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The problem asks to find the probability that player A obtains the first six when A and B alternate rolling a fair six-sided die, with A starting.

### Steps to Solve:
1. **Basic Probabilities**: The probability of rolling a six with a fair six-sided die is \( \frac{1}{6} \). The probability of **not** rolling a six is \( \frac{5}{6} \).

2. **Player A's First Roll**: Since A rolls first, the probability that A rolls a six on the first attempt is:
   \[
   P(A \text{ gets the first six}) = \frac{1}{6}
   \]

3. **Player B's Turn**: If A does **not** roll a six, then it is B's turn to roll. The probability of this happening (A not getting a six) is \( \frac{5}{6} \). Now, for B, the probability that they roll a six is \( \frac{1}{6} \). So, the probability that A misses and B gets a six is:
   \[
   P(A \text{ misses and B gets a six}) = \frac{5}{6} \times \frac{1}{6}
   \]

4. **Repeating Process**: If both A and B fail to roll a six on their first turns, the process repeats with A rolling again. The situation essentially resets, with the same probabilities, after each pair of turns. Hence, this forms a **geometric series** where A’s first chance comes on every odd turn.

### Total Probability that A Obtains the First Six:

The total probability that A rolls the first six can be expressed as an infinite series:

\[
P(A \text{ gets the first six}) = \frac{1}{6} + \left( \frac{5}{6} \times \frac{5}{6} \right) \times \frac{1}{6} + \left( \frac{5}{6} \times \frac{5}{6} \right)^2 \times \frac{1}{6} + \cdots
\]

This is a geometric series with the first term \( \frac{1}{6} \) and the common ratio \( \left( \frac{5}{6} \times \frac{5}{6} \right) = \frac{25}{36} \).

The sum of an infinite geometric series \( S = a + ar + ar^2 + \cdots \) is given by the formula:

\[
S = \frac{a}{1 - r}
\]

where \( a = \frac{1}{6} \) and \( r = \frac{25}{36} \).

Thus, the probability that A gets the first six is:

\[
P(A \text{ gets the first six}) = \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{\frac{1}{6}}{\frac{11}{36}} = \frac{6}{11}
\]

### Final Answer:
The probability that A obtains the first six is \( \frac{6}{11} \).

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Would you like more details on any step or further clarification? Here are some related questions to consider:

1. How would the probability change if the die had more or fewer sides?
2. Can we calculate the expected number of rolls before the first six occurs?
3. What if player B started instead of A—how would that affect the result?
4. How does the geometric series factor into solving this problem?
5. Can this problem be solved using a recursive approach?

**Tip**: Whenever dealing with alternating probabilities, look for patterns that form a geometric series.

Two players, A and B, alternately throw a fair six-sided dice, with A starting, until one of them obtains a six. Find the probability that A obtains the first six.

Discover how to calculate the probability that Player A gets the first six in a dice game where two players alternate rolling a fair six-sided die. The solution uses probability concepts and geometric series, providing a clear step-by-step explanation. Suitable for Grades 9-12 students.

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