Monty Hall problem

Monty Hall problem is an interesting problem that plays with probability.

Problem: There are three doors, say A, B and C. Behind one of the doors there is a prize and two are empty. You have to make a choice and try to win the prize. You choose a door say A, now host opens the door C, which is empty. The question is, should you switch the doors?

Solution: According to one solution proposed, Switching doors will increase probability of winning. This is the explanation behind that

Initially all 3 doors has probability of 1/3 of winning. So when we choose door A, we had probability of winning – 1/3. Now Doors B & C have collective probability of winning as 2/3. So when we know C is empty, probability of prize behind B is 2/3(?).

Frankly the argument does not look very strong, for me the new probability should be 1/2 for each door.

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