We need to get together with a whiteboard or paper.  The value of S at t1
could be 190 cm if S is height and t1 is Nick Thompson.

---
Frank C. Wimberly
140 Calle Ojo Feliz,
Santa Fe, NM 87505

505 670-9918
Santa Fe, NM

On Mon, Dec 6, 2021, 4:23 PM <thompnicks...@gmail.com> wrote:

> Frank,  Thanks for sticking with me on this.
>
>
>
> My comment was not conceptual, merely algebraic:  Can
>
>
>
> Pr(R at t3 | I at t2) = Pr(R at t3 | I at t2 & S at t1).
>
>
>
> …be true if the value of S at t1.is anything other than one?
>
>
>
>
>
>
>
> n
>
> Nick Thompson
>
> thompnicks...@gmail.com
>
> https://wordpress.clarku.edu/nthompson/
>
>
>
> *From:* Friam <friam-boun...@redfish.com> *On Behalf Of *Frank Wimberly
> *Sent:* Monday, December 6, 2021 3:07 PM
> *To:* The Friday Morning Applied Complexity Coffee Group <
> friam@redfish.com>
> *Subject:* Re: [FRIAM] Screening off
>
>
>
> 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
>
>
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