Drawing 3 cards from a deck 3 cards are drawn from a standard deck of 52 cards without replacement. Let $X_1$ be a random variable that is 1 if the 1st card drawn is a King and 0 otherwise $X_2$ be a random variable that is 1 if the 2nd card drawn is a king and 0 otherwise. What is the probability that $X_2 = 1$? 

My thinking is that $P(X_2 = 1) = P(X_2 = 1|X_1 = 1) + P(X_2 = 1|X_1 = 0)$, so $P(X_2 = 1) = \frac{\frac{4}{52} \times \frac{3}{51}}{\frac{4}{52}} + \frac{\frac{48}{52} \times \frac{4}{51}}{\frac{48}{52}} = \frac{7}{51}$.

Is this right way of thinking, or should the answer just be $\frac{4}{52}$ by symmetry?
You made a mistake in the first equation of not multiplying $P(X_1=1)$ and $P(X_1=0)$, it should be $$P(X_2=1)=P(X_1=1)P(X_2=1|X_1=1)+P(X_1=0)P(X_2=1|X_1=0)$$ and your answer is indeed $\frac{4}{52}$. (This equation is the Law of Total Probability)

Note: The calculation is $$P(X_2=1)=\frac{1}{13}\cdot \frac{3}{51}+\frac{12}{13}\cdot\frac{4}{51}=\frac{1}{13}$$
The probability of the first card being a King is 4/52, or 1/13. Should the first card had been a King, the probability of the second card being a King is now 3/51, or 1/17. However, if the first card wasn't a King, then the odds that the second card is a King is 4/51.

If you sum the two scenarios, applying a product of their condition, you get the following:

$P(X_2) = 1/13 * 1/17 + 12/13 * 4/51$

$= 1/221 + 48/663$

$= 1/221 + 16/221$

$= 17/221$

$= 1/13$

Edit:

Fixed fraction error.