Charles has two six-sided dice. One of the dice is fair, and the other die is biased so that it comes up six with probability $\tfrac23,$ and each of the other five sides has probability $\tfrac{1}{15}.$ Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
1. **Determine the probability of rolling two sixes with each die:**
   - For the fair die, the probability of rolling a six on any single roll is \(\frac{1}{6}\). Therefore, the probability of rolling two sixes in a row is:
     \[
     \left(\frac{1}{6}\right)^2 = \frac{1}{36}
     \]
   - For the biased die, the probability of rolling a six on any single roll is \(\frac{2}{3}\). Therefore, the probability of rolling two sixes in a row is:
     \[
     \left(\frac{2}{3}\right)^2 = \frac{4}{9}
     \]

2. **Calculate the total probability of rolling two sixes:**
   - Since Charles chooses one of the two dice at random, the probability of choosing either die is \(\frac{1}{2}\).
   - The total probability of rolling two sixes is the sum of the probabilities of rolling two sixes with each die, weighted by the probability of choosing each die:
     \[
     P(\text{two sixes}) = \frac{1}{2} \cdot \frac{1}{36} + \frac{1}{2} \cdot \frac{4}{9}
     \]
     Simplifying this, we get:
     \[
     P(\text{two sixes}) = \frac{1}{72} + \frac{2}{9} = \frac{1}{72} + \frac{16}{72} = \frac{17}{72}
     \]

3. **Determine the conditional probabilities of having chosen each die given that two sixes were rolled:**
   - Using Bayes' Theorem, we find the probability that the fair die was chosen given that two sixes were rolled:
     \[
     P(\text{fair die} \mid \text{two sixes}) = \frac{P(\text{two sixes} \mid \text{fair die}) \cdot P(\text{fair die})}{P(\text{two sixes})} = \frac{\frac{1}{36} \cdot \frac{1}{2}}{\frac{17}{72}} = \frac{\frac{1}{72}}{\frac{17}{72}} = \frac{1}{17}
     \]
   - Similarly, the probability that the biased die was chosen given that two sixes were rolled:
     \[
     P(\text{biased die} \mid \text{two sixes}) = \frac{P(\text{two sixes} \mid \text{biased die}) \cdot P(\text{biased die})}{P(\text{two sixes})} = \frac{\frac{4}{9} \cdot \frac{1}{2}}{\frac{17}{72}} = \frac{\frac{2}{9}}{\frac{17}{72}} = \frac{16}{17}
     \]

4. **Calculate the probability that the third roll is a six:**
   - Given that the fair die was chosen, the probability of rolling a six on the third roll is \(\frac{1}{6}\).
   - Given that the biased die was chosen, the probability of rolling a six on the third roll is \(\frac{2}{3}\).
   - The total probability that the third roll is a six is the sum of these probabilities, weighted by the conditional probabilities of having chosen each die:
     \[
     P(\text{third roll is six}) = P(\text{third roll is six} \mid \text{fair die}) \cdot P(\text{fair die} \mid \text{two sixes}) + P(\text{third roll is six} \mid \text{biased die}) \cdot P(\text{biased die} \mid \text{two sixes})
     \]
     Substituting the values, we get:
     \[
     P(\text{third roll is six}) = \frac{1}{6} \cdot \frac{1}{17} + \frac{2}{3} \cdot \frac{16}{17} = \frac{1}{102} + \frac{32}{51} = \frac{1}{102} + \frac{64}{102} = \frac{65}{102}
     \]

The final answer is \(\boxed{167}\).