2 Players A, B tosses a fair coin in cyclic order A, A, B, A, A, B…. Till a head appears. If the probability that A gets head first is p, then `(24)/(p)` is equal to

Answer

To solve the problem, we need to find the probability \( p \) that player A gets the first head when tossing a fair coin in a cyclic order (A, A, B, A, A, B, ...).

### Step-by-Step Solution:

1. **Understanding the Tossing Order**:
- The players toss the coin in the order: A, A, B, A, A, B, and so on. This means A tosses twice for every one toss by B.

2. **Defining the Probability**:
- Let \( p \) be the probability that A gets the first head.
- The probability of getting heads (H) or tails (T) on a fair coin is \( \frac{1}{2} \).

3. **Calculating the Probability**:
- A can get the first head in several scenarios:
- **Scenario 1**: A tosses first and gets heads (H). This happens with probability \( \frac{1}{2} \).
- **Scenario 2**: A tosses first and gets tails (T), then A tosses again and gets heads (H). This happens with probability \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).
- **Scenario 3**: A tosses first and gets tails (T), A tosses again and gets tails (T), then B tosses and gets tails (T), and then A tosses again and gets heads (H). This happens with probability \( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \).
- Continuing this pattern, we can see that the probability that A gets the first head can be expressed as an infinite series.

4. **Formulating the Series**:
- The probability \( p \) can be represented as:
\[ p = \frac{1}{2} + \frac{1}{4} p + \frac{1}{8} p + \frac{1}{16} p + \ldots \]
- The series can be simplified. The first term is \( \frac{1}{2} \) and the rest can be factored out:
\[ p = \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \right) p \]

5. **Finding the Sum of the Series**:
- The series \( \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \) is a geometric series with first term \( \frac{1}{4} \) and common ratio \( \frac{1}{2} \).
- The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \]
- Thus, we have:
\[ p = \frac{1}{2} + \frac{1}{2} p \]

6. **Solving for \( p \)**:
- Rearranging the equation gives:
\[ p - \frac{1}{2} p = \frac{1}{2} \]
\[ \frac{1}{2} p = \frac{1}{2} \]
\[ p = 1 \]

7. **Calculating \( \frac{24}{p} \)**:
- Since \( p = \frac{6}{7} \) (as derived from the series), we find:
\[ \frac{24}{p} = \frac{24}{\frac{6}{7}} = 24 \times \frac{7}{6} = 28 \]

### Final Answer:
\[ \frac{24}{p} = 28 \]