Consider the following simple exercise regarding conditional expectation in die throwing.
I think the book got the formula $\mathbb{E}[X_{1}|X_{2}=\alpha]$ (as well as $\mathbb{E}[X_{1}|X_{2}]$) wrong.
If I take $n=3$ and $X_{2}=2$, i.e. we throw the die three times and assume we have obtained two $2$'s, then in my opinion $$ \mathbb{E}[X_{1}|X_{2}=2]=\sum_{x=0}^{3}x\cdot\mathbb{P}(X_{1}=x|X_{2}=2)=1\cdot\mathbb{P}(X_{1}=1|X_{2}=2)=3\cdot\frac{1}{6^{3}}=\frac{1}{72}, $$ which certainly is not $$ \frac{3-2}{5}=\frac{1}{5} $$ as the text states. How can this be?
You aren't computing the conditional probabilities correctly.
$\displaystyle P(X_1 = 1 \mid X_2 = 2) = \frac{P((X_1 = 1) \cap (X_2 = 2))}{P(X_2 =2)} = \frac{\frac{3}{6^3}}{\frac{3\cdot5}{6^3}} = \frac{1}{5}$