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The problem can also be solved by using [[Bayes' theorem]] to evaluate the [[posterior probability]] that the car is behind the initially chosen door, given that the host has opened another door.

The problem can also be solved by using [[Bayes' theorem]] to evaluate the [[posterior probability]] that the car is behind the initially chosen door, given that the host has opened another door.

Let "Prize x" be the event that the prize is behind door x, and let "Open x" be the event that Monty Hall opens door x. Then before the doors are open, P(Prize 1) = P(Prize 2) = P(Prize 3) = 1/3. P(Open 2 | door 1) = 1/2, as if the prize is behind door 1, Monty Hall has two doors he can open, as he must reveal a goat, not the prize behind door 1. P(Open 2) = P(Open 3) = 1/2, as there are two doors Monty Hall can open, both equally likely. Thus, using Bayes' theorem, we get:

Let "Prize x" be the event that the prize is behind door x, and let "Open x" be the event that Monty Hall opens door x. Then before the doors are open, P(Prize 1) = P(Prize 2) = P(Prize 3) = 1/3. P(Open 2 | Prize 1) = 1/2, as if the prize is behind door 1, Monty Hall has two doors he can open, as he must reveal a goat, not the prize behind door 1. P(Open 2) = P(Open 3) = 1/2, as there are two doors Monty Hall can open, both equally likely. Thus, using Bayes' theorem, we get:

<math>P(Prize 1| Open 2) = \frac{P(Open 2| Prize 1)P(Prize 1)}{P(Open 2)} = \frac{\frac{1}{2}*\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3} </math>

<math>P(Prize 1| Open 2) = \frac{P(Open 2| Prize 1)P(Prize 1)}{P(Open 2)} = \frac{\frac{1}{2}*\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3} </math>