Two cards are drawn together from a pack of $52$ cards. What is the probability that one is a spade and one is a heart? I can solve this but I get confused about replacement or not. The textbook gives the answer as $\frac{13}{102}$ which means they count the number of ways to pick a spade ${13\choose1}$ times the number of ways to pick a heart ${13\choose1}$ over the total number of ways to pick $2$ cards ${52\choose2}$. This means they consider the $2$ card draws to be independent. However, if I do probabilities:
$$
P(S \cap H) = P(S)P(H | S) = \frac{1}{4} \times \frac{13}{51} = \frac{13}{204}.
$$
I'm not sure who's right and who isn't. For me, it makes more sense that the answer is $\frac{13}{204}$ because there is no way to draw $2$ cards independently unless you draw one, replace it and reshuffle the deck and draw the other. However, the question does not say this is the case $-$ it just says "two cards are drawn".
 A: Cards are selected together
$$\frac{C(13;1)\times C(13;1)}{C(52;2)}$$
A: The problem is about selecting a random subset with 2 elements. Of all the 2-element subsets $13 \times 13$ comprise a heart and a spade, leading to the textbook answer of $13/102$. You are calculating the probability of getting a heart followed by a spade when you select the cards one after another.
A: 
because there is no way to draw 2 cards independently unless you draw one, replace it and reshuffle the deck and draw the other.

Why do you think the first answer requires independence?  If we did it with independence the probability the first is a heart then a a spade would be $\frac 14\cdot \frac 14$ and the probability of  a spade then a heart would be $\frac 14\cdot \frac 14$ and the probability would be $\frac 14 \frac 14 + \frac 14  \frac 14 = \frac 18$.

$P(S \cap H) = P(S)P(H | S) = \frac{1}{4} \times \frac{13}{51} = \frac{13}{204}.$

That's not quite right.  What does "$S$" mean? Does it mean a specific card is a spade?  Or does it mean at least one of two cards is a spade? Or something else.   If $P(S) = \frac 14$ this seems to imply you mean a specific card is a spade.  But then $P(S\cap H)$ would mean the probability that a specific card is both a heart and a spade and $P(H|S)$ is the probability of a card being a spade given we know it is a heart.  (So $0 = P(S\cap H) = P(S)P(H|S) = \frac 14\cdot 0 = 0$.)
Or maybe $S$ means a specific card is a spade, and $H$ means the other card is a heart.  Then your calculation is correct.....  But you figured the probability that a specific card is a spade and the  other specific card is a heart.... And that was not the question.  The question was that either card is a spade and the other card is a heart.
If $S$ at least one card of two is a spade and $H$ is at least one card of two is a heart then to calculate conditional probability would go like this:
$P(H\cap S) = P(S)P(H|S)$ is $\frac {13*39 + 39*13+13*13}{52*51}\cdot \frac{ 13*13 + 13*13}{13*39 + 39*13 + 13*13}=$
$\frac {13\cdot 26}{52*51}= \frac {13}{102}$
But that's a ridiculously hard way to do it.
Better to either figure there are $2\times 13 \times 13$ (heart, spade) and (spade, heart) pairs where order matters out of $52\times 51$ combos;  or there are $13\times 13$ (heart,spade) pairs were order doesn't matter out of ${52\choose 2}$ combos.
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tl;dr
you figured out the probability of a specific card being a spade and the other not being.  As order does not matter the probability is one half of that.
