Factoring is much harder than dividing, so you can verify A & B in a few seconds with python, but it would take several core-years to verify A on its own. Therefore, if you can perform these verifications, you should put both P(A) and P(A & B) as 1 (minus some epsilon of your tools being wrong). If you can’t perform the validations, then you should not put P(A) as 1- since there was a significant chance that I would put a false statement in A. (in this case, I rolled a D100 and was going to put a false statement of the same form in A if I rolled a 1)
I’m having trouble parsing your second paragraph: inarguably, P(A | B) = P( A & B) = 1, so surely P(A) < P(A | B) implies P(A) < P(A & B) ?
You’ve introduced a third piece of information C: observing the result of a computation that B indeed divides A. With that information, I have P(A|C) = P(A & B|C) = 1. I also have P(A) < P(A | C). But without C, I still have P(A) ≥ P(A & B), not the opposite.
I am having trouble with your second paragraph. Certainly, P(A | B&C) = P( A & B | C) = 1. But when I do not know C, P(A | B) = 1 and P(A & B) < 1.
Factoring is much harder than dividing, so you can verify A & B in a few seconds with python, but it would take several core-years to verify A on its own. Therefore, if you can perform these verifications, you should put both P(A) and P(A & B) as 1 (minus some epsilon of your tools being wrong). If you can’t perform the validations, then you should not put P(A) as 1- since there was a significant chance that I would put a false statement in A. (in this case, I rolled a D100 and was going to put a false statement of the same form in A if I rolled a 1)
I’m having trouble parsing your second paragraph: inarguably, P(A | B) = P( A & B) = 1, so surely P(A) < P(A | B) implies P(A) < P(A & B) ?
You’ve introduced a third piece of information C: observing the result of a computation that B indeed divides A. With that information, I have P(A|C) = P(A & B|C) = 1. I also have P(A) < P(A | C). But without C, I still have P(A) ≥ P(A & B), not the opposite.
I am having trouble with your second paragraph. Certainly, P(A | B&C) = P( A & B | C) = 1. But when I do not know C, P(A | B) = 1 and P(A & B) < 1.