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The new subscripts mean the newest relative times of the latest situations, with large amounts equal to after moments

• $$\ST_0= 1$$ if the Suzy puts, 0 if you don’t
• $$\BT_1= 1$$ in the event the Billy sets, 0 if you don’t
• $$\BS_dos = 1$$ in case your bottle shatters, 0 if you don’t

\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 ## In truth these odds try comparable to$

(Keep in mind that we have additional a little chances towards the bottle to help you shatter due to additional lead to, no matter if none Suzy nor Billy toss the rock. It ensures that the possibilities of the many tasks out of opinions so you’re able to this new details is actually self-confident.) The latest corresponding chart is actually revealed during the 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

## In truth these two odds try comparable to

\]

Carrying fixed you to Billy does not place, Suzys throw enhances the possibilities that package often shatter. Hence the new standards are found to possess $$\ST = 1$$ as a real reason behind $$\BS = 1$$.

• $$\ST_0= 1$$ in the event that Suzy places, 0 otherwise
• $$\BT_0= 1$$ if Billy throws, 0 if you don’t
• $$\SH_1= 1$$ when the Suzys stone strikes the brand new bottle, 0 if not
• $$\BH_1= 1$$ when the Billys material moves the new package, 0 if you don’t
• $$\BS_2= 1$$ if for example the bottle shatters, 0 or even

\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 ## But in fact these odds are equal to$

Since ahead of, we have assigned probabilities alongside, not comparable to, zero and one for many of options. This new graph is actually revealed inside Shape 10.

We want to demonstrate that $$\BT_0= 1$$ is not an actual reason behind $$\BS_2= 1$$ according to F-Grams. We’ll show this by means of a problem: is actually $$\BH_1\for the \bW$$ or is $$\BH_1\during the \bZ$$?

Assume very first you to definitely $$\BH_1\from inside the \bW$$. Up coming, no matter whether $$\ST_0$$ and $$\SH_1$$ are in $$\bW$$ or $$\bZ$$, we have to keeps

\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

## In facts those two odds are equivalent to

\]

95. If we intervene to set $$\BH_1$$ in order to 0, intervening to the $$\BT_0$$ makes no difference on the 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 these two odds is actually equal to

\]

(Another possibilities are only a little huge, as a result of the very small chances you to Billys stone will strike though he does not put it.)

So it doesn’t matter if $$\BH_1\in the \bW$$ or perhaps is $$\BH_1\inside the \bZ$$, position F-G isn’t met, and you can $$\BT_0= 1$$ is not judged to get an authentic factor in $$\BS_2= 1$$. The key idea is that this isn’t sufficient getting Billys throw to raise the possibilities of the bottles shattering; Billys place in addition couples hookup to what will happen after needs to enhance the likelihood of smashing. Because the some thing in reality occurred, Billys stone skipped the latest package. Billys toss together with his material forgotten does not increase the probability of smashing.