You can certainly phrase this question more simply. Without loss of generality you can take $p_i\leq 1/2$, so that the most likely corner is at 0. Then you are looking for the distribution of $\sum_{i=1}^n X_i$ where $X_i$ are independent Bernoulli$(p_i)$.

A simple way to calculate the probabilities you're after is recursively in $n$. Let $a_{r,m}$ be the probability that $\sum_{i=1}^m X_i=r$. Then $a_{0,0}=1$, and for $m\geq 1$

$a_{0,m}=(1-p_m)a_{0,m-1}$

$a_{r,m}=p_m a_{r-1,m-1} + (1-p_m) a_{r,m-1}$ for $1\leq r\leq m.$

This calculates the probabilities with order $n^2$ operations.

Other things that might be relevant if you want to approximate rather than calculate the probabilites: approximation by a normal distribution of mean $\sum p_i$ and variance $\sum p_1(1-p_1)$ (if the mean is reasonably large) or by a Poisson distribution of mean $\sum p_i$ (if the mean is small and each of the $p_i$ is very small). Simulation could also give you a pretty decent answer.

You can certainly phrase this question more simply. Without loss of generality you can take $p_i\leq 1/2$, so that the most likely corner is at 0. Then you are looking for the distribution of $\sum_{i=1}^n X_i$ where $X_i$ are independent Bernoulli$(p_i)$.

A simple way to calculate the probabilities you're after is recursively in $n$. Let $a_{r,m}$ be the probability that $\sum_{i=1}^m X_i=r$. Then $a_{0,0}=1$, and for $m\geq 1$

$a_{0,m}=(1-p_m)a_{0,m-1}$

$a_{r,m}=p_m a_{r-1,m-1} + (1-p_m) a_{r,m-1}$ for $1\leq r\leq m.$

This calculates the probabilities with order $n^2$ operations.

Other things that might be relevant if you want to approximate rather than calculate the probabilites: approximation by a normal distribution of mean $\sum p_i$ and variance $\sum p_i(1-p_i)$ (if the mean is reasonably large) or by a Poisson distribution of mean $\sum p_i$ (if the mean is small and each of the $p_i$ is very small). Simulation could also give you a pretty decent answer.