0
$\begingroup$

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.

$\endgroup$
3
  • 3
    $\begingroup$ What's the probability that the first is a $J$? Given that it was a $J$, what's the probability that the second is a $Q$? And so on. $\endgroup$
    – lulu
    Commented Jun 28, 2022 at 10:31
  • 3
    $\begingroup$ If order matters, the answer would be $\frac{4}{52} \times \frac{4}{51}\times \frac{4}{50}$. If it did not matter then it would be $3!=6$ times this. $\endgroup$
    – Henry
    Commented Jun 28, 2022 at 10:32
  • $\begingroup$ To emphasize, your answer of $\dfrac{\binom{4}{1}^3}{\binom{52}{3}}$ is correct for the question of order not mattering. To make order matter you can multiply by $\frac{1}{3!}$. This of course equals the same thing as what others are saying. You are perfectly allowed to begin with order not mattering and modify the result to make order matter or begin with order mattering and modify that result to make order not matter. There are cases where one will be notably easier than the other. Just be sure you understand how to modify it in either direction to suit your needs. $\endgroup$
    – JMoravitz
    Commented Jun 28, 2022 at 12:56

3 Answers 3

Reset to default
1
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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 $$ \frac{4 \times 4 \times 4}{52 \times 51 \times 50}. $$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ "...to over complicate with combinations." I take issue with that statement. It can be far preferred in some cases to begin with combinations and then choose to order after the fact rather than to start with order mattering. For example, "When dealt seven cards, what is the probability of being dealt three kings, three queens and one jack?" you could go directly and say $\dfrac{\binom{7}{3,3,1}4\cdot 3\cdot 2\cdot 4\cdot 3\cdot 2\cdot 4}{\frac{52!}{45!}}$ or you could say $7!\dfrac{\binom{4}{3}^2\cdot 4}{\binom{52}{7}}$. I far prefer this second expression as it looks much cleaner. $\endgroup$
    – JMoravitz
    Commented Jun 28, 2022 at 12:53
  • $\begingroup$ You have a very good point. I suppose it comes down to how you approach the problem and your previous understanding of combinatorics. The way you have explained it JMoravitz, I would argue is better for someone who has a complete understanding of combinations and hence can go from the general case to the specific. If you are less familiar with combinatorics manually calculating each probability would be simpler. $\endgroup$
    – Johnson
    Commented Jun 29, 2022 at 0:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .