You pick a door. Monty opens another door, revealing a goat. You now get the option to switch from your door to the other two doors, keeping the car if it's behind either one.
Whether Monty opens the door or not doesn't mean much.
That is only equivalent if Monty (as intended but usually not stated clearly) always picks a door without a car using knowledge of where the car is. If, instead, Monty picks randomly from the other two doors and just happens not to have revealed the car this time, odds are the same whether you stick with your door or switch.
Even in the case where your restatement is mathematically equivalent, it's obviously not at all equivalent as a puzzle because what is being requested by the puzzle is seeing that the original situation is mathematically equivalent to that one.
You pick a door. Monty opens another door, revealing a goat. You now get the option to switch from your door to the other two doors, keeping the car if it's behind either one.
Whether Monty opens the door or not doesn't mean much.