Monty hall problem with uneven probability opening door 2 and conditioning on it

Question

I was doing the Monty hall question on harvard's stat 110 strategic pratice 3 Question 1 (b) https://projects.iq.harvard.edu/files/stat110/files/strategic_practice_and_homework_3.pdf

so basically there are 3 doors and Monty favors opening door 2 over door 3 with probability of door 2 being $p$. The question asks what is the probability that switching wins given Monty opens door 2.

I'm having trouble understanding the solution, it saids that $P(W|D_2)$ is the same as $P(C_3|D_2)$ where $W$ is the event that switch wins, $D_i$ is the event Monty opens door $i$ and $C_i$ is $i^{th}$ door contains the car.

the part I don't get is why $P(W|D_2)$ is the same as $P(C_3|D_2)$ shouldn't it be the same as $P(C_3 \cap O_1 | D_2)$ where $O_1$ is the event door 1 is open. in the solution it's just assuming that I always open the wrong door.

Answer 1

It's this bit:

You choose a door, which for concreteness we assume is Door 1.

So if $S_i$ is the event of "you initially select door $i$", then the expression should read:

Note that we are looking for $P(W|D_2,S_1)$, which is the same as $P(C_3|D_2,S_1)$ as if we first choose Door 1, and Monty reveals a goat is behind door 2, then we will win if we switch only if we then switch to Door 3 when the car actually is there.