TITLE: Probability of choosing $n$ numbers from $\{1, \dots, 2n\}$ so that $n$ is 3rd in size
QUESTION [8 upvotes]: We uniformly randomly choose $n$ numbers out of $2n$ numbers from the group $\{1, \dots, 2n\}$ so that order matters and repetitions are allowed. What is the probability that $n$ is the $3^{\text{rd}}$ number in size in the chosen series? (= there are only two bigger than $n$)Note that if a number that is bigger than $n$ was chosen more than once, we still count it as one number bigger than $n$.
Example: $n = 5, \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ 
$7, 1, 5, 9, 9$ - ok
$2, 2, 10, 5, 7$ - ok
$6, 6, 3, 5, 5$ - not ok, 5 is $2^{\text{nd}}$ in size
My lead was to first choose 2 numbers that are bigger than $n$ out of the n numbers that are bigger than $n$ in the group $\{1, \dots, 2n\}$. Then in order to complete a series of $n$ numbers I need to choose $n-3$ more from the numbers that are equal or smaller than $n$, but then I figured this is not right as a series could be formed from the 2 numbers bigger than $n$ and $n$ itself without any additional numbers. At this point I have no additional leads and could very much use your assistance and guidance.

REPLY [2 votes]: 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
$$\frac{1}{(2n)^n} {n\choose 2} 
\left((n+2)^n - 3(n+1)^n + 3n^n - (n-1)^n\right).$$