You are provided with three cups labelled 1, 2 and 3 placed topside down. One of them conceals a ball. You must first choose a cup among the three in random that may contain the ball inside it. After you've made the choice, one of the remaining cups that doesn't contain the ball is revealed. So there are now two cups; one you have chosen first and one other - either of them can contain the ball inside.
Now, you are left with another choice, will you either stay with your first decision assuming that the ball would be under the cup you have chosen first or change your decision otherwise. Which decision is wiser.
The solution may have some probabilistic reasoning towards it. It should involve proper scientific justification towards it.
Hence, what choice will you make? How will you justify it?
You should always switch. Think about it this way: when a cup is revealed, you gain more information about the system. Switching cannot hurt your chance. From a rigorous point of view, you can just count your favourable cases, if you don't switch you have 1/3, if you switch you have 2/3 (in the last case, the only un-favorouble choice is when you initially choose the right one).
Yes, I've found it. It really is the Monty Hall problem, the link is here:
http://math.ucr.edu/~jdp/Monty_Hall/Monty_Hall.html
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