There are $52$ cards in a deck. Find the probability that the $3$ cards add up to $7$. (Note:  Aces are considered a high card, not $1$)
Case 1:  You deal $3$ cards simultaneously, all sets of three cards equally likely.
Case 2:  You deal a card, note the draw, replace it and shuffle and pack, and follow the process $3$ times in all.
a) What is |Omega| or the size of the sample space of all outcomes?
b) Find the probability that the $3$ cards add up to $7$.
Please show details. I have tried several different ways. I do feel confident that Case $1$ sample space as $52$C$3$ and Case $2$ sample space as $52$x$52$x$52$. I am having trouble with the numerator. Help please. Thank you
 A: To do it in a CAS, we can use bivariate ordinary and exponential generating functions.
Case 1b:
\begin{align*}
  G_1(x,y) &= \left(\left(1+x y^A\right) \, \left(1+x y^K\right) \left(1+x y^Q\right) \left(1+x y^J\right)\prod_{i=1}^9 \left(1+x y^i\right)\right)^4
\end{align*}
$[x^3 y^7]G_1(x,y)$ is the required number of ways of choosing the cards without replacement and the probability is $\dfrac{24}{\binom{52}{3}}$
Case 2b:
\begin{align*}
  G_2(x,y) &= \exp\left(4x\left(y^A+y^K+y^Q+y^J+\sum_{i=2}^9 y^i\right)\right)
\end{align*}
$[x^3 y^7]G_2(x,y) \times 3!$ is the required number of ways of choosing the cards with replacement and the probability is $\dfrac{192}{52^3}$
$x$ keeps track of the number of cards chosen and $y$ keeps track of the sum.
A: Case 1:


*

*You select three of fifty-two cards without repetition. There are $^{52}{\rm C}_{3}$ ways to do this. (Order of draw is not important for the probability measure.)

*Arranging the cards from lowest to highest, the faces which add up to seven are: $\{1,1,5\}, \{1,2,4\}, \{1,3,3\}, \{2,2,3\}$. There are $4^3$ ways to select suits for three distinct faces, and there are $4\cdot{^{4}{\rm C}_{2}}$ ways to select suits for each of the remaining three groups (the "pairs").


*

*If Aces are not one, we only need count $\{2,2,3\}$



Case 2:


*

*You select three of thirteen faces with repetition allowed. There are $13^3$ ways to do this.   (We can ignore the suits; but order of draw is important for the probability measure.)

*Arranging the cards from lowest to highest, the faces which add up to seven are: $\{1,1,5\}, \{1,2,4\}, \{1,3,3\}, \{2,2,3\}$. However there are $3$ ways to arrange each of the three "pairs", and $3!$ ways to arrange the singleton hand.


*

*If Aces are not one, we only need count $\{2,2,3\}$


