A simple problem. What am I doing wrong? I am totally new to probability and I am a little bit confused. I have the following homework:

A large group of people are competing for all-expense-paid weekends in Philadelphia. The Master of Ceremonies gives each contestant a well-shuffled deck od cards. The contestant deals two cards off the top of the deck, and wins a weekend if the first card is the ace of hearts or the second card is the king of hearts. What is the probability of wining the weekend?

I tried to solve this exercise in three ways:


*

*Using $P(A ∪ B) = P(A) + P(B) - P(A ∩ B)$. I get:$$\frac{1}{52} + \frac{1}{52} - \frac{1}{52}×\frac{1}{51} = \frac{101}{51×52}.$$

*Using $P(A ∪ B) = P(A) + P(B ∩ A^c)$. I get:$$\frac{1}{52} + \frac{1}{52}×\frac{50}{51} = \frac{101}{51×52}.$$

*Using formula $P(A) = 1 - P(A^c)$ where the opposite is not getting the ace of hearts as the first card and not getting the king of hearts as the second card. In this way I get:$$1 - \frac{51}{52}×\frac{50}{51} = \frac{2}{52} \ne \frac{101}{51×52}.$$
What am I doing wrong in the third way? Thank you in advance for your help.
 A: The problem with your computation in #3 is that you didn't account for the possibility that the first card was the king of hearts.
A: In the third way, the formula is ($A-$ ace of hearts, $B-$ king of hearts):
$$P(A\cup B)=1-P(A^C\cap B^C)=1-\frac{51}{52}\cdot \color{red}{\frac{50}{51}},$$
however, the event $B^C$ depends on the event $A^C$. In other words, you assumed the first card is not king of hearts, however:
$$P(B_2^C|B_1)=1; P(B_2^C|B_1^C)=\frac{1}{51}.$$
Probability tree diagram ($A'=A^C$):
$\hspace{1cm}$
$$B\cup B'=S; A'\cap S=A';\\
P(A'\cap B')=P(\color{blue}{(A'\cap (B\cup B'))}\cap \color{green}{B'})=\\
P(\color{blue}{([A'\cap B]\cup [A'\cap B'])}\cap \color{green}{B'})=\\
P(\{\color{blue}{[A'\cap B]}\cap \color{green}{B'}\}\cup \{\color{blue}{[A'\cap B'])}\cap \color{green}{B'}\})=\\
P(\color{blue}{[A'\cap B]}\cap \color{green}{B'})+P(\color{blue}{[A'\cap B'])}\cap \color{green}{B'})=\\
\frac{1}{52}\cdot \frac{51}{51}+\frac{50}{52}\cdot \frac{50}{51}.$$
Note: The events in blue color are the first card, while in green color are the second card. They can be differentiated by relevant indices (subscripts) as labeled by Graham Kemp in his comment.
Hence:
$$\begin{align}P(A\cup B)&=1-P(A'\cap B')=\\
&=1-\left(\frac{1}{52}\cdot \frac{51}{51}+\frac{50}{52}\cdot \frac{50}{51}\right)=\\
&=\frac{101}{51\cdot 52}.\end{align}$$
