Brand new subscripts suggest the relative times of the fresh events, which have large quantity equal to later on times

• \(\ST_0= 1\) when the Suzy puts, 0 otherwise
• \(\BT_1= 1\) when the Billy throws, 0 otherwise
• \(\BS_2 = 1\) in case the bottle shatters, 0 otherwise

\PP(\BT_1= 1 \mid \ST_0= 1) <> = .1 \\ \PP(\BT_1= 1 \mid \ST_0= 0) <> = .9 \\[1ex] \PP(\BS_2= 1 \mid \ST_0= 1 \amp \BT_1= 1) <> = .95\\ \PP(\BS_2= 1 \mid \ST_0= 1 \amp \BT_1= 0) <> = .5\\ \PP(\BS_2= 1 \mid \ST_0= 0 \amp \BT_1= 1) <> = .9\\ \PP(\BS_2= 1 \mid \ST_0= 0 \amp \BT_1= 0) <> = .01\\ \end

But in facts these two likelihood try comparable to

\]

(Keep in mind that i have additional a little likelihood for the bottle so you’re able to shatter on account of additional produce, even when none Suzy nor Billy toss their rock. That it means the number of choices of the many projects out-of values to help you the brand new details are positive.) Brand new related graph is actually revealed in Shape nine.

\PP(\BS_2= 1 \mid \do(\ST_0= 1) \amp \do(\BT_1= 0)) <> = .5\\ \PP(\BS_2= 1 \mid \do(\ST_0= 0) \amp \do(\BT_1= 0)) <> = .01\\ \end

However in reality both of these probabilities are equivalent to

\]

Holding fixed one to Billy doesnt place, Suzys throw enhances the opportunities that the package commonly shatter. Ergo the fresh requirements is actually found having \(\ST = 1\) getting a real reason for \(\BS = 1\).

• \(\ST_0= 1\) in the event the Suzy throws, 0 if you don’t
• \(\BT_0= 1\) if Billy throws, 0 otherwise
• \(\SH_1= 1\) in the event the hookup apps for married men Suzys stone hits the latest container, 0 otherwise
• \(\BH_1= 1\) in the event the Billys stone strikes the fresh new bottle, 0 if you don’t
• \(\BS_2= 1\) if for example the package shatters, 0 if you don’t

\PP(\SH_1= 1 \mid \ST_0= 1) <> = .5\\ \PP(\SH_1= 1 \mid \ST_0= 0) <> = .01\\[2ex] \PP(\BH_1= 1 \mid \BT_0= 1) <> = .9\\ \PP(\BH_1= 1 \mid \BT_0= 0) <> = .01\\[2ex] \PP(\BS_2= 1 \mid \SH_1= 1 \amp \BH_1= 1) <> = .998 \\ \PP(\BS_2= 1 \mid \SH_1= 1 \amp \BH_1= 0) <> = .95\\ \PP(\BS_2= 1 \mid \SH_1= 0 \amp \BH_1= 1) <> = .95 \\ \PP(\BS_2= 1 \mid \SH_1= 0 \amp \BH_1= 0) <> = .01\\ \end

In truth those two chances are equivalent to

\]

Once the before, i have assigned odds alongside, although not comparable to, zero and one for almost all of one’s choices. New chart are revealed within the Shape 10.

You want to show that \(\BT_0= 1\) is not an authentic factor in \(\BS_2= 1\) considering F-Grams. We are going to inform you which as a dilemma: is actually \(\BH_1\during the \bW\) or is \(\BH_1\in \bZ\)?

Guess basic you to definitely \(\BH_1\within the \bW\). After that, no matter whether \(\ST_0\) and you will \(\SH_1\) have \(\bW\) or \(\bZ\), we need to possess

\PP(\BS_2 = 1 \mid do(\BT_0= 1, \BH_1= 0, \ST_0= 1, \SH_1= 1))\\ \mathbin <\gt>\PP(\BS_2 = 1 \mid do(\BT_0= 0,\BH_1= 0, \ST_0= 1, \SH_1= 1))\\ \end

However in truth these chances is actually equal to

\]

95. When we intervene to set \(\BH_1\) so you’re able to 0, intervening with the \(\BT_0\) makes no difference on likelihood of \(\BS_2= 1\).

\PP(\BS_2 = 1 \mid do(\BT_0= 1, \BH_1= 0, \ST_0= 1, \SH_1= 1))\\ \mathbin <\gt>\PP(\BS_2 = 1 \mid do(\BT_0= 0, \ST_0= 1, \SH_1= 1))\\ \end

In reality both of these odds are equal to

\]

(The next probability is actually a little big, as a result of the tiny chances one Billys stone commonly hit although he does not place it.)

Very whether or not \(\BH_1\when you look at the \bW\) or perhaps is \(\BH_1\inside the \bZ\), condition F-Grams isn’t met, and \(\BT_0= 1\) is not evaluated becoming an authentic reason behind \(\BS_2= 1\). The key idea is that this is not adequate getting Billys toss to improve the probability of the fresh bottle smashing; Billys throw along with what takes place afterwards has to improve the odds of smashing. As anything indeed taken place, Billys stone overlooked the new bottle. Billys place together with his rock missing does not enhance the likelihood of smashing.