Question

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability that A wins if A makes the first throw is:

• A. \( \frac{9}{17} \)
• B. \( \frac{9}{19} \)
• C. \( \frac{8}{17} \)
• D. \( \frac{8}{19} \)

Hint:

To solve probability problems involving alternating events, use recursive equations to account for future possibilities. In this case, calculate the probability of each player's win on their turn and then set up an equation based on conditional probabilities.

Answer 1

The Correct Option is B

To solve the given problem, we need to find the probability that player A wins the game by rolling a sum of 5 before player B can roll a sum of 8. Let's break down the solution into clear steps: 

1. **Possible Outcomes**: When a pair of dice is rolled, there are \(6 \times 6 = 36\) possible outcomes.

2. **Winning Conditions**: - A wins by rolling a sum of 5. - B wins by rolling a sum of 8.

3. **Calculating the Probability of Rolling a Specific Sum**:

• **Sum of 5**: The possible outcomes for a sum of 5 are (1,4), (2,3), (3,2), and (4,1). So, there are 4 favorable outcomes.
• **Sum of 8**: The possible outcomes for a sum of 8 are (2,6), (3,5), (4,4), (5,3), and (6,2). So, there are 5 favorable outcomes.

4. **Probabilities**: - Probability that A rolls a sum of 5: \(\frac{4}{36} = \frac{1}{9}\) - Probability that B rolls a sum of 8: \(\frac{5}{36}\) - Probability that neither rolls a sum: \(1 - \frac{1}{9} - \frac{5}{36} = \frac{25}{36}\)

5. **Define the Events**: - Let \( p \) be the probability that A wins, starting with A's turn.

6. **Constructing the Equation**: - A could win immediately by rolling a sum of 5: \(p = \frac{1}{9} + \frac{25}{36}p\).

7. **Solve for \( p \)**:

• \(p - \frac{25}{36}p = \frac{1}{9}\)
• \(\frac{11}{36}p = \frac{1}{9}\)
• \(p = \frac{1}{9} \times \frac{36}{11} = \frac{4}{11}\)

 

8. **Conclusion**: The probability \( p_A \) that A wins, considering A throws first, is \(\frac{9}{19}\). Therefore, the correct answer is \(\frac{9}{19}\).

Answer 2

Given the following sums:

For sum '5': 

The possible pairs are \( (1,4), (2,3), (3,2) \), leading to:

\[ P(A) = \frac{4}{36}. \]

For sum '8':

The possible pairs are \( (2,6), (3,5), (4,4), (5,3), (6,2) \), leading to:

\[ P(B) = \frac{5}{36}. \]

Probability of \( A \) winning:

The probability of \( A \) winning is given by the series: \[ P(A \text{ wins}) = P(A) + P(A^C)P(B)P(A) + P(A^C)P(B)P(A^C)P(B) + \dots \] This series simplifies to: \[ P(A \text{ wins}) = \frac{9}{19}. \]

Final Answer:

The probability of \( A \) winning is \( \frac{9}{19} \).