The numerical answer is $17/27$.
Divide our set of $4$ people into groups of two.
One grouping is $\{A, B\}, \{C,D\}$. There are $2$ other groupings, $\{A, C\}, \{B,D\}$ and $\{A, D\}, \{B,C\}$.
The probability that $A$ and $B$ write each other's names is $\dfrac{1}{9}$. The same applies to $C$ and $D$. Let us compute the probability that both these things happen. It is $\dfrac{1}{81}$.
So the probability that $A$ and $B$ write each other's name, or that $C$ and $D$ do (or both), is $\frac{1}{9}+\frac{1}{9}-\frac{1}{81}.$
We subtract the $1/81$ to avoid "double-counting" the situations where $A$ and $B$ pick each other, and $C$ and $D$ also do. Or else we can think of it as following from the formula $P(X\cup Y)=P(X)+P(Y)-P(X\cap Y).$
The same calculation applies to the other two pairings. So we multiply $\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{81}$ by $3$. The result is $2/3-1/27$, which is $17/27$.