Math Problem Statement

Suppose that you are offered the following "deal." You roll a six sided die. If you roll a 6, you win $8. If you roll a 4 or 5, you win $4. Otherwise, you pay $8.

a. Complete the PDF Table. List the X values, where X is the profit, from smallest to largest. Round to 4 decimal places where appropriate.

Probability Distribution TableXP(X)

b. Find the expected profit. $ (Round to the nearest cent)

c. Interpret the expected value.

You will win this much if you play a game.

If you play many games you will likely lose on average very close to $1.33 per game.

This is the most likely amount of money you will win.

d. Based on the expected value, should you play this game?

Yes, since the expected value is positive, you would be very likely to come home with more money if you played many games.

No, since the expected value is negative, you would be very likely to come home with less money if you played many games.

Yes, since the expected value is 0, you would be very likely to come very close to breaking even if you played many games, so you might as well have fun at no cost.

Yes, because you can win $8.00 which is greater than the $8.00 that you can lose.

No, this is a gambling game and it is always a bad idea to gamble.

Solution

Step-by-Step Solution

Part (a) Probability Distribution Table

The possible values of profit XX and their probabilities are determined based on the game's rules:

  1. **Profit of -\$8**:** This occurs when you roll a \(1, 22, or 33. The probability is P(X=8)=36=0.5P(X = -8) = \frac{3}{6} = 0.5.
  2. **Profit of \$4**:** This occurs when you roll a \(4 or 55. The probability is P(X=4)=26=0.3333P(X = 4) = \frac{2}{6} = 0.3333.
  3. **Profit of \$8**:** This occurs when you roll a \(6. The probability is P(X=8)=16=0.1667P(X = 8) = \frac{1}{6} = 0.1667.

The completed Probability Distribution Table looks like this:

XX (Profit)P(X)P(X) (Probability)
8-80.50000.5000
440.33330.3333
880.16670.1667

Part (b) Expected Profit

The formula for the expected value E(X)E(X) is:

E(X)=(XP(X))E(X) = \sum \left( X \cdot P(X) \right)

Substitute the values into the formula:

E(X)=(8)(0.5000)+(4)(0.3333)+(8)(0.1667)E(X) = (-8)(0.5000) + (4)(0.3333) + (8)(0.1667)

E(X)=4+1.3332+1.3336E(X) = -4 + 1.3332 + 1.3336

E(X)=1.33321.33(to the nearest cent).E(X) = -1.3332 \approx -1.33 \, \text{(to the nearest cent)}.

Thus, the expected profit is $1.33-\$1.33 per game.


Part (c) Interpretation of Expected Value

The expected value means that if you play the game many times, you are expected to lose an average of $1.33 per game.

This does not mean you will lose $1.33 in every single game. Rather, over a large number of games, your average loss per game will approach $1.33.


Part (d) Should You Play This Game?

Based on the expected value of $1.33-\$1.33, the correct answer is:

No, since the expected value is negative, you would be very likely to come home with less money if you played many games.

This is because the game is biased in favor of the house, and on average, you lose more than you gain.


Questions to Consider:

  1. What is the probability of winning any amount in this game?
  2. How would the expected value change if the payout for rolling a 6 were increased to $10?
  3. What would the expected profit be if the cost of losing were reduced to $6-\$6?
  4. How does the concept of risk factor into the decision to play this game?
  5. What are some alternative strategies to reduce losses in similar gambling setups?

Tip:

When assessing games of chance, always calculate the expected value and probabilities carefully. It helps to avoid losses by identifying whether the game is statistically in your favor.

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Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Decision Making in Games

Formulas

Expected Value: E(X) = Σ(X × P(X))

Theorems

Law of Large Numbers

Suitable Grade Level

Grades 9-12