Intuition behind the expectation of a conditional expectation Let $(\Omega, \cal F, P)$ be a probability space and $X,Y$ two random variables defined on $(\Omega, \cal F, P)$. I stumbled upon the equality $$E[E(X \vert Y)]=E(X)$$
Observing that 
$$\begin{align}
E[E(X \vert Y)] & =\sum_iP(Y=Y_i)E(X\vert Y_i) \\
& = P(Y=Y_1)E(X \vert Y_1)+...+ P(Y=Y_n)E(X\vert Y_n) \\
&=E(X)
\end{align}$$
it "mathematically" makes sense to me, but still not intuitively. How does one intuitively recognise that the expectation of a random variable equal the expectation of an expectation of that random variable given some information?
 A: $E(X)$ is the expectation of $X$, whatever $Y$. You can also express it as a weighted average of the expectations of $X$ for given $Y$, where the weights are the probabilities of the respective values of $Y$.
E.g. we have the drawings
$$(1,1),(1,2),(2,1),(2,2)$$ with respective probabilities $0.1,0.2,0.3,0.4$.
The expectation of $X$ is
$$1\cdot0.1+1\cdot0.2+2\cdot0.3+2\cdot0.4=1.7$$
The marginal probabilities of $Y$ are $0.4, 0.6$.
The conditional probabilities of $X$ are
$$Y=1\to 0.25,0.75,\\Y=2\to 0.33,0.67.$$
The conditional expectations of $X$ are
$$Y=1\to 1\cdot0.25+2\cdot0.75=1.75,\\Y=2\to 1\cdot0.33+2\cdot0.67=1.67.$$
and indeed,
$$1.75\cdot0.4+1.67\cdot0.6=1.7.$$
A: There are many ways of explaining this intuivitely, I will use the equalities you provided, stick to the discrete case and won't go much into the machinery of probability theory.
I claim that the sum
$$\sum_i P(Y=Y_i)E(X|Y_i)$$
is essentially just the expected value of $X$ - a weighted average of all values of $X$ - in disguise. It is only matter of how the space $\Omega$ is partitioned.

$E(X)$
To see that, let's see what $E(X)$ does:
$$E(X) = \sum_i  P(X = X_i)  X_i$$
Think of the event $[X=X_i]$ as a set of all cases (all the $\omega$s) where $X(\omega) = X_i$. These events form a partitioning of $\Omega$ in that they are disjoint sets that add up to $\Omega$.
In $E(X)$ we are therefore looking at the value of $X$ on the set $[X=X_i]$ and weighting it with the size of the set, $P(X=X_i)$.

$E(E(X|Y))$
Now in your case:
$$\sum_i P(Y=Y_i)E(X|Y_i)$$
Here, we are looking at the value of $X$ on the set $[Y=Y_i]$. But we can't just write $X_i$, since $X$ does not have to have the same value on all $\omega \in [Y=Y_i]$. Thus we have to average over all the cases in $[Y=Y_i]$, which is precisely what $E(X|Y_i)$ is. Then we just multiply this with the size of the set, which is $P(Y=Y_i)$.
In both cases, we are just aggregating the $\omega$s in different ways. In $E(X)$, it's over the sets $[X=X_i]$, in $E(E(X|Y))$, it's over the sets $[Y=Y_i]$.
