Monte hall problem

The montee hall show, Let's make a deal, offers an activity that awards prizes. The contestants would choose one of three of doors and one of them would have a prize in it. Once the contestant chose a door, montee hall would open another door that didn't have the prize in it, and he would offer the contestant if he or she wanted to switch doors. The competitor would choose which door from there. 

For several years during that era, it was noticeable that when people switched doors, they got the answer 2/3 of the time while those who didn't got the award 1/3 of the time. Why?

There is actually a bit of math involved. Let's start simple. Picking one door door would give you 1/3 of a chance of getting a prize. Thus you would have 2/3 of a chance picking the wrong door. 1/3+2/3=1. From this you would know that two doors together would have a probability of 2/3 getting the prize. Here's where it gets interesting when he reveals one of the wrong doors. Remember that we already established that the probability of two doors together is 2/3. Staying with your initial choice. Additionally, your initial choice will always be 1/3 of success. If montee hall didn't remove a door, and you changed your mind there are 2 doors to choose from and there is a 2/3 success rate of two doors. Thus, (1/2)(2/3)=2/6=1/3 just like your initial choice. Since he did remove one, you only have one other choice to choose from and remember that two doors has a success rate of 2/3. Therefore 1(2/3)=2/3 probability of success if you switch. Same concept works for four doors. If he didn't remove a door, you have 3 other doors to choose from and a 3/4 success rate of three doors. Thus (1/3)(3/4)=3/12 probability which is just like an initial choice of 1/4 success rate. If he does remove a door there are only two doors to choose from. Thus (1/2)(3/4)=3/8 which is greater than 1/4. Therefore you should always switch doors.

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