How to apply probability with a deck of cards For this question, I'm not sure if I am doing it right. Here is what I have so far? Can anyone please help me out?
Suppose we roll a fair six-sided die and then pick a number of cards from a
well-shuffled deck equal to the number showing on the die. (For example, if the die
shows $4$, then we pick $4$ cards.)
What is the probability that the number of jacks in our hand equals $2$?
$P(\text{rolling a number on die}) = \frac{1}{6}$
$P(\text{number of jacks $= 2$}) = \frac{1}{6}\times{^{52}C_4}\times{^{48}C_2}$
 A: Solution. Let $\mathcal{J}$ denote the event that exactly two jacks are selected in addition let $D_i$ denote the event that the die roll yielded $i\in\{1,2,3,4,5,6\}$. We may compute $\mathbf{P(\mathcal{J})}$ by making use of bayes's formula if you observe that 
$$\mathbf{P}(\mathcal{J}) = \sum_{k=2}^{6}\mathbf{P}(\mathcal{J}|D_k)\mathbf{P}(D_k) = \frac{\binom{4}{2}\binom{48}{0}}{\binom{52}{2}}\cdot\frac{1}{6}+\frac{\binom{4}{2}\binom{48}{1}}{\binom{52}{3}}\cdot\frac{1}{6}+\dots+\frac{\binom{4}{2}\binom{48}{4}}{\binom{52}{6}}\cdot\frac{1}{6}$$
A: Suppose the die shows $n$ (where $n \geq 2$). Then, the probability that among $n$ randomly pocked cards, the number of Jacks is equal to $2$ is 
$$\frac{\binom{4}{2}\binom{48}{n-2}}{\binom{52}{n}}.$$
Namely, there are $\binom{4}{2}$ events that we choose $2$ out of $4$ Jacks and $\binom{48}{n-2}$ events that we choose other $n-2$ cards.  Thus we have the numerator of the expression.  It is easy to see that there are $\binom{52}{n}$ possibilities of choosing $n$ cards randomly, and this is the denominator.  
Next, we are interested in all $n$ numbers that is at least $2$, or $n = \{2,3,4,5,6\}$.  The probability that each of those number show is $\frac{1}{6}$.  Then, we sum up over all $n$ to obtain the final answer:
$$\frac{1}{6}\sum_{n=2}^6 \frac{\binom{4}{2}\binom{48}{n-2}}{\binom{52}{n}}.$$
