You are shown three boxes, and told that one of them contains a prize. You are then asked to pick one box, and if that box is the one with the prize, you will win it. After picking a box, you are shown that one of the other two boxes is empty, and offered a chance to change your selection.

Should you do this? Would changing your choice to the other remaining box affect your odds of winning? Why or why not?

(In reply to Flat out wrong logic in solution. by jduval)

John, you might be interested in this problem, which is quite similar to this one.

Consider the following analogy:

An opaque box has two white marbles and a black marble.  I want to pick the black marble.  After I pick the marble, but before I look at its color, one of the marbles accidentally falls out of the box.  It just so happens to be a white marble.  I realize that the marble in my hand now has a 1/2 probability of being black.

After recounting the incident to my friend, he concocts an ingenious plan to increase the odds of getting a black marble when I play the game again.  After I pick the marble, but before looking at it, my friend would take a random marble out of the box and show it to me.  We realized before long that this didn't work, since there was a chance of picking a black marble out of the box, in which case my probability of winning dropped to 0 rather than going up to 1/2.  The probabilities averaged out to a disappointing 1/3.

So we tried another idea, where my friend would look inside the box and pick a white marble out of the box, not telling me whether he saw any black marbles.  Of course, it was 100% likely that he would find a white marble to show me.  Therefore my chance of winning would be 1/2, right? It didn't happen.  Try it yourself.

Edit: while I'm at it, I might as well link Levik's simulation for this problem, even though it was linked in a below comment.

Edited on March 4, 2005, 3:14 am

Posted by Tristan on 2005-03-03 22:20:52