A, B and C in turn throw a die and one who gets a 6 first, wins the game. A takes the first chance followed by B and C, and the process is repeated till one them who gets a 6, wins the game. Find the probabilities of each for winning the game.

Answer

To find the probabilities of A, B, and C winning the game, we can analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding the Game**:
- A, B, and C take turns throwing a die. The first player to roll a 6 wins.
- The probability of rolling a 6 on a die is \( \frac{1}{6} \).
- The probability of not rolling a 6 is \( \frac{5}{6} \).

2. **Probability of A Winning**:
- A can win in the first round by rolling a 6. The probability of this happening is \( \frac{1}{6} \).
- If A does not win in the first round, the probability of this happening is \( \frac{5}{6} \).
- For A to win in the second round, both B and C must not roll a 6, and then A must roll a 6. The probability of this scenario is \( \left(\frac{5}{6}\right)^2 \times \frac{1}{6} \).
- This pattern continues, leading us to the conclusion that A can win in the \( n \)-th round with the probability:
\[ P(A \text{ wins}) = \frac{1}{6} + \left(\frac{5}{6}\right)^3 \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^6 \cdot \frac{1}{6} + \ldots \]
- This series can be recognized as a geometric series where:
- First term \( a = \frac{1}{6} \)
- Common ratio \( r = \left(\frac{5}{6}\right)^3 \)

3. **Sum of the Infinite Geometric Series**:
- The sum \( S \) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
- Substituting the values:
\[ P(A \text{ wins}) = \frac{\frac{1}{6}}{1 - \left(\frac{5}{6}\right)^3} = \frac{\frac{1}{6}}{1 - \frac{125}{216}} = \frac{\frac{1}{6}}{\frac{91}{216}} = \frac{36}{91} \]

4. **Probability of B Winning**:
- For B to win, A must not win in the first round, and then B must roll a 6. The probability is:
\[ P(B \text{ wins}) = \left(\frac{5}{6}\right) \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^3 \cdot \left(\frac{5}{6}\right) \cdot \frac{1}{6} + \ldots \]
- This can be simplified to:
\[ P(B \text{ wins}) = \left(\frac{5}{6}\right) \cdot \frac{1}{6} \cdot \left(1 + \left(\frac{5}{6}\right)^3 + \left(\frac{5}{6}\right)^6 + \ldots\right) \]
- The inner series is again a geometric series:
\[ P(B \text{ wins}) = \left(\frac{5}{6}\right) \cdot \frac{1}{6} \cdot \frac{1}{1 - \left(\frac{5}{6}\right)^3} = \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{216}{91} = \frac{30}{91} \]

5. **Probability of C Winning**:
- The total probability must equal 1:
\[ P(A \text{ wins}) + P(B \text{ wins}) + P(C \text{ wins}) = 1 \]
- Thus:
\[ P(C \text{ wins}) = 1 - \left(P(A \text{ wins}) + P(B \text{ wins})\right) = 1 - \left(\frac{36}{91} + \frac{30}{91}\right) = 1 - \frac{66}{91} = \frac{25}{91} \]

### Final Probabilities:
- \( P(A \text{ wins}) = \frac{36}{91} \)
- \( P(B \text{ wins}) = \frac{30}{91} \)
- \( P(C \text{ wins}) = \frac{25}{91} \)