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 :

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 probability of player X winning is equal to the probability of player Y winning.

### Step-by-Step Solution:

1. **Define the Probabilities**:
- Let \( P(X) \) be the probability that player X wins.
- Player X has a biased coin with a 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. **Calculate the Probability of X Winning**:
- Player X can win in two scenarios:
1. X wins on the first toss (which happens with probability \( p \)).
2. Both players get tails in their first tosses, and then X wins in the subsequent rounds.

The probability of both getting tails is:
\[ (1 - p) \times \frac{1}{2} \]
After both get tails, the game resets with X starting again. Thus, we can express \( P(X) \) as:
\[ P(X) = p + (1 - p) \cdot \frac{1}{2} \cdot P(X) \]
Rearranging gives:
\[ P(X) - (1 - p) \cdot \frac{1}{2} \cdot P(X) = p \]
\[ P(X) \left(1 - \frac{1 - p}{2}\right) = p \]
\[ P(X) \left(\frac{2 - (1 - p)}{2}\right) = p \]
\[ P(X) \left(\frac{1 + p}{2}\right) = p \]
\[ P(X) = \frac{2p}{1 + p} \]

3. **Calculate the Probability of Y Winning**:
- Player Y can win in two scenarios:
1. Y wins on the first toss (which happens with probability \( \frac{1}{2} \)).
2. Both players get tails in their first tosses, and then Y wins in the subsequent rounds.

The probability of both getting tails is:
\[ (1 - p) \cdot \frac{1}{2} \]
After both get tails, the game resets with X starting again. Thus, we can express \( P(Y) \) as:
\[ P(Y) = \frac{1}{2} + (1 - p) \cdot \frac{1}{2} \cdot P(Y) \]
Rearranging gives:
\[ P(Y) - (1 - p) \cdot \frac{1}{2} \cdot P(Y) = \frac{1}{2} \]
\[ P(Y) \left(1 - \frac{1 - p}{2}\right) = \frac{1}{2} \]
\[ P(Y) \left(\frac{1 + p}{2}\right) = \frac{1}{2} \]
\[ P(Y) = \frac{1}{1 + p} \]

4. **Set the Probabilities Equal**:
Since the probabilities of winning are equal:
\[ \frac{2p}{1 + p} = \frac{1}{1 + p} \]
Cross-multiplying gives:
\[ 2p(1 + p) = 1 \]
\[ 2p + 2p^2 = 1 \]
Rearranging gives:
\[ 2p^2 + 2p - 1 = 0 \]

5. **Solve the Quadratic Equation**:
Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = 2, c = -1 \):
\[ p = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ p = \frac{-2 \pm \sqrt{4 + 8}}{4} \]
\[ p = \frac{-2 \pm \sqrt{12}}{4} \]
\[ p = \frac{-2 \pm 2\sqrt{3}}{4} \]
\[ p = \frac{-1 \pm \sqrt{3}}{2} \]
Since \( p \) must be a probability (between 0 and 1), we take the positive root:
\[ p = \frac{-1 + \sqrt{3}}{2} \]

### Final Answer:
The value of \( p \) is:
\[ p = \frac{-1 + \sqrt{3}}{2} \]