Answer(Monty Hall Show)

(Answer) You should swap!!

This problem is related to Bayes' theorem: P(A|B) =P(A,B)/P(B).

where P(A) and P(B) are prior probabilities, P(A|B) posterior probability, P(A,B) joint probability.

Let A be the state(event) of getting a ball and B the signal of opening a box.

The prob of getting a ball is P(A)=1/3. If you choose one of the three boxes and open another box and see nothing in it, the prob of getting a ball is P(A|B)=2/3.* Then you should swap!

If you choose one of the three boxes and choose another but don't open it, the prob of getting a ball is P(A)=P(A|B)=1/3. It doesn't matter whether you swap or not.

*P(A,B)=1/3 and P(B)=1/2.
Because we open two boxes and P(A,B)=P(A)P(B|A)=1/3×1=1/3. B|A means the signal where we already have got a ball and opened only one box.