I don’t understand where that 1⁄2 comes from. Unless I have made a gross mistake P(A|A ⇒ B) < P(A) even if P(A&B) > P(A¬(B)). In your first example, if I swap P(AB) and P(A¬(B)) so that P(AB) = .5 and P(A¬(B))=.3 then P(A|A=>B) = .5/.7 ~ 0.71 < 0.8 = P(A).
This confused me as well. This being true ensures that the ratio P(A):P(not A) doubles at each step. But part of this comic seems to imply that being less than a half stops the trolling, when it should only stop the trolling from proceeding at such a fast-paced rate.
I don’t understand where that 1⁄2 comes from. Unless I have made a gross mistake P(A|A ⇒ B) < P(A) even if P(A&B) > P(A¬(B)). In your first example, if I swap P(AB) and P(A¬(B)) so that P(AB) = .5 and P(A¬(B))=.3 then P(A|A=>B) = .5/.7 ~ 0.71 < 0.8 = P(A).
This confused me as well. This being true ensures that the ratio P(A):P(not A) doubles at each step. But part of this comic seems to imply that being less than a half stops the trolling, when it should only stop the trolling from proceeding at such a fast-paced rate.