The question isn't whether 1 has occurred but what is the value of 1 (it's a variable) and is the distribution of 3 is independent of that observed value.
--- Frank C. Wimberly 140 Calle Ojo Feliz, Santa Fe, NM 87505 505 670-9918 Santa Fe, NM On Sat, Dec 4, 2021, 5:38 PM Nicholas Thompson <thompnicks...@gmail.com> wrote: > Frank, > > Still need help. Given events 1, 2, and 3, 3 has been screen off by 2 from > 1, if the probability that 3 occurs given that 2 has occurred is equal to > the probability that 3 occurs given that both 2 and one have occurred. > As I understand mathematics this equality requires that the probability of > 1 occurring is 1.00. Another way to say that is that the probability that > 3 occurs if 2 has occurred is the same as the probability that 3 has > occurred if 2 has occurred, and 1 has already occurred. What's the fun in > that? In other words, given the possibility of other causes for 2, the > fact that 2 occurs gives us relatively little evidence that 1 has > occurred. Isn"t this true of all causal abduction? > > N > > .-- .- -. - / .- -.-. - .. --- -. ..--.. / -.-. --- -. .--- ..- --. .- - . > FRIAM Applied Complexity Group listserv > Zoom Fridays 9:30a-12p Mtn UTC-6 bit.ly/virtualfriam > un/subscribe http://redfish.com/mailman/listinfo/friam_redfish.com > FRIAM-COMIC http://friam-comic.blogspot.com/ > archives: > 5/2017 thru present https://redfish.com/pipermail/friam_redfish.com/ > 1/2003 thru 6/2021 http://friam.383.s1.nabble.com/ >
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