Why is the probability of host opening door $3$ in Monty Hall $1 \over 3$?

Question

I have been told that the probability of the host opening door $3$ in the Monty Hall problem is $1 \over 3$ (a priori, given no other information about what has transpired in the game), but when I tried to verify this myself I got $1 \over 2$ and was wondering what I was doing wrong. Let $C_i$ be the event of the contestant choosing door $i$, $P_i$ the event of the prize being behind door $i$, and $H_i$ the event of the host opening door $i$. I have computed $\mathbb{P}[H_3]$ as $$\mathbb{P}[H_3\mid C_1 \cap P_1] \cdot\mathbb{P}[C_1 \cap P_1] + \mathbb{P}[H_3\mid C_1 \cap P_2] \cdot\mathbb{P}[C_1 \cap P_2] + \mathbb{P}[H_3\mid C_2 \cap P_1] \cdot\mathbb{P}[C_2 \cap P_1] + \mathbb{P}[H_3\mid C_2 \cap P_2] \cdot\mathbb{P}[C_2 \cap P_2] = \frac 1 2 \cdot \frac 1 6 + 1 \cdot \frac 1 6 + 1 \cdot \frac 1 6 + \frac 1 2 \cdot \frac 1 6 = \frac 1 2$$ as $C_i$ and $C_j$ are independent so the probability of their intersection is the product of their probabilities, and I didn't think I needed to subtract anything as $C_i$ and $C_j$ are disjoint for different values of $(i,j)$. What have I done wrong?

Answer 1

With this problem it is absolutely essential that you describe exactly what happens. What actions are forced, which are voluntary, which are based on a strategy with some goal, and so on.

If you don’t do that then you will find all kinds of people coming it with a solution for a problem that is slightly different. And if you got an answer from a book without understanding it, then you will be very frustrated because the problem that you stated and the problem your book solved are not the same and the solutions are not the same.

So please describe precisely what you mean by “Monty Hall problem”.