Answer:
a) Expected amount of winnings = -$0.084
Standard deviation = $3.00
b) Expected amount of winnings = -$0.084
Standard deviation = $1.73
c) The expected winnings for both cases is exactly the same, but the spread of the wins about the expected value/mean is low in the case (b) compared to case (a).
So, with the expected winnings using the method in (a) and (b) the same, the risk associated with betting $3 dollar per game is higher because it has a higher standard deviation. This points to higher spread of losses and wins betting $3 once than betting $1 three different times.
Step-by-step explanation:
(a) Suppose you play roulette and bet $3 on a single round. What is the expected value and standard deviation of your total winnings?
It's either one wins or not with a bet of $3.
We can then obtain the probability mass function.
Random variable X represents the amount of winnings.
- If one bets $3 and picks the right colour, the person wins $6.
X = amount of winnings = 6 - 3 = $3
probability of winning = (18/37) = 0.486
- If one bets $3 and picks the wrong colour, then the person wins $0.
X = amount of winnings = 0 - 3 = -$3
Probability of losing = (19/37) = 0.514
The probability mass function is then
X | P(X)
3 | 0.486
-3 | 0.514
E(X) = Σ xᵢpᵢ
xᵢ = each variable
pᵢ = probability of each variable
E(X) = (3×0.486) + (-3×0.514) = -$0.084
Standard deviation = √(variance)
Variance = Var(X) = Σx²p − μ²
μ = E(X) = -0.084
Σx²p = (3²×0.486) + [(-3)² × 0.514] = 9
Variance = 9 - (-0.084)² = 8.993
Standard deviation = √8.993 = 2.999 = $3
(b) Suppose you bet $1 in three different rounds. What is the expected value and standard deviation of your total winnings?
With a bet of $1,
It's either one wins or not with a bet of $1.
We can then obtain the probability mass function.
Random variable X represents the amount of winnings.
- If one bets $1 and picks the right colour, the person wins $2.
X = amount of winnings = 2 - 1 = $1
probability of winning = (18/37) = 0.486
- If one bets $1 and picks the wrong colour, then the person wins $0.
X = amount of winnings = 0 - 1 = -$1
Probability of losing = (19/37) = 0.514
The probability mass function is then
X | P(X)
1 | 0.486
-1 | 0.514
E(X) = Σ xᵢpᵢ
xᵢ = each variable
pᵢ = probability of each variable
E(X) = (1×0.486) + (-1×0.514) = -$0.028
For 3 rounds of $1, the expected amount of winnings = 3 × -0.028 = -$0.084
Variance = Var(X) = Σx²p − μ²
μ = E(X) = -0.028
Σx²p = (1²×0.486) + [(-1)² × 0.514] = 1
Variance = 1 - (-0.028)² = 0.999216
Variance for combining 3 independent distributions = Σ λᵢ²σᵢ²
Variance of 3 rounds of $1 bets = 3[1²× var(X)] = 3 × var(X) = 3 × 0.999216 = 2.997648
Standard deviation = √(variance) = √(2.997648) = $1.73
(c) How do your answers to parts (a) and (b) compare? What does this say about the riskiness of the two games?
The expected winnings for both cases is exactly the same, but the spread of the wins about the expected value/mean is low in the case (b) compared to case (a).
So, with the expected winnings using the method in (a) and (b) the same, the risk associated with betting $3 dollar per game is higher because it has a higher standard deviation. This points to higher spread of losses and wins betting $3 once than betting $1 three different times.
Hope this Helps!!!
) cost $1 per box
) cost $4 per box