In the game of​ roulette, a player can place a ​$44 bet on the number 55 and have a StartFraction 1 Over 38 EndFraction
1
38 probability of winning. If the metal ball lands on 55​, the player gets to keep the ​$44 paid to play the game and the player is awarded an additional ​$140140. ​ Otherwise, the player is awarded nothing and the casino takes the​ player's ​$44. What is the expected value of the game to the​ player? If you played the game 1000​ times, how much would you expect to​ lose?
Question content area bottom
Part 1
The expected value is ​$enter your response here.
​(Round to the nearest cent as​ needed.)
Part 2
The player would expect to lose about ​$enter your response here.
​(Round to the nearest cent as​ needed.)

To solve this, we'll break the problem down into two parts:

### Part 1: Expected Value Calculation
The expected value $ E $ for a game of chance can be calculated using the following formula:

\[
E = (P_{\text{win}} \times \text{win amount}) + (P_{\text{lose}} \times \text{lose amount})
\]

Where:
- $ P_{\text{win}} $ is the probability of winning.
- $ \text{win amount} $ is the net amount won, including the player's original bet.
- $ P_{\text{lose}} $ is the probability of losing.
- $ \text{lose amount} $ is the amount lost when the player loses.

**Step 1: Define the values.**

- The probability of winning is $ P_{\text{win}} = \frac{1}{38} $.
- The probability of losing is $ P_{\text{lose}} = 1 - P_{\text{win}} = \frac{37}{38} $.
- The win amount is $ 140 + 44 = 184 $ (since the player gets their initial bet of \$44 back plus \$140 in winnings).
- The lose amount is $ -44 $ (the player loses their bet).

**Step 2: Apply the values to the formula.**

\[
E = \left( \frac{1}{38} \times 184 \right) + \left( \frac{37}{38} \times (-44) \right)
\]

\[
E = \left( \frac{184}{38} \right) + \left( \frac{-1628}{38} \right)
\]

\[
E = 4.84 + (-42.84)
\]

\[
E = -38
\]

So, the expected value for the game is **-38 dollars**. This means that on average, the player expects to lose \$38 per game.

### Part 2: Expected Loss Over 1000 Games
To find the expected loss over 1000 games, simply multiply the expected value per game by 1000:

\[
\text{Expected loss over 1000 games} = -38 \times 1000 = -38000
\]

So, if you played 1000 games, you would expect to lose about **\$38,000**.

### Final Answers:
1. The expected value of the game is **-38 dollars**.
2. The player would expect to lose about **\$38,000** if they played 1000 times.

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Would you like more details or have any questions about this solution?

Here are 5 related questions to consider:
1. How does increasing the number of bets affect the expected value in gambling games?
2. What would happen to the expected value if the casino paid out \$200 instead of \$140 for a win?
3. Can a player ever expect to win money in a game with a negative expected value?
4. How is expected value used in decision-making in other areas like finance?
5. If the probability of winning changes, how does it affect the overall expected loss?

**Tip:** Understanding expected value helps in evaluating risk in games, investments, and everyday decisions.

In the game of roulette, a player can place a $44 bet on the number 55 and have a 1/38 probability of winning. If the metal ball lands on 55, the player gets to keep the $44 paid to play the game and is awarded an additional $140. Otherwise, the player is awarded nothing and the casino takes the player's $44. What is the expected value of the game to the player? If you played the game 1000 times, how much would you expect to lose? (Round to the nearest cent as needed.)

This guide explains how to calculate the expected value of a $44 roulette bet on the number 55, including step-by-step instructions for determining the expected loss over 1000 games. The problem involves basic probability and the expected value formula. Perfect for students in Grades 10-12 interested in probability and real-world applications of math.

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