Conditional from cards

Question

From a regular deck of 52 cards Let’s say I take 13 cards randomly and uniformly. Then someone takes 13 and then person 3 takes 13 and finally person 4 takes the remaining 13. each card is selected and uniformly. What is the probability each person has a queen if each person has a 4?

Edit: sorry guys for not putting my work I was just finishing up my solution, here it is, please let me know if it’s right or if I’m on the right track thanks. $ (4\binom{44}{11} *3\binom{33}{11}* 2\binom{22}{11})/\binom{48}{12}* \binom{36}{12}* \binom{24}{12}$

Answer 1

To compute the conditional probability, we first find the number of ways that each person gets a $4$. Well, there are $4!$ ways to hand out the $4$s, then $\binom{48}{12}$ ways to pick the remaining cards for the first person, $\binom{36}{12}$ ways to pick the remaining cards for the second person, and $\binom{24}{12}$ ways to pick the remaining cards for the third person. So, the total number of ways is $4! \binom{48}{12} \binom{36}{12} \binom{24}{12}$.

Now, we compute the probability that each person gets one $4$ and one $Q$. There are $(4!)^2$ ways to distribute the $4$ and $Q$ cards to each person. Following this, there are $\binom{44}{11}$ ways to complete the hand for the first person, $\binom{33}{11}$ for the second person, and $\binom{22}{11}$ for the third person. So, the number of successful ways is $(4!)^2 \binom{44}{11} \binom{33}{11} \binom{22}{11}$.

The desired probability is

$$\frac{(4!)^2 \binom{44}{11} \binom{33}{11} \binom{22}{11}}{4! \binom{48}{12} \binom{36}{12} \binom{24}{12}} = \frac{4! \binom{44}{11} \binom{33}{11} \binom{22}{11}}{\binom{48}{12} \binom{36}{12} \binom{24}{12}} = \boxed{\frac{576}{5405}}$$

The second fraction in the line above can be considered as giving out the $4$s initially, and computing the probability of each person getting one $Q$ given a deck with $48$ cards.

Answer 2

An easy way to tackle many conditional probability problems is to simply consider the reduced sample space, here $48$ cards, as it is given that each hand already has a $4$

Furthermore, the computations can be considerably simplified by visualizing $4$ equal groups of $12$ slots each, now available to receive cards.

The first Queen can go anywhere; the next has $36$ slots available out of the $47$ remaining, and so on, thus

$Pr = \large\frac{36}{47}\frac{24}{46}\frac{12}{45} = \frac{576}{5405}$

Answer 3

(posting as a community wiki, as this is a duplicate).

This is a duplicate of this question.

Indeed, given that each player has a four, we can forget about the $4's$ and just consider a $48$ card deck being dealt to four players. Following the logic of the duplicate, there are $\binom {48}4$ ways to place the Queens in the deck and $12^4$ ways to do it such that each player gets one $Q$. Thus the answer is $$\frac {12^4}{\binom {48}4}$$