Given the causal graph ($Z$->$X$, $Z$->$Y$, $X$->$Y$), according to Pearl’s intervention, the effect of intervening $X$ on $Y$ can be estimated as $$P(Y=y|do(X=x)) = \sum_z P(Y=y|X=x,Z=z)P(Z=z)$$ Regarding this formula, I have 2 following questions:

• If for $Z=z^*$,$count(X=x,Z=z^*)=0$, do we set $P(Y=y│X=x,Z=z^* )=0$ ?
• In cases where $P(z^*)$ is large but there is only an instance where $Z=z^*$ and $X=x$ simultaneously, i.e. $count(X=x,Z=z^* )=1$, hence, $P(Y=y│X=x,Z=z^* )=1$.
  Will the term $P(Y=y│X=x,Z=z^* )P(Z=z^*)$ dominate the estimation $P(Y=y│do(X=x))$ ? Is that a problem in this estimation?

The above questions are demonstrated with the following data. The terms related to the questions are bold in the equation. Within this data, $X\in\{0,1\},Y\in\{0,1\}$, and $Z\in\{z_1,z_2,z_3\}$.

To analyze the causal effect of $X$ on $Y$, we estimate the average causal effect (ACE). Have $$P(Y=1│do(X=1))=P(Y=1│X=1,Z=z_1)+P(Y=1│X=1,Z=z_2)+P(Y=1│X=1,Z=z_3 )$$ $$=1/4×5/10+ 0/1×3/10+\mathbf{0/0}×2/10$$ $$P(Y=1│do(X=0) )=P(Y=1│X=0,Z=z_1)+P(Y=1│X=0,Z=z_2)+P(Y=1│X=0,Z=z_3)$$ $$=\mathbf{1/1×5/10}+ 1/2×3/10+2/2×2/10$$