In the 1982 experiment where professional forecasters assigned systematically higher probabilities to “Russia invades Poland, followed by suspension of diplomatic relations between the USA and the USSR” than to “Suspension of diplomatic relations between the USA and the USSR,” each experimental group was only presented with one proposition
Hang on- I don’t think this structure is proof that the forecasters are irrational. As evidence, I will present you with a statement A, and then with a statement B. I promise you- if you are calibrated / rational, your estimate of P(A) will be less than your estimate of P(A and B)
Statement A: (please read alone and genuinely estimate its odds of being true before reading statement B)
186935342679883078818958499454860247034757454478117212334918063703479497521330502986697282422491143661306557875871389149793570945202572349516663409512462269850873506732176181157479 is composite, and one of its factors has a most significant digit of 2
Statement B:
186935342679883078818958499454860247034757454478117212334918063703479497521330502986697282422491143661306557875871389149793570945202572349516663409512462269850873506732176181157479 is divisible by 243186911309943090615130538345873011365641784159048202540125364916163071949891579992800729
I am not getting the intuition that this example is supposed to evoke. B is just one of the many ways (since I cannot factorise the number at sight) that A might be true, so whyever would I think P(A) < P(A & B), or even P(A) ≤ P(A & B) ?
Of course, P(A) < P(A | B), the latter term being 1. Strictly speaking, that is irrelevant, but perhaps the concept of P(A | B) is in the back of the mind of people who estimate P(A) < P(A & B) ?
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.
Before reading statement B, did you estimate the odds of A being true as 1?
I rolled a 100 sided die before generating statement A, and was going to claim that the first digit was 3 in A and then just put “I lied, sorry” in B if I rolled a 1. If, before reading B, you estimated the odds of A as genuinely 1 (not .99 or .999 or .9999) then you should either go claim some RSA factoring bounties, or you were dangerously miscalibrated.
I guess this is evidence that probability axioms apply to probabilities, but not necessarily to calibrated estimates of probabilities given finite computational resources- this is why in my post I was very careful to tale about calibrated estimates of P(...) and not the probabilities themselves.
Hang on- I don’t think this structure is proof that the forecasters are irrational. As evidence, I will present you with a statement A, and then with a statement B. I promise you- if you are calibrated / rational, your estimate of P(A) will be less than your estimate of P(A and B)
Statement A: (please read alone and genuinely estimate its odds of being true before reading statement B)
186935342679883078818958499454860247034757454478117212334918063703479497521330502986697282422491143661306557875871389149793570945202572349516663409512462269850873506732176181157479 is composite, and one of its factors has a most significant digit of 2
Statement B:
186935342679883078818958499454860247034757454478117212334918063703479497521330502986697282422491143661306557875871389149793570945202572349516663409512462269850873506732176181157479 is divisible by 243186911309943090615130538345873011365641784159048202540125364916163071949891579992800729
I am not getting the intuition that this example is supposed to evoke. B is just one of the many ways (since I cannot factorise the number at sight) that A might be true, so whyever would I think P(A) < P(A & B), or even P(A) ≤ P(A & B) ?
Of course, P(A) < P(A | B), the latter term being 1. Strictly speaking, that is irrelevant, but perhaps the concept of P(A | B) is in the back of the mind of people who estimate P(A) < P(A & B) ?
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.
Probability axioms say that P(A)>=P(A and B), not P(A)>P(A and B).
Before reading statement B, did you estimate the odds of A being true as 1?
I rolled a 100 sided die before generating statement A, and was going to claim that the first digit was 3 in A and then just put “I lied, sorry” in B if I rolled a 1. If, before reading B, you estimated the odds of A as genuinely 1 (not .99 or .999 or .9999) then you should either go claim some RSA factoring bounties, or you were dangerously miscalibrated.
I guess this is evidence that probability axioms apply to probabilities, but not necessarily to calibrated estimates of probabilities given finite computational resources- this is why in my post I was very careful to tale about calibrated estimates of P(...) and not the probabilities themselves.