My first thought is the probability that the gift is behind my pick or the other door is equal. So it does not matter if I switch or not. But I was wrong. Here is the solution:
Let A be the event that the door you picked has a gift behind it; B be the event that the opened door has no gift behind it. So what we are interested in is the probability of A given on B, p(A|B).
According to the Bayes' Law:
We know that P(A)=1/3, p(B|A)=1. p(B) is also 1 because the host will make sure the door he opens has no gift behind it. So the answer of p(A|B)=1/3.
An easier way to think about the question is to list out all the situations. Suppose you pick door 1:
Scenario I: 1 has the gift and 2 and 3 have no gift. So don't switch.
Scenario II: 1 has no gift, 2 has the gift, 3 was open. So switch.
Scenario III: 1 has no gift, 3 has the gift, 2 was open. So switch.
It is clear that by switching the door, you only have 1/3 chance not to get the gift. So switching is a smart move if the host reveals one of the remaining doors that has no gift.
p.s Thanks for Charlie who entertained me with this question and provided the solution.