Question

Find the probability of being dealt a bridge hand
of 13 cards containing 5 spades, 2 hearts, 3 diamonds,
and 3 clubs.

Answer 1

Answer on Question #72627, Math / Statistics and Probability

Find the probability of being dealt a bridge hand of 13 cards containing 5 spades, 2 hearts, 3 diamonds, and 3 clubs.

Solution

A bridge deck has 52 cards with 13 cards in each of four suits: spades, hearts, diamonds, and clubs.

Let $X_{1}, X_{2}, X_{3}$ and $X_{4}$ are the random variables which denote the number of spades, hearts, diamonds, and clubs respectively, in a bridge hand of 13 cards.

Thus, $X_{1}, X_{2}, X_{3}$ and $X_{4}$ jointly have a multivariate hypergeometric distribution with parameters, $N = 52, n = 13$ and $a_{i} = 13, \forall i = 1, 2, 3, 4$.

Joint probability mass function of $X_{1}, X_{2}, X_{3}$ and $X_{4}$ is given by

$$\begin{array}{l} h \left( \begin{array}{c} x_{1}, x_{2}, x_{3}, x_{4} \\ a_{1} = 13, a_{2} = 13, a_{3} = 13, a_{4} = 13 \end{array} \right) \\ N = 52, n = 13 \\ = \frac{\binom{a_{1}}{x_{1}} \binom{a_{2}}{x_{2}} \binom{a_{3}}{x_{3}} \binom{a_{4}}{x_{4}}}{\binom{N}{n}} \text{ with } \sum_{i=1}^{4} x_{i} = n \quad \text{and} \quad \sum_{i=1}^{4} a_{i} = N \\ = \frac{\binom{13}{x_{1}} \binom{13}{x_{2}} \binom{13}{x_{3}} \binom{13}{x_{4}}}{\binom{52}{13}} \\ \end{array}$$

We have that $x_{1} = 5, x_{2} = 2, x_{3} = 3, x_{4} = 3$. Then

$$\begin{array}{l} h \left( \begin{array}{c} x_{1} = 5, x_{2} = 2, x_{3} = 3, x_{4} = 3 \\ a_{1} = 13, a_{2} = 13, a_{3} = 13, a_{4} = 13 \end{array} \right) = \\ N = 52, n = 13 \\ = \frac{\binom{13}{5} \binom{13}{2} \binom{13}{3} \binom{13}{3}}{\binom{52}{13}} = \\ = \frac{\frac{13!}{5! (13 - 5)!} \cdot \frac{13!}{2! (13 - 2)!} \cdot \frac{13!}{3! (13 - 3)!} \cdot \frac{13!}{3! (13 - 3)!}}{\frac{52!}{13! (52 - 13)!}} = \\ = \frac{1287(78)(286)(286)}{635013559600} \approx \\ \approx 0.01293 \\ \end{array}$$

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