Alternative Monty Hall Problem So the typical set up for Monty Hall problem, there are 3 doors where 2 have goats and 1 has a car. I, the contestant, get to randomly guess a door looking to get the one with the car, after this the host will open a door that will always be a goat. Thus out the two doors that are left, I have to choose to stay with the original door I chose or switch to the other door. As many analysis of this problem have been done, switching my choice gives me a higher probability of winning. This largely has to do with the fact that since the host always reveals a goat the asking of whether to stay or not, is the same as did you guess right or not, and you have $\frac{2}{3}$ of wrong so you should switch
Now it seems, this "strange" result largely has to do with the fact that the host always reveals a goat. But what if alternatively you had this situation

You are given 3 doors, 2 with a goat and 1 with a car. You randomly choose a door (looking to get one with the car). The host will randomly choose to reveal what is behind one of the 2 doors you haven't chosen. Given that he reveals goat, what is the probability of getting car if you chose to stay with choice? 

My analysis of this problem goes as follows:
Let $D$ be event the door I guessed has car and Let $G$ reprsent the event that host reveals a goat thus what I want to calculate is $P(D|G)$ with this I have
$$P(D|G)=\frac{P(D\cap G)}{P(G)}=\frac{P(G|D)P(D)}{P(G|D)P(D)+P(G|D^{c})P(D^{c})}=\frac{1\left(\frac{1}{3}\right)}{1\left(\frac{1}{3}\right)+\frac{1}{2}\left(\frac{2}{3}\right)}=\frac{1}{2}$$
So it seems it doesn't matter if I choose to switch or not, and this is the result most people come up with when first thinking of problem.
Question: First is my analysis correct for this problem? Second, is it true in general that if you guess out of $n$ doors and host reveals $k$ doors that all have goats, will the probability that car is behind door you choose just $\dfrac{1}{n-k}$?
UPDATE
So I ended up asking my statistics/probability professor about this question and he said the result I got was correct. He explained that the reasoning that the Monty Hall problem inherently causes confusion is because many don't notice the only randomness in the original problem is just in your choice while hosts choice of door is deterministic. Now the problem I asked now has two sets of randomness, your original choice of door and the hosts choice thus the problems are inherently different
 A: Your analysis is correct. Suppose that there are $n$ doors, one of which has a car, the other have goats. The host randomly chooses $k$ doors and opens them. I will use your notation, so $D$ is the event you have chosen the car and $G$ is the event the host reveals $k$ goats.
Then we have
$$
\mathbb{P}(D|G) = \frac{\mathbb{P}(D \cap G)}{\mathbb{P}(G)} = \frac{\frac{1}{n}}{\frac{n-k}{n}} = \frac{1}{n-k}.
$$
This is because


*

*the probability $\mathbb{P}(G)$ that the host only reveals goats is $\frac{n-k}{n}$ (as it is the probability that the car is among one of the other $n-k$ doors),

*the probability $\mathbb{P}(D \cap G)$ that you have chosen the car and the host only reveals goat is $\frac{1}{n}$ as this is the same as the probability $\mathbb{P}(D)$ that you have chosen the car.

A: Your analysis of the intuition looks good. I think your right that people think his way. 
But no. No matter the number of goats revealed, your odds of initially guessing right are always 1 in $n$.
It helps to imagine that there are 100 doors. And that the host reveals 98 goats. 
It is easier to see that you should switch.
A: 1/2 because there are only 2 choices left at that point and one has to be the car.
