If $0 \le p \le 1$ and $0 \le q \le 1$, define $F(p, q)$ by
\[
  F(p, q) = -2pq + 3p(1-q) + 3(1-p)q - 4(1-p)(1-q).
\]
Define $G(p)$ to be the maximum of $F(p, q)$ over all $q$ (in the interval $0 \le q \le 1$).  What is the value of $p$ (in the interval $0 \le p \le 1$) that minimizes $G(p)$?
1. **Express \( F(p, q) \) in terms of \( q \):**
   \[
   F(p, q) = -2pq + 3p(1-q) + 3(1-p)q - 4(1-p)(1-q)
   \]
   Simplify the expression:
   \[
   F(p, q) = -2pq + 3p - 3pq + 3q - 3pq - 4 + 4p + 4q - 4pq
   \]
   Combine like terms:
   \[
   F(p, q) = -2pq - 3pq - 3pq - 4pq + 3p + 4p + 3q + 4q - 4
   \]
   \[
   F(p, q) = -12pq + 7p + 7q - 4
   \]

2. **Evaluate \( F(p, q) \) at \( q = 0 \) and \( q = 1 \):**
   \[
   F(p, 0) = -12p(0) + 7p + 7(0) - 4 = 7p - 4
   \]
   \[
   F(p, 1) = -12p(1) + 7p + 7(1) - 4 = -5p + 3
   \]

3. **Define \( G(p) \) as the maximum of \( F(p, 0) \) and \( F(p, 1) \):**
   \[
   G(p) = \max(7p - 4, 3 - 5p)
   \]

4. **Find the value of \( p \) that minimizes \( G(p) \):**
   To minimize \( G(p) \), we need to find the point where \( 7p - 4 = 3 - 5p \):
   \[
   7p - 4 = 3 - 5p
   \]
   Solve for \( p \):
   \[
   7p + 5p = 3 + 4
   \]
   \[
   12p = 7
   \]
   \[
   p = \frac{7}{12}
   \]

The final answer is \( \boxed{ p = \frac{7}{12} } \)