In American roulette, the wheel contains the numbers 1
 through 36
, alternating between black and red. There are two green spaces numbered 0
 and 00
. A player places a bet of $4.50
 on red to play the game. If the ball lands on red, the player gets a $4.50
 for winning and receives the money back. If the ball does not land on red, then the player simply loses the $4.50
 placed on the bet. Calculate the expected winnings on one game. Round your answer to the nearest cent

In American roulette, there are 38 total spaces on the wheel: 18 red, 18 black, and 2 green spaces (0 and 00).

Here’s the breakdown of the bet:
- **Winning on red**: If the ball lands on red, the player wins $4.50 in addition to receiving the $4.50 bet back, making the total amount returned to the player $9.00.
- **Losing on red**: If the ball does not land on red (i.e., it lands on black or green), the player loses the $4.50 bet.

### Step-by-Step Calculation

1. **Probabilities:**
   - Probability of landing on red: $ P(\text{red}) = \frac{18}{38} $
   - Probability of not landing on red (black or green): $ P(\text{not red}) = 1 - P(\text{red}) = \frac{20}{38} $

2. **Winnings:**
   - If the ball lands on red, the player wins $9.00 (since they receive back the $4.50 bet plus the winnings).
   - If the ball does not land on red, the player loses $4.50.

3. **Expected winnings**:
   The expected value (EV) can be calculated using the formula:

   \[
   EV = (P(\text{red}) \times \text{Winnings on red}) + (P(\text{not red}) \times \text{Winnings on not red})
   \]

   Substituting the values:

   \[
   EV = \left(\frac{18}{38} \times 9.00\right) + \left(\frac{20}{38} \times (-4.50)\right)
   \]

4. **Calculation:**

   \[
   EV = \left( \frac{18}{38} \times 9.00 \right) + \left( \frac{20}{38} \times (-4.50) \right)
   \]

   \[
   EV = \left( \frac{162}{38} \right) + \left( \frac{-90}{38} \right)
   \]

   \[
   EV = \frac{162 - 90}{38}
   \]

   \[
   EV = \frac{72}{38} \approx 1.89
   \]

So, the **expected winnings** on one game are **$1.89**.

### Final Answer:
The expected winnings on one game are **$1.89**.

---

Would you like a more detailed explanation of any part of this? Here are some additional related questions for further exploration:

1. How does the addition of a "0" and "00" affect the probabilities of winning in roulette?
2. How would the expected winnings change in European roulette (with only one "0")?
3. What are the odds of winning a straight-up bet on a single number in American roulette?
4. How would the expected winnings change if the player placed a different bet, such as on black or on a range of numbers?
5. What is the house edge in American roulette, and how does it relate to expected winnings?

**Tip**: Always remember that the house edge in American roulette comes from the presence of the "0" and "00," which are green spaces that don't belong to either the red or black groups, making the game unfavorable for the player in the long run.

In American roulette, the wheel contains the numbers 1 through 36, alternating between black and red. There are two green spaces numbered 0 and 00. A player places a bet of $4.50 on red to play the game. If the ball lands on red, the player gets a $4.50 for winning and receives the money back. If the ball does not land on red, then the player simply loses the $4.50 placed on the bet. Calculate the expected winnings on one game. Round your answer to the nearest cent.

Learn how to calculate the expected winnings of a $4.50 bet on red in American roulette. The problem involves understanding probability, expected value, and basic arithmetic, including the total spaces and the odds of winning or losing. Suitable for students in Grades 9-12.

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