Let’s name the goats Alice and Bob. You pick at random between Alice, Bob, and the Car, each with 1/3 chance. Let’s examine each case.
Case 1: You picked Alice. Monty eliminates Bob. Switching wins. (1/3)
Case 2: You picked Bob. Monty eliminates Alice. Switching wins. (1/3)
Case 3: You picked the Car. Monty eliminates either Alice or Bob. You don’t know which, but it doesn’t matter-- switching loses. (1/3)
It comes down to the fact that Monty always eliminates a goat, which is why there is only one possibility in each of these (equally probable) cases.
From another point of view: Monty revealing a goat does not provide us any new information, because we know in advance that he must always do so. Hence our original odds of picking correctly (p=1/3) cannot change.
In the variant “Monty Fall” problem, where Monty opens a random door, we perform the same analysis:
Case 1: You picked Alice. (1/3)
Case 1a: Monty eliminates Bob. Switching wins. (1/2 of case 1, 1/6 overall)
Case 1b: Monty eliminates the Car. Game over. (1/2 of case 1, 1/6 overall)
Case 2: You picked Bob. (1/3)
Case 2a: Monty eliminates Alice. Switching wins. (1/2 of case 2, 1/6 overall)
Case 2b: Monty eliminates the Car. Game over. (1/2 of case 2, 1/6 overall)
Case 3: You picked the Car. (1/3)
Case 3a: Monty eliminates Alice. Switching loses. (1/2 of case 3, 1/6 overall)
Case 3b: Monty eliminates Bob. Switching loses. (1/2 of case 3, 1/6 overall)
As you can see, there is now a chance that Monty reveals the car resulting in an instant game over-- a 1/3 chance, to be exact. If Monty just so happens to reveal a goat, we instantly know that cases 1b and 2b are impossible. (In this variant, Monty revealing a goat reveals new information!) Of the remaining (still equally probable!) cases, switching wins half the time.