Math Problem Statement

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.)

Solution

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

Part 1: Expected Value Calculation

The expected value EE for a game of chance can be calculated using the following formula:

E=(Pwin×win amount)+(Plose×lose amount)E = (P_{\text{win}} \times \text{win amount}) + (P_{\text{lose}} \times \text{lose amount})

Where:

  • PwinP_{\text{win}} is the probability of winning.
  • win amount\text{win amount} is the net amount won, including the player's original bet.
  • PloseP_{\text{lose}} is the probability of losing.
  • lose amount\text{lose amount} is the amount lost when the player loses.

Step 1: Define the values.

  • The probability of winning is Pwin=138P_{\text{win}} = \frac{1}{38}.
  • The probability of losing is Plose=1Pwin=3738P_{\text{lose}} = 1 - P_{\text{win}} = \frac{37}{38}.
  • The win amount is 140+44=184140 + 44 = 184 (since the player gets their initial bet of $44 back plus $140 in winnings).
  • The lose amount is 44-44 (the player loses their bet).

Step 2: Apply the values to the formula.

E=(138×184)+(3738×(44))E = \left( \frac{1}{38} \times 184 \right) + \left( \frac{37}{38} \times (-44) \right)

E=(18438)+(162838)E = \left( \frac{184}{38} \right) + \left( \frac{-1628}{38} \right)

E=4.84+(42.84)E = 4.84 + (-42.84)

E=38E = -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:

Expected loss over 1000 games=38×1000=38000\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.

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.

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Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Multiplication
Addition
Subtraction

Formulas

E = (P_win * win amount) + (P_lose * lose amount)

Theorems

Expected Value Theorem

Suitable Grade Level

Grades 10-12