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/

Reply via email to