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 when car $C_i$'s preference is taken, $C_i$ backs up $k$-spaces (one by one) and takes the first empty spot. If there are no empty spots in the (up to) $k$ spaces behind $\alpha_i$, they then try 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 $(2t-1)/2^{n-1}$.