Question

Blackjack, or twenty-one as it is frequently called, is a popular gambling game played in Las Vegas casinos. A player is dealt two cards. Face cards (jacks, queens, and kings) and tens have a point value of 10. Aces have a point value of 1 or 11. A 52-card deck contains 16 cards with a point value of 10 (jacks, queens, kings, and tens) and four aces.
a. What is the probability that both cards dealt are aces or 10-point cards (to 4 decimals)?

b. What is the probability that both of the cards are aces (to 4 decimals)?

c. What is the probability that both of the cards have a point value of 10 (to 4 decimals)?

d. A blackjack is a 10-point card and an ace for a value of 21. Use your answers to parts (a), (b), and (c) to determine the probability that a player is dealt blackjack (to 4 decimals). (Hint: Part (d) is not a hypergeometric problem. Develop your own logical relationship as to how the hypergeometric probabilities from parts (a), (b), and (c) can be combined to answer this question.)

Answer 1

Answer:

a, 0.1433

b. 0.0045

c. 0.0905

d. 0.0483

Step-by-step explanation:

We have that the formula for hypergeometric probability is:

          (r) (N - r)

          (x) (n - x)

f (x) = ------------------

               (N)

               (n)

where r is "the number of individuals that y x value that the variable takes.

We have that the population size is 52. N = 52

The number of draws is equal to 2. n = 2

Now, knowing this we solve each one:

a. In this case the values are as follows: r = 20; x = 2

          (20) (52-20)

          (2) (2 - 2)

f (x) = ------------------ = 95/663 = 0.1433

               (52)

               ( 2 )

b. In this case the values are as follows: r = 4; x = 2

          (4) (52-4)

          (2) (2 - 2)

f (x) = ------------------ = 1/221 = 0.0045

               (52)

               ( 2 )

c. In this case the values are as follows: r = 16; x = 2

          (16) (52-16)

          (2) (2 - 2)

f (x) = ------------------ = 20/221 = 0.0905

               (52)

               ( 2 )

d. The probability of blackjack would come being in relation to a, b, c

That is to say:

p (blackjack) = a - b - c

Replacing:

p (blackjack) = 0.1433 - 0.0045 - 0.0905 = 0.0483

Answer 2

Answer:

a) The porbability that both cards dealt are 10-point cards or aces is 0.1433

b) The probability that 2 aces are dealt is 0.0045

c) The probability that the two cards dealt are 10-point cards is 0.0905

d) The probability that a player is dealt a blackjack is 0.0483

Step-by-step explanation:

a) Since there are 52 cards, there are ${52 \choose 2} = 1326$ ways to deal them; also, there are 16+4 = 20 cards that are aces or 10-point cards, thus, there are ${20 \choose 2} = 190$ ways of dealing 2 cards that are 10-point cards or Aces. Since the probability for each individual hand is the same, then, the porbability that both cards dealt are 10-point cards or aces is 190/1326 = 0.1433.

b) There are ${4 \choose 2} = 6$ ways to deal 2 aces. As a result, the probability that 2 aces are dealt is 6/1326 = 0.0045.

c) There are ${16 \choose 2} = 120$ ways to deal two 10-point cards. Therefore, the probability that the two cards dealt are 10-point cards is 120/1326 = 0.0905.

d) There are 3 distinct ways of obtaining a pair of 10-point cards/Aces. Either you obtain two Aces, or you obtain two 10-point cards or you obtain 1 from each (which results in a balck jack). Since each possibility is mutually exclusive from the others, then

P(obtain 2 cards that are Aces or 10-point cards) = P( two 10-point cards) + P(two Aces) + P(Blackjack)

The probability to obtain 2 cards that are Aces or 10-point cards were computed in item (a) and it was 0.1433, and the other 2 probabilities besides blackjack were oobtained earlier in items (b) and (c). As a consequence

0.1433 = 0.0905 + 0.0045 + P(Blackjack)

So we conclude that

P(Blackjack) = 0.1433-0.0905-0.0045 = 0.0483