In American​ roulette, the wheel has the 38​ numbers, 00,​ 0, 1,​ 2, ...,​ 34, 35, and​ 36, marked on equally spaced slots. If a player bets 

​$44 

on a number and​ wins, then the player keeps 

​$44 

and receives an additional 

​$140140. 

​Otherwise, the player is awarded​ nothing, and the casino takes the​ player's 

​$44. 

Find the expected value​ E(x) to the player for one play of the game. If x is the gain to a player in a game of​ chance, then​ E(x) is usually negative. This value gives the average amount per game the player can expect to lose.

The expected value is what?

To calculate the expected value $ E(x) $ for the player in one game of American roulette, we’ll use the following steps:

1. **Identify the possible outcomes and probabilities**:
   - The player wins if the ball lands on the chosen number (1 out of 38 possible slots).
   - The player loses if the ball lands on any of the other 37 slots.

2. **Calculate the probabilities of each outcome**:
   - Probability of winning: $ P(\text{win}) = \frac{1}{38} $
   - Probability of losing: $ P(\text{lose}) = \frac{37}{38} $

3. **Define the outcomes**:
   - **Winning outcome**: The player gains $140140 in winnings and keeps the $44 bet. This makes the total gain $ 140140 $.
   - **Losing outcome**: The player loses their $44 bet, so the total gain is $ -44 $.

4. **Calculate the expected value**:
   The expected value $ E(x) $ is calculated as:
   \[
   E(x) = \left(\frac{1}{38} \times 140140\right) + \left(\frac{37}{38} \times -44\right)
   \]

   Let’s plug in the values:

   - Winning part: $ \frac{1}{38} \times 140140 = 3688.95 $
   - Losing part: $ \frac{37}{38} \times -44 = -42.79 $

   Combining these results:
   \[
   E(x) = 3688.95 - 42.79 = 3646.16
   \]

### Answer:
The expected value $ E(x) $ to the player for one play of the game is approximately **$3646.16**.

Would you like a detailed breakdown of each calculation step, or do you have specific questions?

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**Relative Questions:**

1. How would the expected value change if the payout were different?
2. What happens to $ E(x) $ if there are fewer or more numbers on the wheel?
3. Why is the expected value generally negative in most casino games?
4. How does the house edge affect the probability of winning in roulette?
5. What other betting strategies could impact the expected value in roulette?

**Tip:** Expected values give a long-term average but don’t predict short-term results.

In American roulette, the wheel has the 38 numbers, 00, 0, 1, 2, ..., 34, 35, and 36, marked on equally spaced slots. If a player bets $44 on a number and wins, then the player keeps $44 and receives an additional $140140. Otherwise, the player is awarded nothing, and the casino takes the player's $44. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose.

Explore how to calculate the expected value E(x) in American roulette when betting $44 on a single number. Learn key probability concepts, expected value calculations, and find the average outcome per game. Perfect for students and enthusiasts seeking insight into game theory and casino probabilities.

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