A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If X starts the game, and the probability of winning the game by both the players is equal, then the value of 'p is : (a) `(1)/(3)` (b) `(2)/(5)` (c) `(1)/(4)` (d) `(1)/(5)`

Options

A. `(1)/(3)`
B. `(2)/(5)`
C. `(1)/(4)`
D. `(1)/(5)`

Answer

To solve the problem, we need to find the value of \( p \) such that the probabilities of winning for both players X and Y are equal.

### Step-by-Step Solution:

1. **Define the probabilities**:
- Player X has a biased coin with the probability of heads \( p \) and tails \( 1 - p \).
- Player Y has a fair coin, so the probability of heads is \( \frac{1}{2} \) and tails is also \( \frac{1}{2} \).

2. **Winning probability for Player X**:
- Player X wins immediately if he gets heads on the first toss, which occurs with probability \( p \).
- If he gets tails (with probability \( 1 - p \)), then Player Y gets a chance to toss.
- If Player Y also gets tails (with probability \( \frac{1}{2} \)), the game resets to the original situation where Player X starts again.
- Therefore, the probability \( P_X \) that Player X wins can be expressed as:
\[ P_X = p + (1 - p) \cdot \frac{1}{2} \cdot P_X \]
This equation states that Player X wins either by tossing heads immediately or by both players tossing tails and returning to the same situation.

3. **Rearranging the equation**:
- Rearranging gives:
\[ P_X = p + \frac{(1 - p)}{2} P_X \]
\[ P_X - \frac{(1 - p)}{2} P_X = p \]
\[ P_X \left(1 - \frac{(1 - p)}{2}\right) = p \]
\[ P_X \left(\frac{1 + p}{2}\right) = p \]
\[ P_X = \frac{2p}{1 + p} \]

4. **Winning probability for Player Y**:
- Player Y wins if Player X tosses tails (probability \( 1 - p \)) and then Player Y tosses heads (probability \( \frac{1}{2} \)).
- If Player Y also tosses tails (probability \( \frac{1}{2} \)), the game resets.
- Therefore, the probability \( P_Y \) that Player Y wins can be expressed as:
\[ P_Y = (1 - p) \cdot \frac{1}{2} + (1 - p) \cdot \frac{1}{2} \cdot P_Y \]
\[ P_Y = \frac{(1 - p)}{2} + \frac{(1 - p)}{2} P_Y \]
\[ P_Y - \frac{(1 - p)}{2} P_Y = \frac{(1 - p)}{2} \]
\[ P_Y \left(1 - \frac{(1 - p)}{2}\right) = \frac{(1 - p)}{2} \]
\[ P_Y \left(\frac{1 + p}{2}\right) = \frac{(1 - p)}{2} \]
\[ P_Y = \frac{(1 - p)}{1 + p} \]

5. **Setting the probabilities equal**:
- Since the problem states that the probabilities of winning for both players are equal, we set \( P_X = P_Y \):
\[ \frac{2p}{1 + p} = \frac{(1 - p)}{1 + p} \]
- Cross-multiplying gives:
\[ 2p(1 + p) = (1 - p)(1 + p) \]
\[ 2p + 2p^2 = 1 - p^2 \]
\[ 3p^2 + 2p - 1 = 0 \]

6. **Solving the quadratic equation**:
- Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ p = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
\[ p = \frac{-2 \pm \sqrt{4 + 12}}{6} \]
\[ p = \frac{-2 \pm \sqrt{16}}{6} \]
\[ p = \frac{-2 \pm 4}{6} \]
- This gives two possible solutions:
\[ p = \frac{2}{6} = \frac{1}{3} \quad \text{and} \quad p = \frac{-6}{6} = -1 \text{ (not valid)} \]

7. **Final Answer**:
- The valid solution is \( p = \frac{1}{3} \).

### Conclusion:
The value of \( p \) is \( \frac{1}{3} \).