Math Problem Statement

When playing roulette at a​ casino, a gambler is trying to decide whether to bet

​$1010

on the number

2929

or to bet

​$1010

that the outcome is any one of the

threethree

possibilities

00 comma 0 comma or 100, 0, or 1.

The gambler knows that the expected value of the

​$1010

bet for a single number is

negative 53 cents−53¢.

For the

​$1010

bet that the outcome is

00 comma 0 comma or 100, 0, or 1​,

there is a probability of

StartFraction 3 Over 38 EndFraction338

of making a net profit of

​$4040

and a

StartFraction 35 Over 38 EndFraction3538

probability of losing

​$1010.

a. Find the expected value for the

​$1010

bet that the outcome is

00 comma 0 comma or 100, 0, or 1.

b. Which bet is​ better: a

​$1010

bet on the number

2929

or a

​$1010

bet that the outcome is any one of the numbers

00 comma 0 comma or 100, 0, or 1​?

​Why?

Question content area bottom

Part 1

a. The expected value is

​$enter your response here.

​(Round to the nearest cent as​ needed.)

Part 2

b. Since the expected value of the bet on the number

2929

is

▼   less

greater

than the expected value for the bet that the outcome is

00 comma 0 comma or 100, 0, or 1​,

the bet on

▼   the single number

00, 0, or 1

is better.

Solution

To solve this problem, we calculate the expected value for the second bet and compare it to the first bet.


Part a: Calculating the Expected Value

The expected value (EVEV) is calculated using the formula: EV=(Probability of Outcome×Net Profit/Loss)EV = \sum (\text{Probability of Outcome} \times \text{Net Profit/Loss})

For the $10\$10 bet on the outcome being 0000, 00, or 11:

  • Probability of winning: 338\frac{3}{38}
  • Probability of losing: 3538\frac{35}{38}
  • Net profit if winning: +$40+\$40
  • Loss if losing: $10-\$10

EV=(338×40)+(3538×(10))EV = \left( \frac{3}{38} \times 40 \right) + \left( \frac{35}{38} \times (-10) \right)

  1. Compute each term:

    • 338×40=120383.16\frac{3}{38} \times 40 = \frac{120}{38} \approx 3.16
    • 3538×(10)=350389.21\frac{35}{38} \times (-10) = \frac{-350}{38} \approx -9.21
  2. Add the results: EV3.169.21=6.05EV \approx 3.16 - 9.21 = -6.05

So, the expected value for the bet on 0000, 00, or 11 is approximately: $6.05\boxed{-\$6.05}


Part b: Comparing Bets

For the single-number bet (2929), the expected value is already given as: $0.53-\$0.53

The expected value for the 0000, 00, or 11 bet is: $6.05-\$6.05

Since $0.53-\$0.53 is greater than $6.05-\$6.05, the bet on a single number (29) is better.


Final Answers

a. The expected value for the $10\$10 bet on 0000, 00, or 11 is: $6.05\boxed{-\$6.05}

b. The better bet is: The single-number bet (29)\boxed{\text{The single-number bet (29)}}

Reason: The expected value of the single-number bet ($0.53-\$0.53) is closer to zero than the expected value of the other bet ($6.05-\$6.05), meaning it has a smaller average loss.


Do you want further clarification or calculations?

Five Related Questions:

  1. How is expected value used in determining the fairness of a game?
  2. What does a negative expected value imply for gambling outcomes?
  3. How would the probabilities and expected values change if the casino used a European roulette wheel instead?
  4. Can expected value alone determine whether a bet is worth making?
  5. How does increasing the bet amount affect the expected value in this scenario?

Tip:

When gambling, always compare the expected values of bets to minimize losses and make more informed decisions.

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Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Decision Analysis

Formulas

Expected Value (EV) = Σ(Probability of Outcome × Net Profit/Loss)

Theorems

Law of Total Expectation

Suitable Grade Level

Grades 10-12