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 resolve the stated problem, it is necessary to determine the probability of player A achieving a sum of 5 before player B achieves a sum of 8. The solution is structured into distinct stages:

1. Total Possible Outcomes: The roll of a pair of dice yields \(6 \times 6 = 36\) distinct outcomes.

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

3. Calculation of Probabilities for Specific Sums: * Sum of 5: The combinations resulting in a sum of 5 are (1,4), (2,3), (3,2), and (4,1). This provides 4 favorable outcomes. * Sum of 8: The combinations resulting in a sum of 8 are (2,6), (3,5), (4,4), (5,3), and (6,2). This provides 5 favorable outcomes.

4. Individual Probabilities: * Probability of A rolling a sum of 5: \( \frac{4}{36} = \frac{1}{9} \). * Probability of B rolling a sum of 8: \( \frac{5}{36} \). * Probability of neither player rolling their target sum on a given turn: \( 1 - \frac{1}{9} - \frac{5}{36} = \frac{25}{36} \).

5. Event Definition: Let \( p \) represent the probability that player A wins, commencing with A's turn.

6. Equation Formulation: Player A can win on their first roll if they achieve a sum of 5. If not, the game continues, and the probability of A winning is influenced by the remaining probability of the game continuing. The equation is formulated as: \( p = \frac{1}{9} + \frac{25}{36}p \).

7. Solving 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. Final Result: The probability \( p_A \) of player A winning, given that A initiates the game, is calculated as \( \frac{9}{19} \). Consequently, the definitive answer is \( \frac{9}{19} \).