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.
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
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.
