
In the game of​ roulette, a player can place a ​$4 bet on the number 21 and have a StartFraction 1 Over 38 EndFraction
 probability of winning. If the metal ball lands on 21​, the player gets to keep the ​$4 paid to play the game and the player is awarded an additional ​$140. ​Otherwise, the player is awarded nothing and the casino takes the​ player's ​$4. 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.

To solve this problem, we need to find the expected value $ E(x) $ for one play of the game. The expected value is the weighted average of all possible outcomes, considering the probability of each outcome.

### Step 1: Identify the two possible outcomes.
1. **Winning**: The player wins if the ball lands on 21.
   - The total gain when the player wins is $ 140 + 4 = 144 $ dollars.
   - The probability of this happening is $ \frac{1}{38} $.

2. **Losing**: The player loses if the ball does not land on 21.
   - The loss in this case is the bet amount, $ 4 $ dollars.
   - The probability of this happening is $ 1 - \frac{1}{38} = \frac{37}{38} $.

### Step 2: Set up the expected value formula.
The expected value $ E(x) $ is given by the formula:

\[
E(x) = (\text{{Gain if win}}) \times (\text{{Probability of win}}) + (\text{{Loss if lose}}) \times (\text{{Probability of lose}})
\]

Substitute the values:

\[
E(x) = (144) \times \left(\frac{1}{38}\right) + (-4) \times \left(\frac{37}{38}\right)
\]

### Step 3: Perform the calculations.
First, calculate the two terms:

1. $ 144 \times \frac{1}{38} = \frac{144}{38} = 3.789 $
2. $ -4 \times \frac{37}{38} = \frac{-148}{38} = -3.895 $

Now sum these two results:

\[
E(x) = 3.789 + (-3.895) = -0.106
\]

### Step 4: Interpret the result.
The expected value $ E(x) = -0.106 $ means that, on average, the player can expect to lose approximately 10.6 cents per game.

### Conclusion:
The player can expect to lose about 10.6 cents on average for each play of the roulette game.

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Do you want more details, or have any questions about this? 

Here are 5 follow-up questions you might find interesting:
1. How would the expected value change if the payout for landing on 21 increased?
2. What is the expected value if the game used a European roulette wheel with only 37 slots?
3. How would you calculate the variance in the player's outcomes?
4. What is the probability of losing over 10 consecutive games?
5. How could the casino adjust the payouts to make the game more profitable for them?

**Tip:** Expected value gives a long-term average; a negative value implies a disadvantage for the player over time.

In the game of​ roulette, a player can place a ​$4 bet on the number 21 and have a 1/38 probability of winning. If the metal ball lands on 21​, the player gets to keep the ​$4 paid to play the game and the player is awarded an additional ​$140. ​Otherwise, the player is awarded nothing and the casino takes the​ player's ​$4. Find the expected value​ E(x) to the player for one play of the game.

Find the expected value of a $4 bet on number 21 in roulette with a 1/38 chance of winning. The player gains $144 if they win and loses $4 otherwise. This problem explores probability and expected value calculations, suitable for students in Grades 9-12.

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