Conditional probability problem of three choices I have the following problem where I have difficulties grasping the intuition: 

Lets say we have three boxes, with two of them empty and one
  containing a gold price. Lets say we randomly select one of the boxes.
  After our selection, we are given which one of the remaining two boxes does not contain the price. Now the question is: Should I
  stick with my original selection or select another box from the two
  possible alternatives left. What are the probabilities?

I empirically tried this problem by making a computer program to repeat this experiment 1,000,000 times with first staying with the original choice and then always changing the selection. I got the probabilities to be: 
$$P(golden\; price\;with\;original\;selection)\approx33\%$$
$$P(golden\; price\;with\;changing\;selection)\approx 66\%$$
Intuitively the probabilities seem at first to be 50% for both of these choices, but it seems it's not the case. I can't grasp on why?...
P.S. please let me know if my question is unclear
 A: This is the "Monty Hall problem" if you want to look around for more references. Instead of typing out a solution on my cell, I'll just share this lecture. He does it formally the same way that I like to.
Lecture 6: Monty Hall, Simpson's Paradox | Statis…: http://youtu.be/fDcjhAKuhqQ
For an intuitive approach to the problem, consider that at the beginning of the game there's a $\frac{2}{3}$ chance you picked the wrong box. So when a wrong box is eliminated, there's still a $\frac{2}{3}$ chance you're sitting on a wrong box and hence a $\frac{2}{3}$ chance you'll get the right one by switching.
A: This a very famous problem. For it to really make sense, you need to think of it a little differently. Ask yourself, when do I want to switch? When is it a good idea to switch? Well when I have an empty box of course (I obviously don't want to switch if I have the prize). Now, how often do I have an empty box? What is the probability that I have an empty box? Well that would be $2/3$. So, $2/3$ of the time, it is a good idea to switch!
A: If you stick to your choice then you win if your original choice was correct. Probability $\frac13$.
If you switch then you win if your original choice was wrong. Probability $\frac23$.
