Is the expected value of a probability over an interval meaningful? I am reading an unpublished manuscript and have come across an equation of the following form for the calculation of the probability of an even A,
$$
P[A]=E\Big[P[X>x|Y]\Big]. \tag{1} \label{1}
$$
This is very confusing. The conditional probability $P[X>x|Y]$ must be a number, say $c$ in the interval $[0,1]$ i.e. $P[X>x|Y]=c$. Therefore, \eqref{1} simply reduces to,
$$
P[A]=E[c]=c, \tag{2} \label{2}
$$
as the expected value of a constant is the constant itself. Therefore, using the expected value of the probability seems meaningless, and the following should be sufficient to find the probability of event A,
$$
P[A]=P[X>x|Y]. \tag{3} \label{3}
$$
Is that so or do I miss something? Is it correct and meaningful to use an equation of the form given in \eqref{1}?

The question is incomplete since it does not specify the connection between the event $A$ and the random variables $X$ and $Y$. If we assume that
$$A=\{X>x\}$$then
$$\mathbb P(A)=\mathbb E^X[\mathbb I_{(x,\infty)}(X)]\tag{1}$$
where $\mathbb I_B(\cdot)$ denotes the indicator function. This identity (1) is better understood using integrals and densities, as, for a generic set $A$
$$\mathbb P(X\in A)=\int_A f_X(x)\,\text dx=\int \mathbb I_{A}(x)f_X(x)\,\text dx=\mathbb E^X[\mathbb I_{A}(X)]$$
If $(X,Y)$ is a random vector with joint density $f_{X,Y}(x,y)$,
$$\mathbb E^X[\mathbb I_{A}(X)]=\int \mathbb I_A(X)\,f_{X,Y}(x,y)\,\text d(x,y)=\mathbb E^{X,Y}[\mathbb I_{A}(X)]$$
which can be further decomposed as
\begin{align}
\mathbb E^X[\mathbb I_{A}(X)]& = \int \mathbb I_A(x)\,f_{X,Y}(x,y)\,\text d(x,y)\\
&=\int \mathbb I_A(x)\,f_{Y}(y)f_{X|Y}(x|y)\,\text d(x,y)\\
&=\int\left\{\int \mathbb I_A(x)\,f_{X|Y}(x|y)\,\text dx\right\} f_Y(y)\,\text df_{X|Y}(x|y)\,\text dy\\
&=\mathbb E^{Y}[\mathbb E^{X|Y}[\mathbb I_{A}(X)]]
\end{align}
(known as the law of total expectation).
