Law of iterated expectation - $\mathbb{E}[\mathbb{E}[X\mid Y] \mid Z ] = \mathbb{E}[X]$ I am stuck doing this problem:
$X,Y,Z$ be random variables, then: $$\mathbb{E}[\mathbb{E}[X\mid Y] \mid  Z ] = \mathbb{E}[X]$$
Surely this is from LIE and conditioning, but I can't seem to set it up correctly for it to work out nicely, can anyone point me in the right direction, or correct my error below?
Below is the way I was attacking this, but quickly got stuck and it's making me think I simply didn't set up my summation correctly. The expectation inside the conditional is really throwing me for a loop.
My take (likely incorrect):
\begin{align} \mathbb{E}[\mathbb{E}[X\mid Y]\mid Z] &= \sum_zP(Z=z) (\mathbb{E}[X\mid Y])\\
&=\sum_zP(Z=z)\sum_x xP(X=x\mid Y=y)\\ &= \sum_zP(Z=z)\sum_x x \frac{P(X=x, Y=y)}{P(Y=y)}\end{align}
and I don't see where to go from here. Any help is greatly appreciated.
Edit: Thank you to everyone who commented and pointed out this is generally false. After asking my professor the question should really be
$$\mathbb{E}[\mathbb{E}[\mathbb{E}[X\mid Y] \mid  Z ]] = \mathbb{E}[X]$$
I haven't done the proof yet, but I assume this follows cleanly from LIE.
 A: [We are presuming $X,Y,Z$ are discrete random variables.]
Since $\Bbb E(X\mid Y)$ is a function of random variable $Y$, then we have:
$\qquad\begin{align}\Bbb E(\Bbb E(X\mid Y)\mid Z)&=\left[\sum_y\mathsf E(X\mid Y\,{=}\,y)\,\mathsf P(Y\,{=}\,y\mid Z\,{=}\,z)\right]_{z=Z} \\ &=\left[\sum\limits_y\left(\sum\limits_x x\,\mathsf P(X\,{=}\,x\mid Y\,{=}\,y)\right)\mathsf P(Y\,{=}\,y\mid Z\,{=}\,z)\right]_{z=Z}\\ &=\left[\sum\limits_{x} x\sum\limits_{y}\mathsf P(X\,{=}\,x\mid Y\,{=}\,y)\,\mathsf P(Y\,{=}\,y\mid Z\,{=}\,z)\right]_{z=Z}\end{align}$
Thus it shall not generally be so that $\Bbb E(\Bbb E(X\mid Y)\mid Z)=\Bbb E(X)$ unless, for example, $Y$ is independent of $Z$.

Update
The above expectation is a function of $Z$. The OP has updated the question to be about the expectation of this.  So let us do that, distribute the series and apply the law of total probability...
$\qquad\begin{align}\Bbb E(\Bbb E(\Bbb E(X\mid Y)\mid Z))&=\sum\limits_z\sum\limits_{x} x\sum\limits_{y}\mathsf P(X\,{=}\,x\mid Y\,{=}\,y)\,\mathsf P(Y\,{=}\,y\mid Z\,{=}\,z)\,\mathsf P(Z\,{=}\,z)\\&=\sum\limits_{x} x\sum\limits_{y}\mathsf P(X\,{=}\,x\mid Y\,{=}\,y)\sum\limits_z\mathsf P(Y\,{=}\,y\mid Z\,{=}\,z)\,\mathsf P(Z\,{=}\,z)\\&=\sum\limits_{x} x\sum\limits_{y}\mathsf P(X\,{=}\,x\mid Y\,{=}\,y)\,\mathsf P(Y\,{=}\,y)\\&=\sum\limits_x x\,\mathsf P(X\,{=}\,x)\\&=\Bbb E(X)\end{align}$
