[THOUGHT] To find the expected value of the total prize money $X$, we need to consider the probabilities of drawing a crimson ball on the first and fourth draws. The prize for drawing a crimson ball on the first draw is 100 units, and for the fourth draw is 200 units. The expected value $E[X]$ is the sum of the expected prizes from these two events.

1. First Draw: The probability of drawing a crimson ball on the first draw is $\frac{4}{9}$, and the prize is 100 units. The expected prize from the first draw is $100 \times \frac{4}{9}$.

2. Fourth Draw: The probability of drawing a crimson ball on the fourth draw is more complex because it depends on the outcomes of the first three draws. However, by symmetry, the probability of drawing a crimson ball on any specific draw (including the fourth) is the same as the first draw, which is $\frac{4}{9}$. The expected prize from the fourth draw is $200 \times \frac{4}{9}$.

Thus, the total expected value $E[X]$ is:

The fraction $\frac{400}{3}$ is already in its simplest form, so $p = 400$ and $q = 3$. The value of $p \cdot q^2$ is $400 \times 3^2 = 400 \times 9 = 3600$.

[/THOUGHT] [PYTHON]

Verification of the expected value calculation

from fractions import Fraction

Probability of crimson on first draw

prob_first = Fraction(4, 9) prize_first = 100

Probability of crimson on fourth draw (same as first due to symmetry)

prob_fourth = Fraction(4, 9) prize_fourth = 200

Expected value calculation

E_X = prize_first * prob_first + prize_fourth * prob_fourth p = E_X.numerator q = E_X.denominator result = p * q * q

print(result) [/PYTHON]

[THOUGHT] The Python output confirms that the expected value calculation is correct, yielding $p \cdot q^2 = 3600$.

[/THOUGHT] $\boxed{3600}$