Math Problem Statement

Keith has a deck of 10 cards numbered 1 through 10. He is playing a game of chance. The game is this: Keith chooses one card from the deck at random. He wins an amount of money equal to the value of the card if an odd-numbered card is drawn. He loses $6 if an even-numbered card is drawn. (a) Find the expected value of playing the game. (b) What can Keith expect in the long run, after playing the game many times?

Solution

The image depicts a probability problem where Keith is playing a game of chance with a deck of cards numbered 1 through 10. The problem involves calculating the expected value of the game based on the rules provided.

Problem summary:

  • Deck composition: 10 cards numbered from 1 to 10.
  • Rules:
    • If Keith draws an odd-numbered card, he wins an amount of money equal to the value of the card.
    • If Keith draws an even-numbered card, he loses $6.

Questions:

  1. (a) Find the expected value of playing the game.
  2. (b) What can Keith expect in the long run after playing many times?

Solution to part (a):

To find the expected value, we need to compute the sum of the probabilities multiplied by the outcomes.

Odd-numbered cards:

  • Odd numbers in the deck: 1, 3, 5, 7, 9.
  • Probability of drawing an odd-numbered card: 510=0.5\frac{5}{10} = 0.5.
  • For each odd-numbered card drawn, Keith wins an amount equal to the card value.

The expected value for the odd-numbered cards is calculated by averaging the values of the odd cards: Average value of odd cards=1+3+5+7+95=5\text{Average value of odd cards} = \frac{1 + 3 + 5 + 7 + 9}{5} = 5 So, the expected gain from drawing an odd-numbered card is 0.5×5=2.50.5 \times 5 = 2.5.

Even-numbered cards:

  • Even numbers in the deck: 2, 4, 6, 8, 10.
  • Probability of drawing an even-numbered card: 510=0.5\frac{5}{10} = 0.5.
  • For each even-numbered card drawn, Keith loses $6.

The expected loss from drawing an even-numbered card is: 0.5×(6)=30.5 \times (-6) = -3

Total expected value:

Adding the expected value from odd and even draws: 2.5+(3)=0.52.5 + (-3) = -0.5 Thus, the expected value of playing the game is 0.5-0.5 dollars.

Solution to part (b):

Since the expected value is 0.5-0.5, Keith can expect to lose $0.50 per draw in the long run.


Would you like further clarification or details on this? Here are some related questions you might explore:

  1. How would the expected value change if Keith's win on odd-numbered cards was doubled?
  2. What if the penalty for drawing an even-numbered card was reduced to $4 instead of $6?
  3. How does the law of large numbers relate to Keith’s long-term expectation of losing money?
  4. What would the expected value be if Keith only lost money on even-numbered cards above 6?
  5. How would the expected value change if the deck size increased to 20 cards, maintaining a similar structure?

Tip: Expected value calculations are essential in determining whether a game or investment is favorable in the long run.

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Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Games of Chance

Formulas

Expected Value (E) = Σ [Probability × Outcome]

Theorems

Law of Large Numbers

Suitable Grade Level

Grades 10-12