Al and Bob play Rock Paper Scissors until someone wins a game. What is the probability that this happens on the sixth game?

1. **Identify the probability of a tie and a win/loss in a single game:**
   - The probability of a tie in a single game of Rock Paper Scissors is $\frac{1}{3}$.
   - The probability of not tying (i.e., someone wins) in a single game is $1 - \frac{1}{3} = \frac{2}{3}$.

2. **Determine the sequence of events for the game to end on the sixth game:**
   - For the game to end on the sixth game, the first five games must all be ties, and the sixth game must be a win/loss.
   - The probability of tying five consecutive games is $\left(\frac{1}{3}\right)^5$.
   - The probability of a win/loss on the sixth game is $\frac{2}{3}$.

3. **Calculate the combined probability:**
   - The combined probability of the first five games being ties and the sixth game being a win/loss is:
     \[
     \left(\frac{1}{3}\right)^5 \cdot \frac{2}{3}
     \]

4. **Perform the calculation:**
   \[
   \left(\frac{1}{3}\right)^5 = \frac{1}{3^5} = \frac{1}{243}
   \]
   \[
   \left(\frac{1}{3}\right)^5 \cdot \frac{2}{3} = \frac{1}{243} \cdot \frac{2}{3} = \frac{2}{729}
   \]

Conclusion:
\[
\boxed{\frac{2}{729}}
\]