Here is a question which has boggled me in the understanding of the Law of Iterated Expectations(LIE). Consider a deck of $13$ cards numbered $\left\{1, 2, \cdots, 13\right\}$ being randomly drawn.
Let $X$ denote the value of the first card and $Y$ denote the value of the second card drawn. This is a case of drawing with no replacement so the card drawn at $X$ is not placed back and then the draw at $Y$ occurs. Find $\mathbb E (Y)$.
By LIE I know that $\mathbb E(Y) = \mathbb E[\mathbb E(Y|X)]$ so the working follows:
$$ \begin{align} \mathbb E(Y) &= \mathbb E[\mathbb E(Y|X)]\\ &=\>? \end{align} $$
My attempted working $$ \begin{align} \mathbb E(Y) &= \mathbb E[\mathbb E(Y|X)]\\ &= \mathbb E_X [\mathbb E(Y|X)]\\ &=\sum_{x=1}^{13}\frac{1}{13}\begin{bmatrix}\sum_{y=1 \cap y\neq x}^{13} \frac 1 {12}(Y=y|X=x)\end{bmatrix}\\ &=\frac{1}{13} \begin{bmatrix}\begin{pmatrix}\frac{90}{12}\end{pmatrix}+\begin{pmatrix}\frac{89}{12}\end{pmatrix}+\begin{pmatrix}\frac{88}{12}\end{pmatrix}+\cdots + \begin{pmatrix}\frac{78}{12}\end{pmatrix}\end{bmatrix} \\&=\frac{1092}{13\times 12} \\&= 7 \end{align} $$
$\mathbb E (Y) = \mathbb E (X) = \displaystyle \frac{1}{13} \sum_1^{13} x = 7$ so you seem to have the correct answer.