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

Probability = 0.0483

Percentage = 4.83%

Explanation:

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

1 of the 4 aces =
`(4)/(52)`

So 1 out of the 16 cards worth 10 points will be equal to =
`(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) =
`(4)/(52) * (16)/(51) + (16)/(52) * (4)/(51)` = 0.04827

Percentage of getting blackjack hand = 4.83%

Answer 2

`P =4.83\%`

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=(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 =(52!)/(2!(52-2)!)`

`52C2 = 1326`

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

`P = (4C1 * 16C1)/(52C2)`

`P = (64)/(1326)`

`P = (32)/(663)`

`P =0.0483`

`P =4.83\%`