Bill plays a game in which he rolls two fair standard six-sided dice with sides labeled one through six. He wins if the number on one of the dice is three times the number on the other die. If Bill plays this game three times, compute the probability that he wins at least once.
1. **Identify the winning combinations:**
   Bill wins if the number on one die is three times the number on the other die. The possible pairs that satisfy this condition are:
   - $(1, 3)$
   - $(3, 1)$
   - $(2, 6)$
   - $(6, 2)$

2. **Calculate the probability of winning in a single game:**
   There are 4 winning outcomes out of a total of $6 \times 6 = 36$ possible outcomes when rolling two six-sided dice. Therefore, the probability of winning in a single game is:
   \[
   P(\text{win}) = \frac{4}{36} = \frac{1}{9}
   \]

3. **Calculate the probability of losing in a single game:**
   The probability of losing in a single game is the complement of the probability of winning:
   \[
   P(\text{lose}) = 1 - P(\text{win}) = 1 - \frac{1}{9} = \frac{8}{9}
   \]

4. **Calculate the probability of losing all three games:**
   The probability of losing all three games is the product of the probabilities of losing each individual game:
   \[
   P(\text{lose all three}) = \left(\frac{8}{9}\right)^3 = \frac{8^3}{9^3} = \frac{512}{729}
   \]

5. **Calculate the probability of winning at least once:**
   The probability of winning at least once is the complement of the probability of losing all three games:
   \[
   P(\text{win at least once}) = 1 - P(\text{lose all three}) = 1 - \frac{512}{729} = \frac{729 - 512}{729} = \frac{217}{729}
   \]

Conclusion:
\[
\boxed{\frac{217}{729}}
\]