Respuesta :
Answer:
a) E(X) = -$0.0813 , s.d (X) = 3
b) E(X) = -$0.0813 , s.d (X) = 3
c) expected loss and higher stakes of loosing.
Step-by-step explanation:
Given:
- There are total 37 slots:
Red = 18
Black = 18
Green = 1
- Player on bets on either Red or black
- Wins double the bet money, loss the best is lost
Find:
a) Expected value of earnings X if we place a bet of $3
b) Expected value and standard deviation if we bet $1 each on three rounds
c) compare the two answers in a and b and comment on the riskiness of the two games
Solution:
- Define variable X as the total winnings per round. We will construct a distribution tables for total winnings per round for bets of $3 and $ 1:
- Bet: $3
X -3 3 E(X)
P(X) 1-0.48 = 0.5135 18/37 = 0.4864 3*(.4864-.52135) = -0.08
-The s.d(X) = sqrt(9*(0.5135 + 0.4864) - (-0.08)^2) = 3.0
- Bet: $1
X -1 1 E (X)
P(X) 1-0.48 = 0.5135 18/37 = 0.4864 1*(.4864-.5135) = -0.0271
-The s.d(X) = sqrt(1*(0.5135 + 0.4864) - (-0.0271)^2) = 0.999
- The expected value for 3 rounds is:
E(X_1 + X_2 + X_3) = E(X_1) + E(X_2) + E(X_3)
- The above X winnings are independent from each round, hence:
E(3*X_1) = 3*E(X_1) = 3*(-0.0271) = -0.0813
- The standard deviation for 3 rounds is:
sqrt(Var(X_1 + X_2 + X_3)) = sqrt(Var(X_1) + Var(X_2) + Var(X_3))
- The above X winnings are independent from each round, hence:
sqrt(Var(3*X_1)) = 3*Var(X_1) = 3*(0.999) = 2.9988
- For above two games are similar with an expected loss of $0.0813 for playing the game and stakes are very high due to high amount of deviation for +/- $3 of winnings.
Using expected value and standard deviation concepts, it is found that:
a)
The expected value is of $1.38.
The standard deviation is of $4.5.
b)
The expected value is of $1.38.
The standard deviation is of $2.6.
c)
The more observations there are, the closer the value is to the theoretical mean, hence, the variability and the risk of the game are reduced.
A probability is the number of desired outcomes divided by the number of total outcomes.
In this problem, for the gambler to win, the ball must land on either red or green, according to their bet, so the probability is [tex]\frac{18}{37}[/tex]
Item a:
The expected value is the sum of each outcome multiplied by it's respective probability.
In this problem
- [tex]\frac{18}{37}[/tex] probability of doubling the money, that is, earning $6.
- [tex]\frac{19}{37}[/tex] probability of losing the money, that is, losing $3.
Hence:
[tex]E(X) = 6\frac{18}{37} - 3\frac{19}{37} = \frac{6(18) - 3(19)}{37} = 1.38[/tex]
The expected value is of $1.38.
The standard deviation is the square root of the sum of the difference squared between each outcome and the mean, multiplied by it's respective probability.
Hence:
[tex]\sqrt{V(X)} = \sqrt{(6-1.38)^2\frac{18}{37} + (-3-1.38)^2\frac{19}{37}} = \sqrt{\frac{(6-1.38)^2(18) + (-3-1.38)^2(19)}{37}} = 4.5[/tex]
The standard deviation is of $4.5.
Item b:
For 1 game, with a bet of $1:
[tex]E(X) = 2\frac{18}{37} - 1\frac{19}{37} = \frac{36 - 19}{37} = 0.459459[/tex]
[tex]\sqrt{V(X)} = \sqrt{(2-0.459459)^2\frac{18}{37} + (-1-0.459459)^2\frac{19}{37}} = \sqrt{\frac{(2-0.459459)^2(18) + (-1-0.459459)^2(19)}{37}} = 1.5[/tex]
For 3 games:
[tex]E(X) = 3(0.459459) = 1.38[/tex]
[tex]\sqrt{V(X)} = 1.5\sqrt{3} = 2.6[/tex]
The expected value is of $1.38.
The standard deviation is of $2.6.
Item c:
The more observations there are, the closer the value is to the theoretical mean, hence, the variability and the risk of the game are reduced.
A similar problem is given at https://brainly.com/question/24855677
