Q.55 - A First Course in Probability

Question

Compute the probability that a hand of $13$ cards contains 

(a) the ace and king of at least one suit; 

(b) all $4$ of at least $1$ of the $13$ denominations. 

Step-by-Step Solution

Answer

a). The ace and king of at least one suit is $7.5\%$.

b). All $4$ of at least $1$ of the $13$ denominations are $3.42\%$.

Part (a) Step 1: Given Information

The cards are randomly dealt means that the probability of any of the $\left(\begin{array}{l}52\\ 13\end{array}\right)$ combinations of cards is equal, and since the probability of any event is $1$ .

$P\left(\text{ any specified hand }\right)=\frac{1}{\left(\begin{array}{l}52\\ 13\end{array}\right)} \stackrel{\text{ Axiom }3}{\Rightarrow } P\left(n\text{ specific hands }\right)=\frac{n}{\left(\begin{array}{l}52\\ 13\end{array}\right)}$

Part (a) Step 2: Probability calculation

If we choose the aces and kings that we want, the number of choices is the number of choices for the remaining cards from the rest of the deck.

And $k$ indexes out of $n$ (which is the same as $k$ ordered indexes) can be any of the $\left(\begin{array}{l}n\\ k\end{array}\right)$ combinations.

$$\begin{gathered} P\left(E_1\cup E_2\cup E_3\cup E_4\right)=\left(\begin{array}{l}4\\ 1\end{array}\right)P\left(E_1\right)-\left(\begin{array}{l}4\\ 2\end{array}\right)P\left(E_1\cap E_2\right)+\left(\begin{array}{l}4\\ 3\end{array}\right)P\left(E_1\cap E_2\cap E_3\right)-\left(\begin{array}{l}4\\ 4\end{array}\right)P\left(E_1\cap E_2\cap E_3\cap E_4\right) \\ =4P\left(E_1\right)-6P\left(E_1\cap E_2\right)+4P\left(E_1\cap E_2\cap E_3\right)-P\left(E_1\cap E_2\cap E_3\cap E_4\right) \\ =4\frac{\left(\begin{array}{l}50\\ 11\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)}-6\frac{\left(\begin{array}{c}48\\ 9\end{array}\right)}{\left(\begin{array}{c}52\\ 13\end{array}\right)}+4\frac{\left(\begin{array}{c}46\\ 7\end{array}\right)}{\left(\begin{array}{c}52\\ 13\end{array}\right)}-1\frac{\left(\begin{array}{c}44\\ 5\end{array}\right)}{\left(\begin{array}{c}52\\ 13\end{array}\right)} \\ \approx 0.075 \end{gathered}$$

Part (b) Step 1: Given Information

The cards are randomly dealt means that the probability of any of the $\left(\begin{array}{l}52\\ 13\end{array}\right)$ combinations of cards is equal, and since the probability of any event is $1$ .

$P\left(\text{ any specified hand }\right)=\frac{1}{\left(\begin{array}{l}52\\ 13\end{array}\right)} \stackrel{\text{ Axiom }3}{\Rightarrow } P\left(n\text{ specific hands }\right)=\frac{n}{\left(\begin{array}{l}52\\ 13\end{array}\right)}$

Part (b) Step 2: Probability calculation

The hand which contains $4$ cards of each of the $k$ denominations can vary in the remaining $13-4k$ cards. These can be any of the $\left(\begin{array}{c}52-4k\\ 13-4k\end{array}\right)$ combinations of the remaining $52-4k$ cards (not the ones already chosen)

And $k$ indexes out of $n$ (which is the same as $k$ ordered indexes) can be any of the $\left(\begin{array}{l}n\\ k\end{array}\right)$ combinations.

$$\begin{gathered} P\left(E_1\cup E_2\cup \dots \cup E_4\right)=\left(\begin{array}{c}13\\ 1\end{array}\right)P\left(E_1\right)-\left(\begin{array}{c}13\\ 2\end{array}\right)P\left(E_1\cap E_2\right)+\left(\begin{array}{c}13\\ 3\end{array}\right)P\left(E_1\cap E_2\cap E_3\right)-\left(\begin{array}{c}13\\ 4\end{array}\right)P\left(E_1\cap E_2\cap E_3\cap E_4\right)+\dots \\ =13P\left(E_1\right)-78P\left(E_1\cap E_2\right)+286P\left(E_1\cap E_2\cap E_3\right)-0 \\ =13\frac{\left(\begin{array}{c}48\\ 9\end{array}\right)}{\left(\begin{array}{c}52\\ 13\end{array}\right)}-78\frac{\left(\begin{array}{c}44\\ 5\end{array}\right)}{\left(\begin{array}{c}52\\ 13\end{array}\right)}+286\frac{\left(\begin{array}{c}40\\ 1\end{array}\right)}{\left(\begin{array}{c}52\\ 13\end{array}\right)} \\ \approx 0.0342 \end{gathered}$$