Question

Two persons \(A\) and \(B\) alternately throw a pair of dice. \(A\) wins if he throws a sum of \(4\) before \(B\) throws a sum of \(9\), and \(B\) wins if he throws a sum of \(9\) before \(A\) throws a sum of \(4\). The probability that \(A\) wins if \(B\) makes the first throw is:

• A. \( \dfrac{1}{5} \)
• B. \( \dfrac{2}{5} \)
• C. \( \dfrac{3}{5} \)
• D. \( \dfrac{4}{5} \)

Hint:

For turn-based probability problems, define the probability at a fixed turn and form a recursive equation.

Answer 1

The Correct Option is B

Concept:
Probability of getting a sum of \(4\) with two dice is \( \dfrac{3}{36} = \dfrac{1}{12} \).
Probability of getting a sum of \(9\) with two dice is \( \dfrac{4}{36} = \dfrac{1}{9} \).
When players alternate turns, recursive probability (infinite geometric process) is used.
Step 1: Define the probability Let \(p\) be the probability that \(A\) wins when it is \(B\)'s turn to throw.
Step 2: Consider \(B\)'s throw
With probability \( \dfrac{1}{9} \), \(B\) throws a sum of \(9\) and wins \(\Rightarrow A\) loses.
With probability \( \dfrac{8}{9} \), the game continues to \(A\)'s turn.
Step 3: Consider \(A\)'s throw
With probability \( \dfrac{1}{12} \), \(A\) throws a sum of \(4\) and wins.
With probability \( \dfrac{11}{12} \), the game returns to \(B\)'s turn.
Step 4: Form the equation \[ p = \frac{8}{9}\left(\frac{1}{12} + \frac{11}{12}p\right) \]
Step 5: Solve \[ p = \frac{2}{27}(1 + 11p) \] \[ 27p = 2 + 22p \Rightarrow 5p = 2 \Rightarrow p = \frac{2}{5} \]