a die lands on the number 2, you win $100. Otherwise, you lose $15. hat is he expected value? hould you play the game? = 4.17, yes -4.17, no 16.67, yes -16.67, no
a die lands on the number 2, you win $100. Otherwise, you lose $15. hat is he expected value? hould you play the game? = 4.17, yes -4.17, no 16.67, yes -16.67, no
Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter14: Counting And Probability
Section14.FOM: Focus On Modeling: The Monte Carlo Method
Problem 3P: Dividing a JackpotA game between two players consists of tossing a coin. Player A gets a point if...
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![**Expected Value in a Dice Game**
In this example, you are presented with a dice game where you win $100 if the die lands on the number 2, and you lose $15 otherwise. We need to calculate the expected value and determine whether it is advisable to play the game.
**Game Description:**
- Win $100 if the die lands on the number 2.
- Lose $15 if the die lands on any other number (1, 3, 4, 5, or 6).
**Question:**
What is the expected value of this game?
Should you play the game based on the expected value?
**Options:**
1. 4.17, yes
2. -4.17, no
3. 16.67, yes
4. -16.67, no
**Explanation:**
To find the expected value, we use the following calculation:
\[ \text{Expected Value (EV)} = (P(\text{win}) \times \text{win amount}) + (P(\text{loss}) \times \text{loss amount}) \]
The probability \( P(\text{win}) \) of rolling a 2 on a fair six-sided die is \( \frac{1}{6} \).
The probability \( P(\text{loss}) \) of rolling any other number (1, 3, 4, 5, or 6) is \( \frac{5}{6} \).
Thus, the expected value calculation is:
\[ \text{EV} = \left( \frac{1}{6} \times 100 \right) + \left( \frac{5}{6} \times -15 \right) \]
\[ \text{EV} = \left( \frac{1}{6} \times 100 \right) + \left( \frac{5}{6} \times -15 \right) \]
\[ \text{EV} = \left( \frac{100}{6} \right) + \left( \frac{-75}{6} \right) \]
\[ \text{EV} = \frac{100 - 75}{6} \]
\[ \text{EV} = \frac{25}{6} \]
\[ \text{EV} \approx 4.17 \]
Therefore, the expected value is positive (approximately $4.17),](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F349e5e36-2d82-4a9b-b36b-acebde7ec7cc%2F6d651b40-8c7a-44da-8917-a4e10167d004%2F03s50g_processed.png&w=3840&q=75)
Transcribed Image Text:**Expected Value in a Dice Game**
In this example, you are presented with a dice game where you win $100 if the die lands on the number 2, and you lose $15 otherwise. We need to calculate the expected value and determine whether it is advisable to play the game.
**Game Description:**
- Win $100 if the die lands on the number 2.
- Lose $15 if the die lands on any other number (1, 3, 4, 5, or 6).
**Question:**
What is the expected value of this game?
Should you play the game based on the expected value?
**Options:**
1. 4.17, yes
2. -4.17, no
3. 16.67, yes
4. -16.67, no
**Explanation:**
To find the expected value, we use the following calculation:
\[ \text{Expected Value (EV)} = (P(\text{win}) \times \text{win amount}) + (P(\text{loss}) \times \text{loss amount}) \]
The probability \( P(\text{win}) \) of rolling a 2 on a fair six-sided die is \( \frac{1}{6} \).
The probability \( P(\text{loss}) \) of rolling any other number (1, 3, 4, 5, or 6) is \( \frac{5}{6} \).
Thus, the expected value calculation is:
\[ \text{EV} = \left( \frac{1}{6} \times 100 \right) + \left( \frac{5}{6} \times -15 \right) \]
\[ \text{EV} = \left( \frac{1}{6} \times 100 \right) + \left( \frac{5}{6} \times -15 \right) \]
\[ \text{EV} = \left( \frac{100}{6} \right) + \left( \frac{-75}{6} \right) \]
\[ \text{EV} = \frac{100 - 75}{6} \]
\[ \text{EV} = \frac{25}{6} \]
\[ \text{EV} \approx 4.17 \]
Therefore, the expected value is positive (approximately $4.17),
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