Q.2.2 - A First Course in Probability

Question

Suppose that you are playing blackjack against a dealer. In a freshly shuffled deck, what is the probability that neither you nor the dealer is dealt a blackjack?

Step-by-Step Solution

Answer

The probability that neither you nor the dealer is dealt a blackjack is $0.952178$.

Step 1 Given Information.

Given that you are playing blackjack against a dealer. In a freshly shuffled deck,

Step 2 Explanation.

The card values are as follows: Cards $2-10$are valued at the face value of the card. Face cards such as the King, Queen, and Jack are valued $10$each. The Ace card is valued at$11$.

Both the player and the dealer are dealt a pair of cards each. A blackjack is when the total value of a pair of cards is exactly$21$. This happens only when one of the cards is an ace and the other card is a value $10$card.

Let's calculate the probability, $P$that at least one of them gets a blackjack. Then the probability that none of them gets a blackjack is$1-P$.

Consider the following events:

$A:$ the player gets a blackjack. The number of aces is 4 and the number of cards with a value$10$ is$16$. Therefore the number of ways this can happen is${41650}^{*}49$. (There are four ways for the player to get an ace, $16$ways for him to get a$10$ value card, and then the dealer can get any )

$B:$the dealer gets a blackjack. Proceeding the same way as$A$ again we get the number of ways to be$41650 * 49$.

$A\cap B$: i.e. both of them get a blackjack. there are${4163}^{*}15$ ways the player can get an ace in $4$some ways and a $10$value card in $16$ways. The dealer has to get one of the $3$remaining aces and one of the $15$remaining$10$ value cards.

A total number of card configurations$={525150}^{*}49$.

$$\begin{gathered} P =P(A \cup B) \\ =P\left(A\right)+P\left(B\right)-P(A\cup B) \\ =\frac{4*16*50*49}{52*51*50*49}+\frac{4*16*50*49}{52*51*50*49}-\frac{4*16*3*15}{52*51*50*49} \\ =0.0478222 \end{gathered}$$

The required probability is$1-P=1-0.0478222=0.952178$.