Question

A blackjack player at a Las Vegas casino learned that the house will provide a free room if play is for four hours at an average bet of $50. The player’s strategy provides a probability of .49 of winning on any one hand, and the player knows that there are 60 hands per hour. Suppose the player plays for four hours at a bet of $50 per hand.
a. What is the player’s expected payoff?

b. What is the probability the player loses $1000 or more?

c. What is the probability the player wins?

d. Suppose the player starts with $1500. What is the probability of going broke?

Answer 1

Answer:

a) player’s expected payoff is $ 240

b) probability the player loses $1000 or more is 0.1788

c)  probability the player wins is 0.3557

d) probability of going broke is 0.0594

Step-by-step explanation:

Given:

Since there are 60 hands per hour and the player plays for four hours then the sample size is:

n = 60 * 4 = 240

The player’s strategy provides a probability of .49 of winning on any one hand so the probability of success is:

p = 0.49

a)

Solution:

Expected payoff is basically the expected mean

Since the bet is $50 so $50 is gained when the player wins a hand and $50 is lost when the player loses a hand. So

Expected loss =  μ

                        = ∑ x P(x)

                        = 50 * P(win) - 50 * P(lose)

                        = 50 * P(win) + (-50) * (1 - P(win))

                         = 50 * 0.49 - 50 * (1 - 0.49)

                        = 24.5 - 50 ( 0.51 )

                        = 24.5 - 25.5

                        = -1

Since n=240 and expected loss is $1 per hand then the expected loss in four hours is:

240 * 1 = $ 240

b)

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

  = 110.5 - 117.6 / √117.6(1-0.49)

  = −7.1/√117.6(0.51)

  = −7.1/√59.976

  = −7.1/7.744417

  =−0.916789

Here the player loses 1000 or more when he loses at least 130 of 240 hands so the wins is 240-130 = 110

Using normal probability table:

P(X≤110) = P(X<110.5)

             = P(Z<-0.916)

             = 0.1788

c)

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

  = 120.5 - 117.6 / √117.6(1-0.49)

  = 2.9/√117.6(0.51)

  = 2.9/√59.976

  = 2.9/7.744417

  =0.374463

Here the player wins when he wins at least 120 of 240 hands

Using normal probability table:

P(X>120) = P(X>120.5)

              = P(Z>0.3744)  

             =  1 - P(Z<0.3744)

             = 1 - 0.6443

             = 0.3557

d)

Player goes broke when he loses $1500

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

  = 105.5 - 117.6 / √117.6(1-0.49)

  = -12.1/√117.6(0.51)

  = -12.1/√59.976

  = -12.1/7.744417

  =−1.562416

Here the player loses 1500 or more when he loses at least 135 of 240 hands so the wins is 240-135 = 105

Using normal probability table:

P(X≤105) = P(X<105.5)

             = P(Z<-1.562)

             = 0.0594