Probability of choosing $n$ numbers from $\{1, \dots, 2n\}$ so that $n$ is 3rd in size

You can use the inclusion-exclusion principle.

Let $S:=$ The event of getting a valid sequence of numbers ($n$ numbers $\le n$ and 2 numbers $\gt n$)

$A_1 :=$ The event that $n$ does not appear in the sequence

$A_2 :=$ The event that one of the larger numbers than $n$ does not appear in the sequence

$A_3 :=$ The event that the second number that is larger than $n$ does not appear in the sequence

Now, the probability we are looking for is the probability that S occurs but the union of $A_1,\, A_2,\, A_3$ does not occur.

We have the formula

$$ \Pr \left[S\setminus \bigcup_{i=1}^{3}A_i \right] = \Pr \left[ S \right] - \sum_{i=1}^{3} \Pr\left [ A_i \right ] + \sum_{1\leq i< j\leq 3} \Pr\left [ A_i\cap A_j \right ] - \sum_{1\leq i<j<k\leq 3} \Pr\left [ A_i\cap A_j\cap A_k \right ] $$

Let's calculate the probabilities of the events defined above.

First of all, our sample space $\Omega$ is all sequences of $n$ numbers from the group $\{1, 2, \ldots, 2n\}$.

So $\left| \Omega \right| = \left( 2n \right)^n $.

Now, $ \Pr \left[S \right] = \frac{\left ( n+2 \right )^n}{\left ( 2n \right )^n} $ (we have $n$ numbers $\le n$ and 2 numbers greater than $n$)

Similarly, $ \Pr \left[A_1 \right] = \Pr \left[A_2 \right] = \Pr \left[A_3 \right] = \frac{\left ( n+1 \right )^n}{\left ( 2n \right )^n} $

We also have to calculate the intersections.

$ \Pr\left [ A_i\cap A_j \right ] = \frac{n^n}{\left ( 2n \right )^n} \; ($for $ 1\leq i<j\leq 3 )$ - notice there are 3 such intersections. And also there is the intersection of all three:

$ \Pr\left [ A_1\cap A_2\cap A_3 \right ] = \frac{\left ( n-1 \right )^n}{\left ( 2n \right )^n} $

One last thing - there are $ \binom{n}{2} $ ways to choose those 2 numbers that are greater than $n$.

Now, we'll plug everything in the inclusion-exclusion formula and multiply by $ \binom{n}{2} $, and we'll get:

$$ \Pr \left[S\setminus \bigcup_{i=1}^{3}A_i \right] = \binom{n}{2} \left( \frac{\left ( n+2 \right )^n}{\left ( 2n \right )^n} - 3\frac{\left ( n+1 \right )^n}{\left ( 2n \right )^n} +3\frac{n^n}{\left ( 2n \right )^n} - \frac{\left ( n-1 \right )^n}{\left ( 2n \right )^n} \right) = \binom{n}{2} \frac{\left ( n+2 \right )^n - 3\left ( n+1 \right )^n + 3n^n - \left ( n-1 \right )^n}{\left ( 2n \right )^n} $$

Choose the two values from $[n+1,2n]$ above $n$ to get

$${n\choose 2}.$$

Let $q$ be the additional values from below $n$ that are present (other than the three we have selected). We then have $0\le q\le n-1.$

Counting these configurations we obtain

$${n\choose 2} \sum_{q=0}^{n-1} {n-1\choose q} \times {n\brace q+3} \times (q+3)!$$

Recall the species of set partitions which is $$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$ so that the bivariate generating function is $$G(z, u) = \exp(u(\exp(z)-1))$$

which finally yields $${n\brace q} = n! [z^n] \frac{(\exp(z)-1)^q}{q!}.$$

This yields for our sum

$${n\choose 2} n! [z^n] \sum_{q=0}^{n-1} {n-1\choose q} \times \frac{(\exp(z)-1)^{q+3}}{(q+3)!} \times (q+3)! \\ = {n\choose 2} n! [z^n] \sum_{q=0}^{n-1} {n-1\choose q} \times (\exp(z)-1)^{q+3} \\ = {n\choose 2} n! [z^n] (\exp(z)-1)^3 \exp((n-1)z) \\ = {n\choose 2} n! [z^n] (\exp((n+2)z)-3\exp((n+1)z)+3\exp(nz)-\exp((n-1)z)).$$

Extracting coefficients we have

$${n\choose 2} \left((n+2)^n - 3(n+1)^n + 3n^n - (n-1)^n\right)$$

for a probability of

$$\frac{1}{(2n)^n} {n\choose 2} \left((n+2)^n - 3(n+1)^n + 3n^n - (n-1)^n\right).$$