It may not be fun but it's essential in algorithms for inferring causation
from correlation.

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

505 670-9918
Santa Fe, NM

On Sat, Dec 4, 2021, 5:53 PM Frank Wimberly <wimber...@gmail.com> wrote:

> The point is if you know 2 has occurred you know the probability that 3
> will occur.  2 occurring makes 1 irrelevant.
>
> Very formally  1 is independent of 3 given 2.  Please use my variable
> names to avoid my making an error.
>
> Frank
> ---
> 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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>
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