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 need to calculate the expected value for a single game and then multiply it by 1000 to find the expected total loss over 1000 games. Let's go through each part.

### Part 1: Calculating the Expected Value of a Single Game

1. **Probability of Winning:** The probability of winning on a single spin is $\frac{1}{38}$.
2. **Probability of Losing:** The probability of losing is $1 - \frac{1}{38} = \frac{37}{38}$.
3. **Winnings and Losses:**
   - If the player wins, they receive a total payout of $\$140 + \$44 = \$184$ (since they get their $44 back plus $140 in winnings).
   - If the player loses, they lose their $44 bet.

Using these values, we can calculate the expected value $E$ of a single game:

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

Let's compute each part of this sum.

#### Step 1: Calculate Winning Term
\[
\frac{1}{38} \times 184 = \frac{184}{38} = 4.84
\]

#### Step 2: Calculate Losing Term
\[
\frac{37}{38} \times -44 = -42.84
\]

#### Step 3: Add the Terms
\[
E = 4.84 - 42.84 = -38.00
\]

The expected value of a single game to the player is $-\$38.00$.

### Part 2: Expected Loss Over 1000 Games

Since the expected value per game is $-\$38.00$, over 1000 games the player would expect to lose:

\[
1000 \times -38.00 = -38000
\]

So, the player would expect to lose about **$38,000** if they played the game 1000 times.

---

### Final Answer

- **The expected value of a single game is:** $-\$38.00$
- **Expected loss over 1000 games:** $-\$38,000$

Would you like any further details or have any other questions?

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Here are five questions to expand on this:

1. What is the expected value if the payout changes to $200 for the same bet?
2. How would the expected value change if the probability of winning was $\frac{2}{38}$ instead of $\frac{1}{38}$?
3. What is the break-even payout for a game with a $\frac{1}{38}$ probability of winning?
4. How would the expected loss over 1000 games change if the bet was $50 instead of $44?
5. How does the expected value compare between different casino games like roulette and blackjack?

**Tip:** Understanding expected value helps assess the long-term profitability or loss of a gambling game, showing the advantage or disadvantage to the player.

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 the player 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?

Understand how to calculate the expected value of a $44 bet in roulette with a 1/38 probability of winning. The solution includes step-by-step calculations for expected value and estimated loss over 1000 games. Perfect for high school students learning about probability and expected value.

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