Monty Hall Problem Overall, I think I understand the Monty Hall problem, but there's one particular part that I either don't understand, or I don't agree with, and I'm hoping for an intuitive explanation why I'm wrong or confirmation that I'm right.
The variant that I'm working on is the one with 3 doors. Prize is uniformly located behind 1 of the doors. You choose Door 3. The host opens door 2 and reveals that it's empty. Do you switch to door 1 or stay with your original choice of door 3?
Okay so, if the problem was instead that the host opens either door 1 or door 2 (which ever is empty), then I would switch.
But I think this problem is different because the problem statement states explicitly that the host opens door 2 and reveals that it's empty. So therefore the prize is behind either door 1 or door 3, each with equal probability. So switching is not expected to be advantageous.
The author of the book I am studying from states this as a solution:
Suppose you play the game repeatedly and always
choose Door 3. If you look at all the times the host reveals Door 2 empty, you
will find that two-thirds of the time the prize lies behind Door 1, and one-third
of the time it is behind Door 3. Seeing Door 2 empty is thus a stronger signal
that Door 1 has the prize than it is that Door 3 has it. This argument is
more general, of course. Whichever door you choose, seeing the host reveal an
empty door is a signal that you should switch.
I disagree with this reasoning because of my intuition above. Am I wrong?
Edit: I agree with the last sentence, which doesn't constrain the host to only opening door 2, but I disagree with the first few sentences where the host is constrained to opening door 2.
Another edit: The original problem statement is (Verbatim):
You choose Door 3. He opens Door 2 and reveals that it is empty. You now know that the prize lies behind either Door 3 or Door 1. Should you switch your choice to Door 1?
I feel that this problem is equivalent to just eliminating the host and the second door, and asking the question "If the prize is uniformly distributed behind doors 1 and 3, and you choose door 3, should you switch to door 1."
 A: The main question is why the host opens door #2. If he opens it because he knows it is empty, then indeed it sends a signal about door #1. But if he opens it because he wants to open door #2 and doesn't know it is empty, then the other two doors have equal probabilities and it doesn't matter.
In the latter case, in one-third of the time he would accidentally reveal the prize behind door #2.
In the rest of the time, he will open the empty door #2, so the fact that the door is empty doesn't imply anything about the other doors, and the ratio between their prior probabilities remains.
I think that this is why Monty Hall is such a difficult question. The wording matters a lot, and small inaccuracies change the problem significantly.
A: If Monty tells you in advance that, after you've made your initial selection, he will open one of the other two doors, revealing it to be empty, and then give the option of switching to the door he didn't open, he is, in effect, telling you that you can either stick with what's behind your initial selection or else have what's behind both the other two doors. Viewed this way, it's obvious you have a $1/3$ probability of winning the prize by sticking and a $2/3$ probability by taking the sum of the other two doors, i.e., by switching after the reveal.
The key is that the rules are established in advance. If Monty is allowed the option of sometimes making the offer and sometimes not, all bets are off.
A: I prefer following explanation:
From scratch I have probability $\frac{1}{3}$ to win and $\frac{2}{3}$ against me. Because Monty should open door with empty choice, then that $\frac{2}{3}$ which was against me now work for me, if I change selection, because in these cases I'll win now.
