, 2024

Solution:

We solve for the expected number of total socks drawn and subtract two at the end.
Let $E_{n}$ be the expected number of socks drawn for $n$ pairs of socks, so that $E_{1}=2$. Suppose there are $n$ pairs of socks, Rishabh continued to draw socks until the drawer was empty, and without loss of generality let the last sock drawn be red. If we ignore the two red socks, the process is equivalent to drawing from a drawer with $n-1$ pairs of socks. Let $k$ be the number of socks drawn until a pair of identical socks is found, after ignoring the two red socks. Then the first red sock has probability $\frac{k}{2 n-1}$ of being before this stopping point, so the expected value is $k+\frac{k}{2 n-1}=k \cdot \frac{2 n}{2 n-1}$. Since the expected value of $k$ is $E_{n-1}$, we have
$$E_{n}=\frac{2 n}{2 n-1} \cdot E_{n-1}$$
Applying this recurrence, we get
$$E_{n}=\frac{(2 n)!!}{(2 n-1)!!}=\frac{2^{n} \cdot n!}{(2 n-1)!!}=\frac{4^{n} \cdot(n!)^{2}}{(2 n)!}=\frac{4^{n}}{\binom{2 n}{n}}$$
Subtracting two and plugging in $n=2024$ gives a final answer of $\frac{4^{2024}}{\binom{4048}{2024}}-2$.

Solution:

Let $P(k)$ denote the probability that Rishabh draws more than $k$ socks. We compute $P(k)$ for all $0 \leq k \leq 2024$ (and note $P(k)=0$ for larger $k$).
The number of ways to draw $k$ socks, none identical to each other, is
$$4048 \cdot 4046 \cdots (4050-2k) = 2^{k} \cdot \frac{2024!}{(2024-k)!}$$
while the total number of ways to draw $k$ socks is
$$4048 \cdot 4047 \cdots (4049-k) = \frac{4048!}{(4048-k)!}$$
Thus,
$$P(k) = \frac{2^{k} \cdot \frac{2024!}{(2024-k)!}}{\frac{4048!}{(4048-k)!}} = \frac{2024!}{4048!} \cdot \frac{2^{k}(4048-k)!}{(2024-k)!} = \frac{1}{\binom{4048}{2024}} \cdot 2^{k} \binom{4048-k}{2024}$$
The expected number of socks drawn is
$$P(0)+P(1)+\cdots+P(2024)=\frac{1}{\binom{4048}{2024}} \sum_{k=0}^{2024} 2^{k}\binom{4048-k}{2024}$$
This sum is equivalent to Putnam 2020 A2. We claim that it is equal to $4^{2024}$. We do this via a counting argument: we count how many ways there are to choose at least half of the elements from the set $\{1,2, \ldots, 4049\}$. On the one hand that is $\frac{2^{4049}}{2}=4^{2024}$. On the other hand, letting $k+1$ be the 2025th largest element chosen, there are $\binom{4048-k}{2024}$ ways to choose the elements larger than it, and $2^{k}$ ways to choose the elements smaller than it. Varying $k$, we get
$$\sum_{k=0}^{2024} 2^{k}\binom{4048-k}{2024}=4^{2024}$$
This means the expected number of socks is
$$\frac{4^{2024}}{\binom{4048}{2024}}$$
and subtracting two for the matching pair gives $\frac{4^{2024}}{\binom{4048}{2024}}-2$.