Respuesta :
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
The Player’s expected payoff is $ 240.
Probability the player loses $1000 or more is 0.1788.
Probability the player wins is 0.3557.
Probability of going broke is 0.0594.
Given that,
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
- 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(loss)
= 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
- By 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.
To Compute z-score:
[tex]z = \frac{x - np}{\sqrt{no (1-p )} } \\\\z = \frac{110.5 - 117.6}{\sqrt{117.6 ( 1-0.49)} }\\\\z = \frac{- 7.1}{\sqrt{117.6 (0.51) } } \\\\z = \frac{-7.1}{\sqrt{59.97} } \\\\z = \frac{-7.1}{7.74}\\\\z = - 0.916[/tex]
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
- By using normal approximation of binomial distribution: Both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution,
To compute z-score:
[tex]z = \frac{x - np}{\sqrt{no (1-p )} } \\\\z = \frac{120.5 - 117.6}{\sqrt{117.6 ( 1-0.49)} }\\\\z = \frac{2.9}{\sqrt{117.6 (0.51) } } \\\\z = \frac{2.9}{\sqrt{59.97} } \\\\z = \frac{2.9}{7.74}\\\\z = 0.34[/tex]
Here the player wins when he wins at least 120 of 240 hands.
By using normal probability table:
P(X>120) = P(X>120.5)
= 1 - P(Z<0.3744)
= 1 - 0.6443
= 0.37
- The player starts with $1500 the probability of going broke,
Both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution.
To Compute z-score:
[tex]z = \frac{x - np}{\sqrt{no (1-p )} } \\\\z = \frac{105.5 - 117.6}{\sqrt{117.6 ( 1-0.49)} }\\\\z = \frac{-12.1}{\sqrt{117.6 (0.51) } } \\\\z = \frac{-12.1}{\sqrt{59.97} } \\\\z = \frac{-12.1}{7.74}\\\\z = -1.56[/tex]
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)
= 0.54
For the more information about Probability distribution click the link given below.
https://brainly.com/question/12905909
