Jacob and Aaron are playing a game in which Aaron is trying to guess the outcome of an unfair coin which shows heads $\tfrac{2}{3}$ of the time.  Aaron randomly guesses ``heads'' $\tfrac{2}{3}$ of the time, and guesses ``tails'' the other $\tfrac{1}{3}$ of the time.  If the probability that Aaron guesses correctly is $p$, compute $9000p$.

[i]Proposed by Aaron Lin[/i]
1. Let's denote the probability of the coin showing heads as \( P(H) = \frac{2}{3} \) and the probability of the coin showing tails as \( P(T) = 1 - P(H) = \frac{1}{3} \).

2. Aaron guesses heads with probability \( P(G_H) = \frac{2}{3} \) and guesses tails with probability \( P(G_T) = 1 - P(G_H) = \frac{1}{3} \).

3. We need to find the probability \( p \) that Aaron guesses correctly. This can happen in two scenarios:
   - The coin shows heads and Aaron guesses heads.
   - The coin shows tails and Aaron guesses tails.

4. The probability that the coin shows heads and Aaron guesses heads is:
   \[
   P(H \cap G_H) = P(H) \cdot P(G_H) = \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}
   \]

5. The probability that the coin shows tails and Aaron guesses tails is:
   \[
   P(T \cap G_T) = P(T) \cdot P(G_T) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}
   \]

6. The total probability that Aaron guesses correctly is the sum of the probabilities of the two scenarios:
   \[
   p = P(H \cap G_H) + P(T \cap G_T) = \frac{4}{9} + \frac{1}{9} = \frac{5}{9}
   \]

7. We are asked to compute \( 9000p \):
   \[
   9000p = 9000 \cdot \frac{5}{9} = 1000 \cdot 5 = 5000
   \]

The final answer is \( \boxed{5000} \).