in the game of roulette, a player can place a $8 bet on the number 6 and have a 1/38 probability of winning. if the metal ball lands on 6, the player gets to keep the $8 paid to play the game and the player is awarded an additional $280. otherwise, the player is awarded nothing and the casino takes the player’s $8. 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?

To find the expected value of the game to the player, we'll analyze the outcomes, their probabilities, and the associated payoffs. Here's the step-by-step solution:

### Step 1: Define the Payoffs and Probabilities
- **Probability of winning (landing on 6):** $ \frac{1}{38} $
- **Probability of losing (not landing on 6):** $ \frac{37}{38} $
- **Payoff if the player wins:** The player keeps the $8 bet and wins an additional $280, for a total of:
  \[
  \text{Total winnings} = 8 + 280 = 288
  \]
- **Payoff if the player loses:** The player loses their $8 bet, so:
  \[
  \text{Total loss} = -8
  \]

### Step 2: Calculate the Expected Value (EV)
The expected value is calculated using the formula:
\[
\text{EV} = (\text{Probability of winning} \times \text{Payoff when winning}) + (\text{Probability of losing} \times \text{Payoff when losing})
\]

Substitute the values:
\[
\text{EV} = \left(\frac{1}{38} \times 288\right) + \left(\frac{37}{38} \times -8\right)
\]

### Step 3: Compute the Expected Value
Now, calculate each part:
1. Calculate the expected gain when winning:
   \[
   \frac{1}{38} \times 288 = \frac{288}{38} \approx 7.58
   \]
2. Calculate the expected loss when losing:
   \[
   \frac{37}{38} \times -8 = -\frac{296}{38} \approx -7.79
   \]

### Step 4: Combine the Results
\[
\text{EV} = 7.58 - 7.79 = -0.21
\]

The expected value of the game to the player is approximately **$-0.21$**, meaning the player is expected to lose about 21 cents per game.

### Step 5: Calculate Expected Loss for 1000 Games
If the player plays 1000 times, the expected total loss can be found by multiplying the expected value by the number of games:
\[
\text{Total expected loss} = 1000 \times (-0.21) = -210
\]

So, if the player played this game 1000 times, they would expect to lose **$210** on average.

---

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

Here are 5 related questions that could deepen understanding:

1. What is the variance of the player's earnings in this game?
2. How does the house edge affect the expected value for different types of bets in roulette?
3. What would be the expected value if the payout for winning was changed to $300?
4. How would the expected value change if the game were played with a European roulette wheel (1/37 probability of winning)?
5. Can the expected value ever become positive for the player in standard roulette games?

**Tip:** Understanding expected value is crucial for evaluating whether a game is fair or skewed in favor of the house. It can help make more informed decisions in probabilistic scenarios.

In the game of roulette, a player can place a $8 bet on the number 6 and have a 1/38 probability of winning. If the metal ball lands on 6, the player gets to keep the $8 paid to play the game and the player is awarded an additional $280. Otherwise, the player is awarded nothing and the casino takes the player’s $8. 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?

Learn how to calculate the expected value of a roulette game when betting $8 on the number 6. This detailed solution includes the probability of winning, payoffs, and expected loss over 1000 games. Explore key probability concepts and the impact of the house edge in gambling scenarios.

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