TITLE: Monty Hall Problem confusion
QUESTION [0 upvotes]: I have recently learned of the Monty Hall problem, where given three doors, you choose one with a 1/3 probability of choosing the correct door, and then switching to the other door changing the probability of choosing the correct door to 1/2. I am confused because wouldn't staying at the same door have an equal probability at 1/2 because by staying at that door you are effectively choosing that door in the smaller sample space?
Thanks for the help!

REPLY [0 votes]: You have two strategies: Switch doors or not to switch. The strategy can be chosen beforehand as nothing in the game dynamics will add any valuable information to you. Monty will always open one of the others doors and it will be empty. Thus, there are two possible situation: 
Under switching strategy. Using the law of total probability you have that
$$
P(\text{Win}) = P(\text{Win}|\text{Right})P(\text{Right}) + P(\text{Win}|\text{Wrong})P(\text{Wrong}) = \frac{1}{3}\times0 + \frac{2}{3}\times1=\frac{2}{3}.
$$
While under non-switching strategy. Using the law of total probability you have that
$$
P(\text{Win}) = P(\text{Win}|\text{Right})P(\text{Right}) + P(\text{Win}|\text{Wrong})P(\text{Wrong}) = \frac{1}{3}\times 1 + \frac{2}{3}\times0=\frac{1}{3}.
$$
As such, by switching you are maximizing your probability to win.
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