2
$\begingroup$

The four jacks, queens, and kings from a standard deck of cards are shuffled, and three cards are dealt to each of four players. Compute the probability that each player gets one jack, one queen, and one king.

I know that there are $12$ cards total and since every player gets one jack, one queen, and one king, would it be $\binom{12}3$ for the first player?

$\endgroup$
5
  • $\begingroup$ Have you calculated the denominator? $\endgroup$ Jul 13 '20 at 18:31
  • $\begingroup$ No, you have to distribute three of the twelve cards to the first player, three of the remaining nine cards to the second player, three of the remaining six cards to the third player, and all three of the remaining three cards to the fourth player. $\endgroup$ Jul 13 '20 at 18:38
  • $\begingroup$ You add when events are mutually exclusive. Since there are $\binom{9}{3}$ ways to distribute three cards to the second player for each of the $\binom{12}{3}$ ways you could distribute three cards to the second player, you need to multiply. This is an application of the Multiplication Principle rather than the Addition Principle. $\endgroup$ Jul 13 '20 at 18:44
  • $\begingroup$ So multiplying them is 369600 $\endgroup$
    – qs13
    Jul 13 '20 at 18:45
  • $\begingroup$ That is correct. Do you have any ideas on how to calculate the numerator. For instance, in how many ways can you distribute the kings to the four players so that each player receives one king? $\endgroup$ Jul 13 '20 at 18:47
3
$\begingroup$

Let's write up what we discussed in chat.

Since we have to distribute three of the twelve cards to the first player, three of the remaining nine cards to the second player, three of the remaining six cards to the third player, and give the fourth player all three of the remaining three cards, there are $$\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$ ways to distribute the twelve cards to four players so that each player receives three cards each.

If each player receives one king, then there are four ways to give one of the four kings to the first player, three ways to give one of the remaining three kings to the second player, two ways to give one of the remaining two kings to the third player, and one way to give the remaining king to the fourth player. Hence, there are $4! = 4 \cdot 3 \cdot 2 \cdot 1$ ways to distribute the four kings so that each player receives one. By symmetry, there are also $4!$ ways to distribute the queens so that each player receives one and $4!$ ways to distribute the jacks so that each player receives one. Hence, the number of favorable cases is $$4!4!4!$$ Therefore, the probability that each player receives one king, one queen, and one jack when the twelve face cards are distributed to four players when each player is dealt three cards is $$\frac{4!4!4!}{\dbinom{12}{3}\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}$$

$\endgroup$
1
$\begingroup$

Taking the probabilities of each card dealt we get this product of fractions:

$$\frac{12}{12}\frac{8}{11}\frac{4}{10}\frac{9}{9}\frac{6}{8}\frac{3}{7}\frac{6}{6}\frac{4}{5}\frac{2}{4}\frac{3}{3}\frac{2}{2}\frac{1}{1}=\frac{72}{1925} \approx 0.03740$$

I notice that this can be expressed as: $$\frac{(3!)^4(4!)^3}{12!}$$

I can't explain why though.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.