I have a problem regarding a modified Monty Hall problem.
In this monty hall problem we have $n$ doors and $m$ cars, where $n ≥ 5$ and $1 ≤ m ≤ n − 3$. The rule of the game is also changed: you choose two doors then Monty reveals another door where there is a goat and then you are allowed to take 3 different actions
- not switching: stick to your original choices.
- switch-one door: randomly pick one of the two doors that you originally chose and switch it with another door.
- switch-two doors: switch both doors with another two doors.
Compute the probabilities of getting at least one car with the not switching, switch-one door, and switch-two doors. Express your answers by $m$ and $n$. What is the best choice?
I have tried to make out the probability of each case but it seemed to be too rigorous? or perhaps that is the way it is to be answered? can anyone help me with this problem?
