The "Monty Hall Problem" is:
[Y]ou are given the opportunity to select one closed door of three, behind one of which there is a prize. The other two doors hide “goats” (or some other such “non–prize”), or nothing at all. Once you have made your selection, Monty Hall will open one of the remaining doors, revealing that it does not contain the prize. He then asks you if you would like to switch your selection to the other unopened door, or stay with your original choice. Here is the problem: Does it matter if you switch?The obvious answer is "No", since while it was originally a 1 in 3 change of picking correctly, after Monty opens one of the remaining doors the odds merely improve to 50/50, so it doesn't matter if you switch or not.
The thinking that leads you to that conclusion is in fact wrong, and by switching you actually increase your odds of winning to 2 in 3. The link above gives a nice readable explanation of how this works, but I just want to cheat here and give you an example that I think will help you more quickly grasp how making the switch improves your odds.
Suppose that instead of three doors there were a hundred, so your odds of originally picking the right door would be 1/100. So now Monty has Carol or Vanna or whoever open all but one of the remaining doors, leaving just two: the one you chose and one other. The chance that you originally chose the right door was 1 in a 100, but now that there's only two unseen doors do you really think that your odds have somehow magically improved to 50/50? It's the same principle whether you started with a hundred doors, fifty, twenty, or...three.
So, are you going to switch?
Deal or no deal?
(Unrelated side note: Looking at Carol Merrill's bio I see that she was born about 30 miles from where I grew up. Small world.)