[jncfhnb]

Suppose Car is C.

The only possible options are (equally likely)

You choose A and get shown B You choose B and get shown A You choose C and get shown A You choose C and get shown B

4 options. You lose 50% of the time.

The 2 options where you would normally get the 2/3 odds are explicitly ignored when you are told Monty chosen randomly and randomly got a goat.

You choose A and get shown C or you choose B and get shown C are forbidden.

50/50 probability on the dot.

If Monty chooses with intention then the options where you choose goat and are shown goat double in probability so you get back to 2/3