Sure, here's the simulation. There are only six distinct cases, so we can "simulate" it exhaustively, and a Monte Carlo will approximate this.

Let "A", "B", and "C" represent the prizes; A is valuable, B and C are goats. They're shuffled behind doors, so the player and Monty choose "1" or "2" or "3", and after the fact we'll map ABC to 123. As such, we can assume without loss of generality that the player always chooses "1" and Monty always chooses "2".

What happens?

  123
  ABC - player was right at first, so they lose if they switch
  ACB - player was right at first, so they lose if they switch
  BAC - player was wrong, Monty chooses 2 which is the car, so the game is ruined
  BCA - player was wrong, Monty finds a goat, player wins if they switch.
  CAB - player was wrong, Monty chooses 2 which is the car, so the game is ruined
  CBA - player was wrong, Monty finds a goat, player wins if they switch.

Huh. Of the 4 non-ruined games, half of them are improved by switching, half are worsened.

I believed you were wrong, but now I believe you.