If I randomly choose my sample space, and then choose from that sample space, how would I find the number of possible values? I'm currently developing my own card game with probabilities, and I am currently working out the expected values and outcomes for it. The game involves choosing 12 cards from a standard 52-card deck, and then choosing 4 cards from the 12 cards (without replacement) that I picked previously, and then, after the 4 cards are picked, I replace them with 4 other cards from the 40 cards that I have left, until the 52 cards run out. 
My question is: How would I calculate the number of outcomes for the first "trial"? I have two possible answers:
${52 \choose 12}*{12 \choose 4}$, 
which creates approximately 1.0215781e+14 combinations (which I find worrying, due to the possibility of overcounting), or simply 
${52 \choose 4}$,
which would essentially take out the middleman. The problem is, I can only pick 4 cards out of the 12-card sample space, and not out of the whole 52-card deck. However, I am randomly picking the 12 cards from the 52-card deck, which leads me to believe that the ${52 \choose 12}$ is necessary in the equation.
Sorry if my question is mathematically poorly written, as I'm a high school student studying combinatorics and don't really know that much high-level math. 
Thanks in advance!
 A: As always in combinatorics, it depends on what you consider "equivalent".  
${52 \choose 12} {12 \choose 4}$ counts the number of ways to divide the deck into $3$ sets: of size $40$ (unselected in the first choice), size $8$ (unselected in the second choice), and size $4$ (selected into your hand).
$${52 \choose 12} {12 \choose 4} = {52! \over 40! \times 12!} \times {12! \over 8! \times 4!} = {52! \over 40! ~8! ~4!}$$
${52 \choose 4}$ counts the number of ways to divide the deck into $2$ sets: of size $48$ (unselected into your hand), and size $4$ (selected into your hand).
$${52 \choose 4} = {52! \over 48! ~ 4!}$$
So the question is: do you care about the split of the $48$ cards into $40+8$ (first model), or do you care only about what's in your hand (second model)?
As far as what's in your hand, all probabilities you can calculate will be equal, e.g. prob of all spades, one of each suit, one pair, two pairs, etc.  This is because even though the denominator changed, the numerator would change accordingly (if you do the math right) depending on which model your adopt.
