Finding a King and a Queen in 3 Cards

Question

I'm looking at a problem for a card game where you already have a king in hand and need to find a king and a queen when drawing 3 cards. Right now I'm looking at something like:

$3! \times \frac{3}{51} \times \frac{4}{50}$

But I am pretty sure that this is not correct, as the order matters if the king or queen is found first because the chance of finding either is not the same.

EDIT: For simplicity's sake, let's assume that the only card not in the deck is the one king in hand. Also, the third card does not matter and is allowed to be another king or queen or any other card.

Answer 1

You are close but you double count all draws of $KKQ$ or $KQQ$ twice, once for each of the paired cards. You need to subtract those once, so it becomes $$3!\cdot \frac 3{51}\cdot \frac 4{50}-3\cdot \frac 3{51}\cdot \frac 4{50}\cdot \frac 2{49}-3\cdot \frac 3{51}\cdot \frac 4{50}\cdot \frac 3{49}$$

Answer 2

Do you want the probability that at least one from seven favoured cards will be among the next three cards drawn from the fifty-one cards remaining?   The complement of this event is that none from the seven will be drawn when drawing three from fifty one.

$$1-\frac{\binom 7 0\binom {51-7}3}{\binom {51}3}$$

Do you want the probability that at least one from three(remaining) kings and at least one from four queens will be among the next three cards drawn from the fifty-one cards remaining?   The principle of inclusion and exclusion will be used.

$$1-\frac{\binom 3 0\binom {51-3}3}{\binom {51}3}-\frac{\binom 4 0\binom {51-4}3}{\binom {51}3}+\frac{\binom 7 0\binom {51-7}3}{\binom {51}3}$$

Answer 3

There are 6 in total- (KQN,KNQ,NKQ)x2 with K,Q interchanged.

Now we have to avoid double counting. So -

$$P(KQN) = \frac{3}{51}. \frac{4}{50} $$ $$P(QKN) = \frac{4}{51}. \frac{3}{50} $$ $$P(KNQ) = \frac{3}{51} .\frac{46}{50}. \frac{4}{49}$$ $$P(QNK) = \frac{4}{51} .\frac{47}{50}. \frac{3}{49}$$ Any of 50 card except the queens/kings for 2nd pick. $$P(NKQ) = \frac{44}{51}.\frac{3}{50} .\frac{4}{49}$$ $$P(NQK) = \frac{44}{51}.\frac{4}{50} .\frac{3}{49}$$ Any of 51 cards excpet kings and queens for 1st pick.

$$P(total) = \sum_{all} P = \frac{279*3*4}{51*50*49}$$

(This turns to be same as Ross's answer)