Math Problem Statement

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?

Solution

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): 138\frac{1}{38}
  • Probability of losing (not landing on 6): 3738\frac{37}{38}
  • Payoff if the player wins: The player keeps the $8 bet and wins an additional $280, for a total of: Total winnings=8+280=288\text{Total winnings} = 8 + 280 = 288
  • Payoff if the player loses: The player loses their $8 bet, so: Total loss=8\text{Total loss} = -8

Step 2: Calculate the Expected Value (EV)

The expected value is calculated using the formula: EV=(Probability of winning×Payoff when winning)+(Probability of losing×Payoff when losing)\text{EV} = (\text{Probability of winning} \times \text{Payoff when winning}) + (\text{Probability of losing} \times \text{Payoff when losing})

Substitute the values: EV=(138×288)+(3738×8)\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: 138×288=288387.58\frac{1}{38} \times 288 = \frac{288}{38} \approx 7.58
  2. Calculate the expected loss when losing: 3738×8=296387.79\frac{37}{38} \times -8 = -\frac{296}{38} \approx -7.79

Step 4: Combine the Results

EV=7.587.79=0.21\text{EV} = 7.58 - 7.79 = -0.21

The expected value of the game to the player is approximately 0.21-0.21, 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: Total expected loss=1000×(0.21)=210\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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Statistics

Formulas

Expected Value (EV) = (Probability of winning × Payoff when winning) + (Probability of losing × Payoff when losing)

Theorems

Law of Large Numbers

Suitable Grade Level

Grades 10-12