Susan plays a game in which she rolls two fair standard six-sided dice with sides labeled one
through six. She wins if the number on one of the dice is three times the number on the other
die. If Susan plays this game three times, compute the probability that she wins at least once.
1. **Identify the winning pairs:**
   Susan wins if the number on one die is three times the number on the other die. The possible winning pairs are:
   \[
   (1, 3), (3, 1), (2, 6), (6, 2), (3, 9), (9, 3)
   \]
   However, since the dice are standard six-sided dice, the pairs \((3, 9)\) and \((9, 3)\) are not possible. Therefore, the valid winning pairs are:
   \[
   (1, 3), (3, 1), (2, 6), (6, 2)
   \]

2. **Calculate the probability of winning on a single roll:**
   There are 4 winning outcomes out of a total of \(6 \times 6 = 36\) possible outcomes when rolling two dice. Thus, the probability of winning on a single roll is:
   \[
   P(\text{victory}) = \frac{4}{36} = \frac{1}{9}
   \]

3. **Calculate the probability of losing on a single roll:**
   The probability of losing on a single roll is the complement of the probability of winning:
   \[
   P(\text{loss}) = 1 - P(\text{victory}) = 1 - \frac{1}{9} = \frac{8}{9}
   \]

4. **Calculate the probability of losing all three games:**
   Since each game is an independent event, the probability of losing all three games is:
   \[
   P(\text{no victory in 3 games}) = \left( \frac{8}{9} \right)^3 = \frac{8^3}{9^3} = \frac{512}{729}
   \]

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:
   \[
   P(\text{at least one victory}) = 1 - P(\text{no victory in 3 games}) = 1 - \frac{512}{729} = \frac{729 - 512}{729} = \frac{217}{729}
   \]

The final answer is \(\boxed{\frac{217}{729}}\)