A and B throw alternately with a pair of dice. A wins if he throws 6 before B throws 7, and B if he throws 7 before A throws 6. If A begins, find his chance of winning.

Explanation

To solve the problem of A and B throwing dice alternately, we need to calculate the probability of A winning by throwing a sum of 6 before B throws a sum of 7.

### Step-by-Step Solution:

1. **Determine the Total Outcomes**:
When throwing two dice, the total number of outcomes is \(36\) (since each die has 6 faces, \(6 \times 6 = 36\)).

2. **Calculate the Probability of A Winning (Throwing a 6)**:
The combinations that yield a sum of 6 are:
- (1,5)
- (2,4)
- (3,3)
- (4,2)
- (5,1)

Thus, there are \(5\) favorable outcomes for A. Therefore, the probability \(P(A \text{ wins})\) is:
\[ P(A \text{ wins}) = \frac{5}{36} \]

3. **Calculate the Probability of B Winning (Throwing a 7)**:
The combinations that yield a sum of 7 are:
- (1,6)
- (2,5)
- (3,4)
- (4,3)
- (5,2)
- (6,1)

Thus, there are \(6\) favorable outcomes for B. Therefore, the probability \(P(B \text{ wins})\) is:
\[ P(B \text{ wins}) = \frac{6}{36} = \frac{1}{6} \]

4. **Calculate the Probability of Neither Winning**:
The probability that neither A nor B wins in a single throw is:
\[ P(\text{neither}) = 1 - P(A \text{ wins}) - P(B \text{ wins}) = 1 - \frac{5}{36} - \frac{6}{36} = \frac{25}{36} \]

5. **Establish the Winning Probability for A**:
Let \(P\) be the probability that A wins the game. A can win in the first throw, or if neither wins, the game resets with A's turn again. Therefore, we can express \(P\) as:
\[ P = P(A \text{ wins}) + P(\text{neither}) \cdot P \]
Substituting the values we have:
\[ P = \frac{5}{36} + \frac{25}{36} P \]

6. **Solve for P**:
Rearranging gives:
\[ P - \frac{25}{36} P = \frac{5}{36} \]
\[ \frac{11}{36} P = \frac{5}{36} \]
\[ P = \frac{5}{11} \]

Thus, the probability that A wins is:
\[ \boxed{\frac{5}{11}} \]