Question

In Exercises 5–36, express all probabilities as fractions.

Blackjack In the game of blackjack played with one deck, a player is initially dealt 2 different cards from the 52 different cards in the deck. A winning “blackjack” hand is won by getting 1 of the 4 aces and 1 of 16 other cards worth 10 points. The two cards can be in any order. Find the probability of being dealt a blackjack hand. What approximate percentage of hands are winning blackjack hands?

Short answer

The probability of getting a blackjack hand is $\frac{32}{663}$.

The percentage of getting a blackjack hand is 4.9%.

Step-by-step solution

Given information

A blackjack hand is obtained when out of the two given cards, one is ace the other is one of the 16 cards worth 10 points.

Define combination

To count ways to select a certain number of units from the specified total, the combination rule is applied. The selections are done without replacement and their order is not important.

The formula for r selections from n is shown below:

${}^n{C}_r=\frac{n!}{\left(n-r\right)!r!}$

Compute the counts of getting the blackjack

Let A be the event of getting a blackjack hand.

Case 1: Getting one ace

The total number of cards is 52.

The number of aces is 4.

The number of ways in which one out of four aces can be selected without replacement (in any order) is

$$\begin{gathered} {}^4{C}_1=\frac{4!}{\left(4-1\right)!1!} \\ =4. \end{gathered}$$

The probability of selecting an ace out of 52 cards is

$P\left(\text{getting}\text{an}\text{ace}\right)=\frac{4}{52}$

Case 2: Getting one out of 16 cards worth points

The total number of cards is 52.

The number of cards worth points is 16.

The number of ways in which one card can be chosen from the 16 cards worth points is

$$\begin{gathered} {}^{16}{C}_1=\frac{16!}{\left(16-1\right)!1!} \\ =16. \end{gathered}$$

The probability of selecting one card worth 10 points out of the 52 cards is

$P\left(\text{getting}\text{1}\text{card}\text{worth}\text{points}\right)=\frac{16}{52}.$

Compute the probability of getting the blackjack

Here, selections are made without replacement. So, the selection of the second card will be out of 51 remaining cards.

Also, the selections are made in any order.

The probability of choosing first an ace and then a card worth points is

$\frac{4}{52}\times \frac{16}{51}=\frac{64}{2652}$

The probability of choosing first a card worth points and then an ace is

.$\frac{16}{52}\times \frac{4}{51}=\frac{64}{2652}$

The probability of getting a blackjack hand is the sum of the two probabilities computed above:

$$\begin{gathered} P\left(A\right)=\frac{64}{2652}+\frac{64}{2652} \\ =\frac{32}{663} \\ =0.049 \end{gathered}$$

Therefore, the probability of getting the blackjack hand is 0.049.

Compute the percentage equivalent probability

The percentage of getting a blackjack hand is

$0.049\times 100=4.9\%$

Therefore, the percentage of getting a blackjack hand is 4.9%.