A, B and C play a game and chances of their winning it in an attempt are `(2)/(3),(1)/(2)` and `(1)/(4)` respectively. A has the first chance, followed by B and then by C. The cycle is repeated till one of them wins the game. Find their respective chances of winning the game.

Answer

To solve the problem, we need to calculate the probability of each player (A, B, and C) winning the game based on their respective chances of winning in each attempt.

### Step-by-Step Solution:

1. **Define the Winning Probabilities:**
- Let \( P(A) = \frac{2}{3} \) (Probability of A winning)
- Let \( P(B) = \frac{1}{2} \) (Probability of B winning)
- Let \( P(C) = \frac{1}{4} \) (Probability of C winning)

2. **Calculate the Losing Probabilities:**
- Probability of A losing: \( P(A') = 1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3} \)
- Probability of B losing: \( P(B') = 1 - P(B) = 1 - \frac{1}{2} = \frac{1}{2} \)
- Probability of C losing: \( P(C') = 1 - P(C) = 1 - \frac{1}{4} = \frac{3}{4} \)

3. **Calculate the Total Probability of Each Player Winning:**
- A can win in the first round: \( P(A) = \frac{2}{3} \)
- A can win in the second round if both B and C lose in the first round:
\[ P(A \text{ wins in 2nd round}) = P(A') \times P(B') \times P(C) = \frac{1}{3} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{24} \]
- A can win in the third round if A loses in the first two rounds and then wins:
\[ P(A \text{ wins in 3rd round}) = P(A') \times P(B') \times P(C') \times P(A) = \frac{1}{3} \times \frac{1}{2} \times \frac{3}{4} \times \frac{2}{3} = \frac{1}{4} \]

4. **Sum the Probabilities for A Winning:**
- The total probability of A winning is:
\[ P(A \text{ wins}) = P(A) + P(A \text{ wins in 2nd round}) + P(A \text{ wins in 3rd round}) + \ldots \]
- This forms a geometric series:
\[ P(A \text{ wins}) = \frac{2}{3} + \frac{1}{24} + \frac{1}{4} + \ldots \]

5. **Calculate the Total Probability of B Winning:**
- B can win in the first round:
\[ P(B \text{ wins}) = P(A') \times P(B) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \]
- B can win in the second round if A and C lose in the first round:
\[ P(B \text{ wins in 2nd round}) = P(A') \times P(B') \times P(C) = \frac{1}{3} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{24} \]

6. **Calculate the Total Probability of C Winning:**
- C can win in the first round:
\[ P(C \text{ wins}) = P(A') \times P(B') \times P(C) = \frac{1}{3} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{24} \]

7. **Final Calculation:**
- The total probabilities for A, B, and C can be summed up and normalized to find their respective chances of winning.

### Final Result:
Let \( P_A, P_B, P_C \) be the probabilities of A, B, and C winning respectively. After calculating and normalizing, we can find their respective chances.