Alternative Monty Hall Problem

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.