A chess match between two players `A` and `B` is won by whoever first wins a total of two games. Probability of `A's` winning, drawng and losing any particular game are `1/6, 1/3` and `1/2` respectively. (The game are independent). If the probability that `B` wins the match in the `4^(th)` gams is `p`, then `6p` is equal to

Explanation

To solve the problem, we need to find the probability \( p \) that player \( B \) wins the match in the 4th game. The match is won by the first player to win 2 games.

### Step-by-step Solution:

1. **Understanding the Game Structure**:
- Player \( A \) can win, draw, or lose a game with probabilities:
- \( P(A \text{ wins}) = \frac{1}{6} \)
- \( P(\text{Draw}) = \frac{1}{3} \)
- \( P(B \text{ wins}) = \frac{1}{2} \)

Since the game can end in a draw, we need to consider the scenarios where \( B \) wins the match in the 4th game.

2. **Possible Scenarios for \( B \) to Win in the 4th Game**:
- For \( B \) to win in the 4th game, the following scenarios can occur:
- \( A \) wins 1 game, \( B \) wins 1 game, and the other 2 games are draws.
- \( B \) wins 2 games and \( A \) wins 1 game (not possible since we need \( B \) to win in the 4th game).

The valid sequence of outcomes for \( B \) to win in the 4th game can be:
- \( A, B, D, B \)
- \( D, A, B, B \)
- \( B, D, A, B \)
- \( D, D, A, B \)
- \( D, B, D, B \)

3. **Calculating the Probability**:
- The probability that \( B \) wins in the 4th game can be calculated as follows:
- Let’s denote:
- \( W_A = \frac{1}{6} \) (A wins)
- \( W_B = \frac{1}{2} \) (B wins)
- \( D = \frac{1}{3} \) (Draw)

The probability that \( B \) wins in the 4th game can be calculated by considering the valid sequences:
- For the sequence \( A, B, D, B \):
\[ P(A, B, D, B) = W_A \cdot W_B \cdot D \cdot W_B = \frac{1}{6} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{72} \]
- For the sequence \( D, A, B, B \):
\[ P(D, A, B, B) = D \cdot W_A \cdot W_B \cdot W_B = \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{72} \]
- For the sequence \( B, D, A, B \):
\[ P(B, D, A, B) = W_B \cdot D \cdot W_A \cdot W_B = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{1}{2} = \frac{1}{72} \]
- For the sequence \( D, D, A, B \):
\[ P(D, D, A, B) = D \cdot D \cdot W_A \cdot W_B = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{1}{2} = \frac{1}{108} \]
- For the sequence \( D, B, D, B \):
\[ P(D, B, D, B) = D \cdot W_B \cdot D \cdot W_B = \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{36} \]

4. **Combining the Probabilities**:
- The total probability \( p \) that \( B \) wins in the 4th game is:
\[ p = P(A, B, D, B) + P(D, A, B, B) + P(B, D, A, B) + P(D, D, A, B) + P(D, B, D, B) \]
\[ p = \frac{1}{72} + \frac{1}{72} + \frac{1}{72} + \frac{1}{108} + \frac{1}{36} \]

To combine these fractions, we find a common denominator, which is \( 216 \):
\[ p = \frac{3}{216} + \frac{3}{216} + \frac{3}{216} + \frac{2}{216} + \frac{6}{216} = \frac{17}{216} \]

5. **Calculating \( 6p \)**:
\[ 6p = 6 \cdot \frac{17}{216} = \frac{102}{216} = \frac{17}{36} \]

### Final Answer:
Thus, \( 6p = \frac{17}{36} \).