving a vague counterargument along the same lines, I just ran the math directly. You can argue that P(C|E) decreases all you want, but since I found that the actual value is still too high, it clearly doesn’t decrease fast enough.
If you want the vague counterargument, it’s simple: The probability that it’s a lie approaches unity. It just doesn’t approach it fast enough. It’s a heck of a lot less likely that someone who threatens 3^^^3 lives is telling the truth than someone who’s threatening one. It’s just not 3^^^3 times less likely.
If you mean what I think you mean, I’m ignoring it because I’m going with worst case. Rather than tracking how the probability of someone making the threat reduces slower than the probability of them carrying it out (which means a lower probability of them carrying it out), I’m showing that even if we assume that the probability is one, it’s not enough to discount the threat.
P(Person is capable of carrying out the threat) is high enough for you to pay it off on its own. The only way for P(Person is capable of carrying out the threat | Person makes the threat) to be small enough to ignore is if P(Person makes the threat) > 1.
The prior probability of X is 2^-(K-complexity of X). There are more possible universes where they carry out smaller threats, so the K-complexity is lower. What I showed is that, even if there were only a single possible universe where the threat was carried out, it’s still simple enough that the K-complexity is small enough that it’s worth paying the threatener.
You gave a vague argument. Rather than giving a vague counterargument along the same lines, I just ran the math directly. You can argue that P(C|E) decreases all you want, but since I found that the actual value is still too high, it clearly doesn’t decrease fast enough.
If you want the vague counterargument, it’s simple: The probability that it’s a lie approaches unity. It just doesn’t approach it fast enough. It’s a heck of a lot less likely that someone who threatens 3^^^3 lives is telling the truth than someone who’s threatening one. It’s just not 3^^^3 times less likely.
What you’re ignoring is the comparison probability. See philH’s comment.
I’m not sure what you mean.
If you mean what I think you mean, I’m ignoring it because I’m going with worst case. Rather than tracking how the probability of someone making the threat reduces slower than the probability of them carrying it out (which means a lower probability of them carrying it out), I’m showing that even if we assume that the probability is one, it’s not enough to discount the threat.
P(Person is capable of carrying out the threat) is high enough for you to pay it off on its own. The only way for P(Person is capable of carrying out the threat | Person makes the threat) to be small enough to ignore is if P(Person makes the threat) > 1.