When selecting $r$ shoes from $n$ pairs of shoes, the probability that there are exactly $k$ matching pairs amongst the $r$ selected shoes is given by $$ \mathfrak{p}_{n}^{(r)}: \left\{0, \ldots, \left\lfloor\frac{r}{2}\right\rfloor\right\} \to [0,1], \ k \mapsto \binom{n}{k} \binom{n - k}{r - 2k} 2^{r - 2k} \binom{2n}{n}^{-1}, $$ where $r \le n$ and $r, n \in \mathbb{N}$, as detailed i.e. here.
Upon stumbling about this post I wondered if $\mathfrak{p}$ is a PMF and what its expected value, etc. are and if it can be characterised by any off the "standard" probability distributions.
Concretely, I want to
Verify $ \sum_{k = 0}^{\left\lfloor\frac{r}{2}\right\rfloor} \mathfrak{p}_{n}^{(r)} = 1, $ which is equivalent to $$ \sum_{k = 0}^{\left\lfloor\frac{r}{2}\right\rfloor}\binom{n}{k} \binom{n - k}{r - 2k} 2^{r - 2k} = \binom{2n}{r}. $$
Find $$ \mathbb{E}[X] = \sum_{k = 0}^{\left\lfloor\frac{r}{2}\right\rfloor} \frac{k \cdot \binom{n}{k} \binom{n - k}{r - 2k} 2^{r - 2k}}{\binom{2n}{r}}, $$ where $X \sim P$ and $P$ is the distribution induced by the PMF $\mathfrak{p}_{n}^{(r)}$.
Unfortunately, the above mentioned post didn't really help me since I want to to solve / calculate those in an elementary fashion, without generating functions or integrals (I am aware an able to use the probabilty generating function, though).
I'm not sure what your goal is with (1); $\mathfrak{p}_n^{(r)}$ is a PMF by construction, assuming your formula is right. If you want to be more explicit/rigorous about this, that can be done by defining some random variables.
Let $S = [n] \times \{1, 2\}$ be the set of all shoes, and let $T$ be a random subset of $S$ of size $r$, so $T$ is chosen uniformly from all $\binom{2n}{r}$ subsets of $S$ of size $r$. Now let $P \subset [n]$ be the set of pairs of shoes in $T$, so $P = \{i \in [n] : (i, 1), (i, 2) \in T\}$. Then $\mathfrak{p}_n^{(r)}$ is just the PMF of the random variable $|P|$.
To prove this, note that $\mathbb{P}[|P| = k]$ is the fraction of sets $T$ which have $k$ pairs and $r - 2k$ unpaired shoes. Such a $T$ is determined uniquely by the corresponding set $P$ of pairs, the set $U \subset [n]$ of unpaired shoe types, and the parities of all the unpaired shoes (i.e. whether they are left or right). There are $\binom{n}{k}$ ways to choose $P$, $\binom{n-k}{r-2k}$ ways to choose $U$ given $P$, and $2^{r-2k}$ ways to choose parities of the unpaired shoes, hence $\binom{n}{k}\binom{n-k}{r-2k} 2^{r-2k}$ such $T$, out of a total of $\binom{2n}{r}$ possible $T$, giving $$\mathbb{P}[|P| = k] = \binom{n}{k}\binom{n-k}{r-2k} 2^{r-2k} \binom{2n}{r}^{-1} = \mathfrak{p}_n^{(r)}$$ as desired. It immediately follows that $$1 = \sum_{k=1}^\infty \mathbb{P}[|P| = k] = \sum_{k=1}^{\lfloor r/2 \rfloor} \binom{n}{k}\binom{n-k}{r-2k} 2^{r-2k} \binom{2n}{r}^{-1}$$ hence $$\sum_{k=1}^{\lfloor r/2 \rfloor} \binom{n}{k}\binom{n-k}{r-2k} 2^{r-2k} = \binom{2n}{r}$$
To answer (2), for each $i$ let $A_i$ be the event that the $i$-th pair of shoes is in $T$, i.e. that $(i, 1), (i, 2) \in T$, so letting $\mathbf{1}_{A_i}$ be the indicator variable of $A_i$, we have $|P| = \sum_{i=1}^n \mathbf{1}_{A_i}$, hence by linearity of expectation, $$\mathbb{E}[|P|] = \sum_{i=1}^n \mathbb{E}[\mathbf{1}_{A_i}] = \sum_{i=1}^n \mathbb{P}[A_i].$$ But the number of $T$ containing $(i, 1), (i, 2)$ is just $\binom{2n-2}{r-2}$, hence for each $i$ we have $$\mathbb{P}[A_i] = \frac{\binom{2n-2}{r-2}}{\binom{2n}{r}} = \frac{r(r-1)}{2n(2n-1)}$$ which gives $$\mathbb{E}[|P|] = n \cdot \frac{r(r-1)}{2n(2n-1)} = \frac{r(r-1)}{4n-2}.$$ Note that this last formula is just $\frac{1}{2n-1} \binom{r}{2}$, which suggests another (arguably easier) approach we could have taken to compute $\mathbb{E}[|P|]$.