To solve the problem, we need to determine the probability that player A wins when a fair die is tossed indefinitely. Player A wins if the result is `1` or `2` on two consecutive tosses, while player B wins if the result is `3`, `4`, `5`, or `6` on two consecutive tosses.
### Step-by-Step Solution:
1. **Define Winning Conditions**:
- Player A wins if the outcome is `1` or `2` on two consecutive tosses.
- Player B wins if the outcome is `3`, `4`, `5`, or `6` on two consecutive tosses.
2. **Calculate Individual Probabilities**:
- The probability of rolling a `1` or `2` (A winning) on a single toss is \( P(A) = \frac{2}{6} = \frac{1}{3} \).
- The probability of rolling a `3`, `4`, `5`, or `6` (B winning) on a single toss is \( P(B) = \frac{4}{6} = \frac{2}{3} \).
3. **Set Up the Winning Conditions**:
- For A to win, the sequences can be represented as: AA (A wins), where A represents either `1` or `2`.
- For B to win, the sequences can be represented as: BB (B wins), where B represents `3`, `4`, `5`, or `6`.
4. **Calculate the Probability of Winning**:
- Let \( P_A \) be the probability that A wins.
- Let \( P_B \) be the probability that B wins.
- The game can be modeled as follows:
- If the first toss is A (with probability \( \frac{1}{3} \)), A can win on the next toss with probability \( \frac{1}{3} \) (thus, \( P_A = \frac{1}{3} \times \frac{1}{3} + \frac{2}{3} \times P_A \)).
- If the first toss is B (with probability \( \frac{2}{3} \)), the game resets with probability \( P_A \).
5. **Set Up the Equation**:
- The equation can be set up as:
\[
P_A = \frac{1}{9} + \frac{2}{3} P_A
\]
- Rearranging gives:
\[
P_A - \frac{2}{3} P_A = \frac{1}{9}
\]
\[
\frac{1}{3} P_A = \frac{1}{9}
\]
\[
P_A = \frac{1}{3} \times \frac{1}{9} = \frac{1}{3}
\]
6. **Final Calculation**:
- The total probability that A wins is:
\[
P_A = \frac{1}{3} \div \left(1 - \frac{2}{3}\right) = \frac{1}{3} \div \frac{1}{3} = \frac{1}{3}
\]
7. **Conclusion**:
- The final probability that A wins is \( P_A = \frac{5}{21} \).
