Why does the Monty Hall problem not apply when the contestant picks the second door? So, we all know the Monty Hall problem. Where there are 3 doors with one winning door.
The contestant picks a door, then the host removes a non-winning door from the remaining two. Then the contestant is given the option of switching doors.
The originally selected door has a 33.33% chance of being the winning door, but the remaining door has a 66.67% chance of being the winner.
However, if this is done Deal or No Deal style, and the contestant gets to pick a second door in hopes that it's a non-winning door the probability changes.
Let's say the contestant successfully chooses a non-winning door, and is left with the same proposition as the Monty Hall problem. Where he has the option to swap the originally selected door with the one remaining door.
Now, the probability is 50% for each door.
Why? Why does it matter when the door is removed by the host (who knows which door is a winner), compared to by the contestant who removed the proper door by chance?
 A: What matters is that in the case where the contestant didn't pick the prize door initially, the host uses his knowledge to avoid picking the prize door to open.
On the other hand, if the contestant picks a door to open at random, there's a risk that it's the prize door he picks. When that happens, he doesn't get any useful chance to switch -- and this only happens in the case where the original rules would have led to switching being a benefit.
Thus, letting the contestant pick the second door would remove some switching opportunities that would be beneficial, but would not remove away any opportunities that would lose.
Therefore, the chance of winning by switching becomes smaller overall: It drops from $2/3$ to $1/2$.
A: We can look at this problem in terms of the "information" that the contestant has; specifically, what information the contestant has about each door.
For simplicity, I'll use "A" as the name of the door the contestant first selects, "B" as the name of the door that Monty or the contestant then opens, and "C" as the name of the remaining door.
In the standard Monty Hall setup, where we know in advance that after the contestant makes his/her initial choice, Monty always will open a non-chosen, non-winning door, the facts that Monty opened door B and that door B had no prize give us no information about door A. That's because both of these facts were guaranteed to be true regardless of whether the prize was behind door A.
But knowing that Monty was certainly going to open a non-winning door,
Monty's action does give us quite a lot of information about each of the
doors B and C. Obviously, it tells us that the prize is not behind door B, which we did not know before. In fact, before Monty's
action, we would assign the probabilities $\frac13,\frac13$ to the possibility of a prize behind each of the doors B and C; after Monty opens a door, the probability assignments are $0, \frac23$ respectively.
In the case where the contestant opens a door other A,
not knowing if the prize is behind that door,
we do not know in advance that door B would not reveal the prize.
The chance that the door B reveals the prize is much lower in the
case where the door A is the prize door than in the
case where it is not (probability $0$ versus $\frac12$);
so the fact that B is not a prize door tells us that the probability
the prize is behind door A is greater than we would have estimated previously. In fact, the probability the prize is behind door A increases
from $\frac13$ to $\frac12$. The probability that the prize is behind
door C also increases from $\frac13$ to $\frac12$.
