4
$\begingroup$

A regular deck of 52 playing cards has 13 ranks in 4 suits. The ranks of Jack, Queen, King and Ace of each suit are top cards. Suppose you are randomly dealt seven cards. What is the probability of getting three top cards in the same suit and any four cards in another suit (but all four in one suit)?

This is what I was thinking of calculating but am completely unsure:

$$ { \begin{pmatrix} 4 \\ 3 \\ \end{pmatrix} × \begin{pmatrix} 39 \\ 4 \\ \end{pmatrix} × \begin{pmatrix} 9 \\ 4 \\ \end{pmatrix} \over \begin{pmatrix} 52 \\ 7 \\ \end{pmatrix} } = 0.30986...$$

Any help appreciated. Thanks.

$\endgroup$
2
  • $\begingroup$ That answer seems rather high, no? $\endgroup$
    – lulu
    Commented Feb 25, 2018 at 18:57
  • $\begingroup$ Yes, that's what I thought when I compared it to previous questions $\endgroup$
    – uahimself
    Commented Feb 25, 2018 at 19:06

1 Answer 1

1
$\begingroup$

There are $\binom{4}{1}$ ways to choose the suit from which the three top cards are drawn, $\binom{4}{3}$ ways to choose three top cards of that suit, $\binom{3}{1}$ ways to choose a suit from which four cards are drawn, and $\binom{13}{4}$ ways to choose four cards of that suit. Hence, the number of favorable cases is $$\binom{4}{1}\binom{4}{3}\binom{3}{1}\binom{13}{4}$$ Since there are $\binom{52}{7}$ ways to select seven cards from the deck, the desired probability is $$\frac{\dbinom{4}{1}\dbinom{4}{3}\dbinom{3}{1}\dbinom{13}{4}}{\dbinom{52}{7}}$$

$\endgroup$

You must log in to answer this question.

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