(i) A and B toss a coin alternately till one of them tosses a head and wins the game. If A starts the game, find their respective probability of winning.
(ii) A and B throw a coin alternately till one of them gets a 'head' and wins the game. If A starts the game, find the probability of his winning at his third throw.

Answer

To solve the given problem, we will break it down into two parts as stated in the question.

### Part (i): Probability of A and B Winning

1. **Understanding the Game**: A and B toss a coin alternately until one of them gets a head. A starts first.
2. **Probabilities**:
- The probability of getting a head (H) on any toss is \( P(H) = \frac{1}{2} \).
- The probability of getting a tail (T) on any toss is \( P(T) = \frac{1}{2} \).

3. **Winning Scenarios for A**:
- A can win on his first toss: \( P(A \text{ wins on 1st toss}) = \frac{1}{2} \).
- If A gets a tail (T), then B tosses. For A to win on his second turn, both A and B must get tails on their first tosses:
- \( P(A \text{ wins on 2nd toss}) = P(T) \times P(T) \times P(H) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \).
- For A to win on his third turn, both A and B must get tails on their first two tosses, and A must get a head on his third toss:
- \( P(A \text{ wins on 3rd toss}) = P(T) \times P(T) \times P(T) \times P(H) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16} \).

4. **Total Probability of A Winning**:
- The total probability \( P(A \text{ wins}) \) is the sum of the probabilities of winning on the first, second, third, etc. turns:
\[ P(A \text{ wins}) = P(A \text{ wins on 1st toss}) + P(A \text{ wins on 2nd toss}) + P(A \text{ wins on 3rd toss}) + \ldots \]
\[ P(A \text{ wins}) = \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \ldots \]
- This is a geometric series where the first term \( a = \frac{1}{2} \) and the common ratio \( r = \frac{1}{4} \).
- The sum of an infinite geometric series is given by \( S = \frac{a}{1 - r} \):
\[ S = \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3} \]

5. **Conclusion for Part (i)**:
- The probability of A winning is \( \frac{2}{3} \).
- The probability of B winning is \( 1 - P(A \text{ wins}) = 1 - \frac{2}{3} = \frac{1}{3} \).

### Part (ii): Probability of A Winning on His Third Throw

1. **Winning on the Third Throw**: For A to win on his third throw, the sequence of events must be:
- A must toss a tail (T) on his first toss.
- B must toss a tail (T) on his first toss.
- A must toss a tail (T) on his second toss.
- B must toss a tail (T) on his second toss.
- Finally, A must toss a head (H) on his third toss.

2. **Calculating the Probability**:
- The probability of this specific sequence is:
\[ P(T) \times P(T) \times P(T) \times P(T) \times P(H) = \left(\frac{1}{2}\right)^4 \times \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^5 = \frac{1}{32} \]

3. **Conclusion for Part (ii)**:
- The probability of A winning on his third throw is \( \frac{1}{32} \).

### Final Answers:
(i) Probability of A winning = \( \frac{2}{3} \), Probability of B winning = \( \frac{1}{3} \).
(ii) Probability of A winning on his third throw = \( \frac{1}{32} \).