# Probability in deck of only face cards.

I want to find the probability that in a deck of only face cards (jacks, queens, kings, each with their standard 4 suits), you draw exactly 2 queens and exactly 2 kings. Only 4 cards are being drawn.

Also, the same question again but where each queen has a king of the same matching suit. I was thinking we use counting for this so would the first one just be 5 divided by 13 choose 4? Where 5 is the permutations of the four cards (2 queens and 2 kings). For the second I am not so sure, any pointers?

• Where did you get 5 from? Oct 3 '14 at 12:37
• Wait. @user180652, how many cards are being drawn? Oct 3 '14 at 12:43

Our mini-deck has $12$ cards. There are $\binom{12}{4}$ equally likely ways to choose $4$ cards from this deck.

Now we count the favourables, the hands that have $2$ Queens and $2$ Kings.

The $2$ Queens can be chosen from the $4$ available in $\binom{4}{2}$ ways. For each way of choosing $2$ Queens, there are $\binom{4}{2}$ ways to choose the $2$ Kings. Thus the total number of favourables is $\binom{4}{2}\binom{4}{2}$.

For the probability, divide.

For the second problem, there are $\binom{4}{2}$ ways to choose the $2$ Queens. Once this is done, the Kings are determined, since they must be of the same suits as the chosen Queens. Thus the number of favourables in the "suits must match" problem is just $\binom{4}{2}$.

There are 12 cards in the deck. You draw four of them into a hand. The probability space is ${12\choose 4}$ equally possible outcomes.

The count of outcomes for the event of drawing 2 of 4 queens and 2 of 4 kings is : $\underline{\qquad}$?

The count of outcomes for the event of drawing 2 of 4 queens and 2 kings of the same suit as the queens is: $\underline{\qquad}$?

On the other hand if you're drawing 5 cards into a hand, the probability space is ${12\choose 5}$ outcomes.

The count of outcomes for the event of drawing 2 of 4 queens, 2 of 4 kings, and 1 of 4 jacks is : $\underline{\qquad}$?

The count of outcomes for the event of drawing 2 of 4 queens and 2 kings of the same suit as the queens, and 1 of 4 jacks is: $\underline{\qquad}$?

You are starting with $3$*$4$ = $12$ cards ($3$ ranks and $4$ suits for each).

To get $2$ queens and $2$ kings, that would be ${4 \choose 2} {4 \choose 2} / {12 \choose 4}$ = $36$ / $495$ = about $7.3$%

For the 2nd part of your question where the suits of the queens and kings must match, it would be ${4 \choose 2} 1 / {12 \choose 4}$ = $6/495$ = about $1.2$%. The $1$ on the left side is because once you choose the $2$ queens, there is only $1$ pair of kings that match those same $2$ suits of the pair of queens. I put it there for clarity. Notice it makes the probability $1/6$th that of allowing any $2$ kings as in the first part of your question because there is only $1$ pair of kings that match suits (of the $2$ queens) vs. $6$ pair of any kings.