What is the probability of choosing a king for the second card drawn if the first card drawn without replacement was a queen?

If 2 cards are selected from a standard deck of 52 cards without replacement, in order to find the probability that both are the same suit, start with the first card...The probability that the first card is any suit is 52 in 52, or 1.Now, consider the second card. There are 12 cards remaining in the same suit, and 39 cards remaining in the other three suits...The probability that the second card is the same suit as the first card is 12 in 51, or 4 in 17, or 0.235.The probability of both events occurring is the product of those two probabilities. That is still 4 in 17, or 0.235.

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The probability of getting drawing a king and queen from a deck of 52 cards without replacement is $\frac{4}{52} \frac{4}{51}$. I'm confused why it's not twice of this. We could achieve a king and queen in 2 different ways. First drawn card is a king and second drawn card is a queen, or first drawn card is a queen and second drawn card is a king. These are disjoint events, so wouldn't the probability actually be $\frac{4}{52} \frac{4}{51} + \frac{4}{52} \frac{4}{51}$?

I saw the first result left by the comment by tpb261 in Probability in cards that $4$ people each get queen and king?.