The Monty Hall Problem
The situation is like a classic cups problem. There are three cups, and under one of them is a nut.
You pick a cup, and the presenter then proceeds to turn over one of the "wrong" cups, under which there is nothing.
At this point, you can either stay with your choice, or pick the third cup.
Question: are you better off staying with your choice or switching to the third cup, that neither of you has chosen?
Intuitively, you might think that you're better off staying with your choice: there's a 50/50 chance you're right,isn't there?
The answer is no. There's only a 33% chance that you've picked the right cup, and there's a 67% chance that the third cup is the right one.
So you're much better off choosing the third cup.
You don't believe me.
To see how it works, here are two ways to visualize the problem differently:
1. When you make your first choice, there is a 67% chance you picked the wrong cup.
Nothing changes that. So, when you are in the second round, with two cups to choose from, you know that *still*, there is a 67% chance your first choice was wrong.
It follows logically that there is a 67% chance that the other cup is the right one.
2. Changing the scale of the problem makes things more obvious.
Let's say that there are 1000 cups.
You choose a cup, then the presenter takes away 998 other cups.
That leaves the cup you chose, and one other one. Do you still think that there's a 50/50 chance your choice is right, just because there are two cups?
Isn't it obvious that although you chose at random, the presenter will usually be certain that of the 1000 cups, only that one he left is the right one?
So you can see that choosing one out of two cups doesn't at all imply that you have a 50/50 chance of being right.
The fact that the presenter left you a cup implies that there is a better than even chance that it's the right one.
For more information, see "The Monty Hall Problem and Monkeys."

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