Question

Two persons A and B throw a pair of dice alternately until one of them gets the sum of the numbers appeared on the dice as 4 and the person who gets this result first is declared as the winner. If A starts the game, then the probability that B wins the game is:

• A. \( \frac{11}{23} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{5}{11} \)
• D. \( \frac{8}{17} \)

Hint:

For solving problems involving alternating events, break the problem into stages, find the probability of each event happening, and use geometric series when the events repeat in cycles.

Answer 1

The Correct Option is A

Step 1: Find the probability of getting a sum of 4. The possible outcomes for the sum of two dice to be 4 are: \[ (1, 3), (2, 2), (3, 1). \] Thus, the probability of getting a sum of 4 in one roll is: \[ P(\text{sum} = 4) = \frac{3}{36} = \frac{1}{12}. \] Step 2: Calculate the probability of B winning. The probability that B wins is the sum of the probabilities for B winning on subsequent turns. This is a geometric series, where: \[ P(\text{B wins}) = \frac{11}{12} \times \frac{1}{12} + \left( \frac{11}{12} \right)^2 \times \frac{1}{12} + \left( \frac{11}{12} \right)^3 \times \frac{1}{12} + \cdots \] Step 3: Sum the geometric series. Using the formula for the sum of an infinite geometric series, we get: \[ P(\text{B wins}) = \frac{\frac{11}{144}}{1 - \frac{11}{12}} = \frac{11}{23}. \]

Answer 2

Given:
- Two persons A and B throw a pair of dice alternately.
- The first to get a sum of 4 on the dice wins.
- A starts the game.
Find the probability that B wins.

Step 1: Probability of getting sum 4 on a pair of dice:
Number of outcomes with sum 4: (1,3), (2,2), (3,1) = 3
Total outcomes = 36
\[ p = \frac{3}{36} = \frac{1}{12} \] Probability of **not** getting sum 4:
\[ q = 1 - p = \frac{11}{12} \]

Step 2: The game proceeds with A starting first.
B wins if A fails on first turn and B succeeds, or both fail and then B wins later.

Step 3: Probability B wins:
\[ P = q \times p + q^2 \times q \times p + q^4 \times q \times p + \dots \] Where:
- \( q \times p \): A fails, B succeeds on first turn.
- \( q^2 \times q \times p = q^3 p \): both fail once, then B succeeds.
- \( q^4 \times q \times p = q^5 p \): both fail twice, then B succeeds.

Step 4: Sum the infinite geometric series:
\[ P = q p (1 + q^2 + q^4 + \dots) = q p \sum_{k=0}^\infty q^{2k} = q p \times \frac{1}{1 - q^2} \]

Step 5: Substitute values:
\[ P = \frac{11}{12} \times \frac{1}{12} \times \frac{1}{1 - \left(\frac{11}{12}\right)^2} = \frac{11}{144} \times \frac{1}{1 - \frac{121}{144}} = \frac{11}{144} \times \frac{1}{\frac{23}{144}} = \frac{11}{144} \times \frac{144}{23} = \frac{11}{23} \]

Therefore,
\[ \boxed{\frac{11}{23}} \]