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.
Best Answer
There are $^{52}C_4$ possible sets of four positions in the deck where the four aces may be located. We can take this as a uniform sample space.
Then the question becomes, what is the probability that after taking a sample from this sample space, position $n$ will be one of the four positions selected, one position will be before $n,$ and the other two will be after $n$?
There is only one way to select position $n,$ but there are $^{n-1}C_1 = n-1$ ways to choose one position before $n$ and $^{52-n}C_2 = \frac{(52-n)!}{2!(50-n)!} = \frac{(52-n)(51-n)}{2}$ ways to choose two positions after $n.$ So the probability should be \begin{align} P &= \frac{{^{n-1}C_1} {^{52-n}C_2}}{^{52}C_4}\\[.5ex] &= \frac{(n-1)\dfrac{(52-n)(51-n)}{2}}{\dfrac{52!}{4!48!}}\\[.5ex] &= \frac{4!48!(n-1)(52-n)(51-n)}{2\cdot52!}\\[.5ex] &= \frac{4\cdot3(n-1)(52-n)(51-n)}{52\cdot51\cdot50\cdot49}. \end{align}
On the other hand, \begin{align} \frac{^{48}C_{n-2}\cdot4\cdot3}{^{52}C_{n}} &= \frac{\left(\dfrac{48!}{(n-2)!(50-n)!}\right)\cdot4\cdot3} {\left(\dfrac{52!}{n!(52-n)!}\right)} \\[.3ex] &= \frac{4\cdot3\cdot48!n!(52-n)!}{52!(n-2)!(50-n)!}\\ &= \frac{4\cdot3n(n-1)(52-n)(51-n)}{52\cdot51\cdot50\cdot49}. \end{align}
This differs by a factor of $n$ from the result I got, but it's $n$ times larger than my result. If it gets multiplied by $n$ again, as in the formula that you say is supposed to be the "correct" answer, it would be wrong by a factor of $n^2$ according to my calculations.
Let's try this for $n=2.$ What is the probability that the second ace will be the second card? It's just the probability that the first two cards are both aces, which is $$ P_2 = \frac{4\cdot3}{52\cdot51}.$$
Now observe that if $n=2,$ then \begin{align} \frac{4\cdot3(n-1)(52-n)(51-n)}{52\cdot51\cdot50\cdot49} &= \frac{4\cdot3(2-1)(52-2)(51-2)}{52\cdot51\cdot50\cdot49}\\ &= \frac{4\cdot3\cdot1\cdot50\cdot49}{52\cdot51\cdot50\cdot49}\\ &= \frac{4\cdot3}{52\cdot51}, \end{align} so the formula I derived gives the correct result when $n=2.$ On the other hand, again setting $n=2,$ the allegedly "correct" answer comes out to \begin{align} \frac{^{48}C_{n-2}\cdot n\cdot 4\cdot 3}{^{52}C_n} &= \frac{^{48}C_{2-2}\cdot 2\cdot 4\cdot 3}{^{52}C_2} \\ &= \frac{^{48}C_0\cdot 2\cdot 4\cdot 3}{\dfrac{52!}{2!50!}} \\ &= \frac{4\cdot 4\cdot 3}{52\cdot51}, \end{align} which is exactly $4$ times the correct result. So this formula is incorrect.
Now the question still remains why your first attempt did not get a correct answer. (For the case $n=2$ you are still off by a factor of $2.$) It seems to me that the problem is you are taking a sample space of all the combinations of $n$ cards out of $52,$ where the same $n$ cards in a different sequence are considered the same outcome in the sample space. But mixed in with this are two events that depend on the sequence of the cards, namely, which of the two aces came first (which you assume are distinguishable events, although in the sample space of $^{52}C_n$ outcomes they are not), and the requirement that the second ace must be the $n$th card drawn.
Once you start mixing an "order doesn't matter" probability space with "order does matter" events, it can get hard to tease out what's actually going on. But we can make one small adjustment to your sample space: count it as $n\cdot{^{52}C_n}$ equally likely outcomes, each outcome defined by the $n$ cards that were drawn and the identity of the one card among the $n$ that was drawn last.
Now that is a suitable sample space within which you can count $^{48}C_{n-2}\cdot 4$ possible combinations of cards that were not the last card drawn, multiply by the $3$ remaining possibilities for the card that was drawn last, and thereby obtain the number of favorable outcomes that can be divided by $n\cdot{^{52}C_n}.$