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?
Best Answer
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 $1/2$.