Probability of getting a jack, a queen and a king

Question

You randomly choose 3 cards without replacement from a deck of 52 cards. The question is what is the chance of choosing a jack, a queen and a king, where the order is important, but the color doesn't matter. Here I thought maybe using combinatorics. First there are ${4\choose 1}^3$ ways of choosing a jack, a queen and a king, because there are 4 different colours of each card and you only need one. My problem is I don't know what to do with the order of the cards, in other words how I can choose a jack first then a queen and lastly a king. Finally I suppose you divide ${4\choose 1}^3$ by ${52\choose 3}$ because you're choosing 3 cards out of a deck of 52 cards. So without knowing how to get the right answer my naive solution would be $\dfrac{{4\choose 1}^3}{{52\choose 3}}$.

I suppose if the order didn't matter then this would be the right answer, but I'm also not so sure about that. Any help would be greatly appreciated.

Answer 1

Taking three cards one after another, in this order Jack, Queen and King

$\text{P}=\frac{\binom{4}{1}}{\binom{52}{1}}\times \frac{\binom{4}{1}}{\binom{51}{1}}\times \frac{\binom{4}{1}}{\binom{50}{1}}= \frac{4}{52}\times \frac{4}{51}\times \frac{4}{50}$

If order is not important, Jack, Queen and King can be arranged in $3!$ different ways.

Probability will be

$\text{P}=3! \times \frac{4}{52}\times \frac{4}{51}\times \frac{4}{50}$

Answer 2

This all seems awfully complicated. It also feels unnatural to use $\binom{a}{b}$ when it says "order is important". The probability is just $$ \frac{\text{the number of (ordered) ways of choosing a jack, then a queen, then a king}}{\text{the number of (ordered) ways of choosing 3 cards}} $$ which is $$ \frac{4 \times 4 \times 4}{52 \times 51 \times 50}. $$

Answer 3

In cases where order matters, there is no need to over-complicate the answer with combinations. Simply find the joint probability of the desired number of jacks, queens, kings, etc exactly how Rebecca J. Stones has done.

If you want/need to extent the scope of your question, then consider using hyper geométrics for probabilities without replacement and where order does not matter. But for a single permutation, manually listing the order is a very viable solution especially for your proposed problem.