Two people, P and Q, decide to independently roll two identical dice, each with 6 faces, numbered 1 to 6. The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by P and Q. Assume that all 6 numbers on each dice are equi-probable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is ______.

In a single trial where P and Q each roll a fair 6-sided die, there are $6 \times 6 = 36$ equally likely possible outcomes.

1. Probability of a tie in one trial: A tie occurs when both roll the same number: $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$.

   $$P(\text{Tie}) = \frac{6}{36} = \frac{1}{6}$$

2. Probability of a win (someone wins) in one trial:

   $$P(\text{Win}) = 1 - P(\text{Tie}) = 1 - \frac{1}{6} = \frac{5}{6}$$

3. Probability that one of them wins on the third trial: This requires a tie on Trial 1, a tie on Trial 2, and a win on Trial 3:

   $$P = P(\text{Tie}) \times P(\text{Tie}) \times P(\text{Win}) = \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) \times \left(\frac{5}{6}\right) = \frac{5}{216} \approx 0.02315$$

Rounded to 3 decimal places, the probability is 0.023 (acceptable range: $0.021$ to $0.024$).