Math Problem Statement

Suppose that you and a friend are playing cards and decide to make a bet. If you draw three fives in succession from a standard deck of 52 cards with replacement, you win $70. Otherwise, you pay your friend $10. What is the expected value of your bet? Round your answer to the nearest cent, if necessary.

Solution

To calculate the expected value (EV) of the bet, we need to consider the possible outcomes, their probabilities, and their associated payouts.

Step 1: Probabilities

  • Drawing a 5 from a deck of 52 cards with replacement: There are 4 fives in a deck of 52 cards, so the probability of drawing a 5 on each draw is: P(draw a 5)=452=113P(\text{draw a 5}) = \frac{4}{52} = \frac{1}{13}

  • Drawing three fives in a row (since the draws are independent and with replacement): P(three fives in a row)=(113)3=12197P(\text{three fives in a row}) = \left(\frac{1}{13}\right)^3 = \frac{1}{2197}

  • The probability of not drawing three fives in a row is: P(not three fives in a row)=112197=21962197P(\text{not three fives in a row}) = 1 - \frac{1}{2197} = \frac{2196}{2197}

Step 2: Payouts

  • If you draw three fives in a row, you win $70.
  • If you don’t draw three fives in a row, you lose and pay $10.

Step 3: Expected Value Calculation

The expected value is given by: EV=(probability of winning)×(payout for winning)+(probability of losing)×(payout for losing)EV = (\text{probability of winning}) \times (\text{payout for winning}) + (\text{probability of losing}) \times (\text{payout for losing}) Substitute the known values: EV=(12197)×70+(21962197)×(10)EV = \left(\frac{1}{2197}\right) \times 70 + \left(\frac{2196}{2197}\right) \times (-10)

Now, let's compute each term:

  • 12197×70=7021970.0319\frac{1}{2197} \times 70 = \frac{70}{2197} \approx 0.0319
  • 21962197×(10)=2196021979.9886\frac{2196}{2197} \times (-10) = \frac{-21960}{2197} \approx -9.9886

Thus, the expected value is: EV0.0319+(9.9886)9.96EV \approx 0.0319 + (-9.9886) \approx -9.96

Conclusion:

The expected value of the bet is approximately -9.96. This means that on average, you would lose about $9.96 per game in the long run.

Would you like further details or have any questions?

Here are 5 related questions to consider:

  1. How would the expected value change if you drew cards without replacement?
  2. What if the payout for winning was increased to $100? How would the expected value shift?
  3. How would the probability of winning change if you had to draw four fives in succession instead of three?
  4. What is the probability of winning if you draw cards from a smaller deck?
  5. How can expected value help in making strategic decisions in gambling scenarios?

Tip: In games of chance, calculating expected value helps you understand whether a bet is in your favor in the long term!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability
Expected Value

Formulas

P(drawing a 5) = 4/52 = 1/13
P(three fives in a row) = (1/13)^3 = 1/2197
Expected Value = (P(win) × payout) + (P(loss) × loss)

Theorems

Probability of independent events

Suitable Grade Level

Grades 9-12