[jncfhnb]

You confused yourself. There are four possible outcomes at the end, but they are not equally likely. So not 50/50.

If Monty always shows a goat, it is undeniably a 2/3 chance to win on switch.

The nuance here being discussed is whether you can assume Monty would have shown you the goat had you chosen a different initial door just because he showed you one this time. If you don’t know that, then you don’t learn anything.

Edit: actually I misread. You just chose to ignore the cases where Monty revealed a car. Which is correct although most people chalk that up as a win or loss.

[LegibleCrimson]

I'm covering the problem statement. I'm ignoring the cases where Monty revealed a car because it didn't happen. You can't chalk it up as a win or a loss, because it didn't happen.

It's always "You choose a door, the host reveals a goat behind another door. Do you switch?" The scenario does not include him revealing the car.

If he intentionally chooses a goat, switching gets the car 2 times out of 3.

If he chose randomly and just happened to choose a goat, switching doesn't matter. It's 50/50.

Of course, these aren't the only scenarios. As others have mentioned, if Monty can decide whether to reveal what's behind a door, he can act maliciously and only reveal a goat if you've already selected a car, and switching will always make you lose. Without knowing these constraints, a single answer isn't knowable, as you mentioned before.

[jncfhnb]

Yes you’re correct. You’re just the first person I’ve ever seen cover the true random Monty variety who doesn’t force the situation where he shows a car in as a win or loss.