Find the probability of getting the second ace as the $n^{th}$ card is picked from $52$ cards. 
Cards are drawn one by one at random from a well shuffled full pack of $52$ cards. Find the probability of  the second ace being the $n^{th}$ card.

For this what I did was, for the $n-1^{th}$ positions, we need to make a selection of $n-2$ card from $48$ cards and then choose $1$ ace out of $4$. This can be done in $^{48}C_{n-2} × 4$ . Now for $n^{th}$ position, I can have an ace in $3$ ways. So total no of favourable case is $^{48}C_{n-2} × 4× 3$. For total cases I select n card from  $52$ in $^{52}C_{n}$. So $$P=\frac{^{48}C_{n-2} × 4× 3}{^{52}C_{n}}$$
But this is not quite the answer.  The answer instead is $$P=\frac{^{48}C_{n-2} ×\color{red}{n} × 4× 3}{^{52}C_{n}}$$, which is quite similar but differs only by an $n$ . Can someone please explain what have I missed.

We must multiply the probability that exactly one ace is selected among the first $n - 1$ cards by the probability that the $n$th card selected is an ace.  
We can select $n - 1$ of the $52$ cards in the deck in $\binom{52}{n - 1}$ ways.  If these cards include exactly one ace, we must select one of the four aces and $n - 2$ of the remaining $48$ cards.  Hence, the probability that the first $n - 1$ cards selected contain exactly one ace is 
$$\frac{\dbinom{4}{1}\dbinom{48}{n - 2}}{\dbinom{52}{n - 1}}$$
Since $n - 1$ cards have been selected, there are $52 - (n - 1) = 53 - n$ cards left in the deck.  Three of these are aces.  Hence, the probability that the $n$th card is an ace given that exactly one ace has been selected among the first $n - 1$ cards is 
$$\frac{3}{53 - n}$$
Therefore, the probability that the second ace is the $n$th card selected is 
\begin{align*}
\Pr(\text{second ace is $n$th card selected}) & = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}}{\dbinom{52}{n - 1}} \cdot \frac{3}{53 - n}\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}}{\dfrac{52!}{(n - 1)![52 - (n - 1)]!}} \cdot \frac{3}{53 - n}\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}}{\dfrac{52!}{(n - 1)!(53 - n)!}} \cdot \frac{3}{53 - n}\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}}{\dfrac{52!}{(n - 1)!(52 - n)!}} \cdot 3\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}\dbinom{3}{1}}{\dfrac{52!n}{n(n - 1)!(52 - n)!}}\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}\dbinom{3}{1}}{\dfrac{52!n}{n!(52 - n)!}}\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}\dbinom{3}{1}}{n\dbinom{52}{n}}
\end{align*}
so the stated answer is incorrect. That factor of $n$ should be in the denominator.
Your error was not accounting for the fact that the second ace must be in the $n$th position.  That requires two separate events.  There must be one ace among the first $n - 1$ cards and the $n$th card selected must be an ace.  While you accounted for this in your numerator, you did not do so in your denominator.
