Monty hall problem with uneven probability opening door 2 and conditioning on it I was doing the Monty hall question on harvard's stat 110 strategic pratice 3 Question 1 (b) < 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.

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 rst 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_.