Question

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?

Answer 1

Answer:

$P =4.83\%$

Step-by-step explanation:

First we calculate the number of possible ways to select 2 cards an ace and a card of 10 points.

There are 4 ace in the deck

There are 16 cards of 10 points in the deck

To make this calculation we use the formula of combinations

$nCr=\frac{n!}{r!(n-r)!}$

Where n is the total number of letters and r are chosen from them

The number of ways to choose 1 As is:

$4C1 = 4$

The number of ways to choose a 10-point letter is:

$16C1 = 16$

Therefore, the number of ways to choose an Ace and a 10-point card is:

$4C1 * 16C1 = 4 * 16 = 64$

Now the number of ways to choose any 2 cards from a deck of 52 cards is:

$52C2 =\frac{52!}{2!(52-2)!}$

$52C2 = 1326$

Therefore, the probability of obtaining an "blackjack" is:

$P = \frac{4C1 * 16C1}{52C2}$

$P = \frac{64}{1326}$

$P = \frac{32}{663}$

$P =0.0483$

$P =4.83\%$

Answer 2

Answer:

Probability = 0.0483

Percentage = 4.83%

Step-by-step explanation:

We know that a blackjack hand played with one deck consists of:

1 of the 4 aces = $\frac{4}{52}$

So 1 out of the 16 cards worth 10 points will be equal to = $\frac{16}{52}$

Finding the probability of getting a blackjack hand assuming that the cards were not replaced:

P (blackjack hand) =  P(1st ace) × P(2nd 10 point card) +  P(1st 10 point card) × P(2nd ace)

P (blackjack hand) = $\frac{4}{52} \times \frac{16}{51} + \frac{16}{52} \times \frac{4}{51}$ = 0.04827

Percentage of getting blackjack hand = 4.83%