It's right depending on your assumptions. If the host has rules of only picking the right most door the contestant doesn't pick, and it happens to be a goat then the answer is 50%. Basically if the host is just as willing to show a car as show a goat and it is a goat which is shown, the answer is 50%.

There is a confirmation bias that people expect people who are aware of the Monty Hall Problem expect others to be wrong about it and attribute it to people not understanding. I think it's perfectly acceptable for someone to read the vague "the host, who is well-aware of what’s going on behind the scenes" and not assume that means he chooses to pick a goat. They're probably right, and the people saying 50% are reasoning along the lines of "there are two doors so 50/50".

For rigor, I think the host's rules need to be more strict: "The host knows what is behind the doors and will never reveal the car".

Or a 50% answer is correct if you lay out the principles of how the host acts given his knowledge.

Intention doesn’t matter. The problem states that the host opens another door and reveals a goat.

It doesn’t matter that the host meant to do that. Now that the information is revealed, the probabilities have changed.

Obviously if the host revealed the car, that would also affect the probabilities!

No. Intention does matter. If the host chooses the goat by chance ( with the possibility to choose the car instead ) then my chances to win are 50% ( switching or not ).

But we are talking about Monty Hall after all. So I could be wrong :P

Nope. If the host shows the goat _for any reason_, then the 2/3 chance that the doors you didn't choose contain the car now applies to the one unchosen door remaining -- and you should switch.

Staying is an equally good option as switching in this version if the host revealed a goat, since the host randomly revealing a goat now makes it more likely that your initial choice was correct.

P(initial choice is correct) = 1/3

P(host shows goat) = 2/3

P(host shows goat | initial choice is correct) = 1

and vice versa P(host shows goat | initial choice is wrong) = 1/2

Applying Bayes:

P(initial choice is correct | host shows goat) = P(host shows goat | initial choice is correct) * P(initial choice is correct) / P(host shows goat) = 1 * 1/3 / (2/3) = 1/2

So, initial choice now has 1/2 odds for being right. The host revealing a goat gave us information on whether he was playing the P=1/2 or the P=1 game.

Intuitively applying this with the 100 door version: the host getting lucky and revealing 98 goats in a row makes it fairly likely that the initial door was correct all along and they didn't get super lucky avoiding the car 98 times, as there's a good chance the contestant helped them by hiding the car.

Perfect case study right here. It’s trivial to simulate and y’all aren’t budging. In this case, simulating, I see 2/3 for switching after the host shows a goat on purpose, and 1/2 for switching when the host happens to have shown a goat on accident.

It does matter if he meant to or not, because if he chooses randomly from the remaining doors, the probability of him revealing a goat depends on whether you have already chosen the car. Whereas if you know he could only ever reveal a goat, then the fact that he revealed a goat tells you nothing about your current selection.

If you are interested in more details, Wikipedia lists this as the "Ignorant Monty" variant.