  Consider $n$ cars $C_1, C_2, \ldots, C_n$ that want to park in a parking lot
with parking spaces $1,2,\ldots,n$ that appear in order. Each car $C_i$ has a
parking preference $\alpha_i \in \{1,2,\ldots,n\}$. The cars appear in order,
if their preferred parking spot is not taken, they take it, if the parking spot
is taken, they move forward until they find an empty spot. If they do not find
an empty spot, they do not park. An $n$-tuple $(\alpha_1, \alpha_2, \ldots,
\alpha_n)$ is said to be a parking function, if this list of preferences allows
every car to park under this algorithm. For an integer $k$, we say that an
$n$-tuple is a $k$-Naples parking function if the cars can park with the
modified algorithm, where park $C_i$ backs up $k$-spaces (one by one) if their
spot is taken before trying to find a parking spot in front of them. We
introduce randomness to this problem in two ways: 1) For the original parking
function definition, for each car $C_i$ that has their preference taken, we
decide with probability $p$ whether $C_i$ moves forwards or backwards when
their preferred spot is taken; 2) For the $k$-Naples definition, for each car
$C_i$ that has their preference taken, we decide with probability $p$ whether
$C_i$ backs up $k$ spaces or not before moving forward. In each of these
models, for an $n$-tuple $\alpha\in\{1,2,\ldots,n\}^n$, there is now a
probability that the model ends in all cars parking or not. For each of these
random models, we find a formula for the expected value. Furthermore, for the
second random model, in the case $k =1$, $p=1/2$, we prove that for any $1\le
t\le 2^{n-2}$, there is exactly one $\alpha\in\{1,2,\ldots,n\}^n$ such that the
probability that $\alpha$ parks is $(2t-1)/2^{n-1}$.
