Question

Two players A and B are alternately throwing a coin and a die together. A player who first throws a head and a 6 wins the game. If A starts the game, then the probability that B wins the game is:

• A. \( \frac{12}{23} \)
• B. \( \frac{11}{23} \)
• C. \( \frac{5}{119} \)
• D. \( \frac{12}{119} \)

Hint:

In problems involving alternating turns and infinite sequences, identify the probabilities of the sequence of events and use the sum of an infinite geometric series to solve for the desired outcome.

Answer 1

The Correct Option is B

To address this problem, we first compute the probability of obtaining a head on the coin and a 6 on the die concurrently in a single trial. This probability is calculated as: \[ P(\text{Head and 6}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}. \] Subsequently, we ascertain the probability of Player B winning the game. Player B achieves victory if Player A fails to win on the first, third, fifth, and subsequent odd-numbered throws, and Player B then secures a win on a subsequent throw. Consequently, our analysis must encompass the sequence of events where Player A does not win initially, followed by Player B's victory.

Step 1: Compute the probability of Player B winning on the second, fourth, sixth, and other even-numbered throws. The probability of Player A not winning on the initial throw is: \[ P(\text{A does not win in 1st throw}) = \frac{11}{12}. \] The probability of Player B winning on the second throw is: \[ P(\text{B wins in 2nd throw}) = \frac{1}{12}. \] Therefore, the probability that Player B wins on the second throw is: \[ P(\text{B wins in 2nd throw}) = \frac{11}{12} \times \frac{1}{12}. \] Similarly, for Player B to win on the fourth throw: \[ P(\text{B wins in 4th throw}) = \frac{11}{12} \times \frac{11}{12} \times \frac{1}{12}. \]

 Step 2: Formulate the infinite series. A discernible pattern emerges where the probabilities of Player B winning on the 2nd, 4th, 6th, and other even-numbered throws constitute an infinite geometric series: \[ P(\text{B wins}) = \frac{11}{12} \times \frac{1}{12} + \frac{11}{12}^2 \times \frac{1}{12} + \frac{11}{12}^3 \times \frac{1}{12} + \dots \] The sum of this infinite geometric series is determined by the formula: \[ P(\text{B wins}) = \frac{1}{12} \cdot \frac{\frac{11}{12}}{1 - \left(\frac{11}{12}\right)^2}. \]

 Step 3: Simplify the derived expression. Upon simplification of the aforementioned expression: \[ P(\text{B wins}) = \frac{1}{12} \cdot \frac{\frac{11}{12}}{1 - \frac{121}{144}} = \frac{1}{12} \cdot \frac{\frac{11}{12}}{\frac{23}{144}}. \] \[ P(\text{B wins}) = \frac{1}{12} \times \frac{11}{12} \times \frac{144}{23} = \frac{11}{23}. \] 

Final Answer: \[ \boxed{\frac{11}{23}}. \]