Answer: $\quad \frac{9}{5}$ We first compute the expected number of checks between a single knight-king pair. If the king is located at any of the 4 corners, the knight has 2 possible checks. If the king is located in one of the 8 squares on the side of the board but not in the corner, the knight has 3 possible checks. If the king is located in any of the 4 central squares, the knight has 4 possible checks. Summing up, $4 \cdot 2+8 \cdot 3+4 \cdot 4=48$ of the $16 \cdot 15$ knight-king positions yield a single check, so each pair yields $\frac{48}{16 \cdot 15}=\frac{1}{5}$ expected checks. Now, note that each of the 9 knight-king pairs is in each of $16 \cdot 15$ possible positions with equal probability, so by linearity of expectation the answer is $9 \cdot \frac{1}{5}=\frac{9}{5}$.
4. [6] Joe B. is frustrated with chess. He breaks the board, leaving a $4 \times 4$ board, and throws 3 black knights and 3 white kings at the board. Miraculously, they all land in distinct squares! What is the expected number of checks in the resulting position? (Note that a knight can administer multiple checks and a king can be checked by multiple knights.)