In Monty Hall problem generalized to $n$ doors, the Monty hall's problem was generalized to $n$ doors with Monty opening $m$ of them, where $m < n - 1$.
The probability of winning by switching is $\frac{n - 1}{n(n - m - 1)}$. I am confused about what the probability of winning by not switching is in this case. Is it $\frac{1}{n}$, $\frac{1}{n - m}$, or $1- \frac{n - 1}{n(n - m - 1)}$? I think it's $\frac{1}{n}$, but this doesn't seem to incorporate monty opening $M$ doors.
The probability of winning by not switching is just the probability of picking the door with the car at your initial guess.
Since there are $n$ doors, the car is behind one unique door and you only choose one door, the probability of winning without switching is just $$\frac{1}{n}.$$