Question

If a gambler rolls two dice and gets a sum of 4, he wins $20, and if he gets a sum of 11, he wins $50. The cost to play the gam If a gambler rolls two dice and gets a sum of 4, he wins $20, and if he gets a sum of 11, he wins $50. The cost to play the game is $10. What is the expectation of this game? e is $10. What is the expectation of this game?

Answer 1

To find the expectation of the game, we'll calculate the expected value or expected payoff, denoted as \(E(X)\).

The formula for the expectation \(E(X)\) of a random variable is calculated by summing the products of each outcome's probability and its respective payoff:

\[ E(X) = p_1 \cdot x_1 + p_2 \cdot x_2 + \ldots + p_n \cdot x_n \]

Where:
- \( p_1, p_2, \ldots, p_n \) = probabilities of each outcome.
- \( x_1, x_2, \ldots, x_n \) = payoffs associated with each outcome.

Given:
- The gambler wins $20 if the sum of the dice is 4.
- The gambler wins $50 if the sum of the dice is 11.
- The cost to play the game is $10.
- The probabilities of getting a sum of 4 or 11 need to be determined.

Let's calculate the probabilities of getting a sum of 4 or 11 when rolling two dice:
The possible combinations that result in a sum of 4 are (1, 3), (2, 2), and (3, 1), which are 3 out of 36 possible combinations when rolling two dice.
The possible combinations that result in a sum of 11 are (5, 6), (6, 5), which are 2 out of 36 possible combinations when rolling two dice.

The probabilities are:
- Probability of getting a sum of 4: \( P(\text{sum = 4}) = \frac{3}{36} = \frac{1}{12} \)
- Probability of getting a sum of 11: \( P(\text{sum = 11}) = \frac{2}{36} = \frac{1}{18} \)

Now, let's calculate the expectation of the game:
\[ E(X) = P(\text{sum = 4}) \times \text{payoff for sum = 4} + P(\text{sum = 11}) \times \text{payoff for sum = 11} - \text{cost to play} \]

\[ E(X) = \frac{1}{12} \times 20 + \frac{1}{18} \times 50 - 10 \]
\[ E(X) = \frac{5}{3} + \frac{25}{18} - 10 \]
\[ E(X) = \frac{90}{18} + \frac{25}{18} - \frac{180}{18} \]
\[ E(X) = \frac{90 + 25 - 180}{18} \]
\[ E(X) = \frac{-65}{18} \]

Therefore, the expectation of this game is approximately \(-\$3.61\).

Answer 2

The expectation of the dice game is calculated by considering the probabilities of winning outcomes (sums of 4 and 11), their associated rewards, and the cost to play the game.

Step-by-step explanation:

To calculate the expectation of the dice game, you first need to consider all the possible outcomes when rolling two dice and their associated probabilities. Then you calculate the expected gains or losses for each outcome, factoring in the cost to play the game. For rolling a sum of 4, there are three combinations that can result in a 4: (1,3), (2,2), and (3,1). The probability of getting a sum of 4 is therefore 3/36 or 1/12. If the gambler wins $20 in this case, the expected gain is (1/12)*20. For rolling a sum of 11, there are two combinations: (5,6) and (6,5), which gives a probability of 2/36 or 1/18. The expected gain for a sum of 11 is (1/18)*50. Since the gambler has to pay $10 to play the game, this cost needs to be subtracted from the expected gains. Finally, the total expectation for the game is calculated by adding the expectations for each winning outcome and subtracting the cost to play.