Ian Griffiths blogs about the Monty Hall problem.
The problem, named after the host of a game show on which it sometimes appeared, is as follows:
There are three doors, behind one of which is a valuable prize, but you don’t know which door. Choose a door. You are not told straight away whether you’ve made the right choice. Instead, the host of the game will then open one of the doors you did not pick, showing you that there is no prize behind it. You are now offered the chance to change your mind. This effectively narrows down your choice - the prize is behind one of two doors, either the one you picked, or the door that neither you nor the host picked.
What should you do to maximize the probability of winning the prize? Should you stick with your first choice, or switch to the other door? Or does it not matter?
I love this problem as an example of a very straightforward but non-intuitive result. I once presented this problem to a group of young kids who were in a summer math and science enrichment program. They were floored by the result. I demonstrated the proof to the kids via both the logical proof (as Ian does) as well as by running a Monte Carlo simulation. I had two teams play the game over and over, one choosing to switch every time, and one choosing to stay. Like mathematical magic, over a series of 20 or so trials it becomes quite clear that always switching is indeed the better strategy. Argue with me as they did, they could not argue with their own eyes.
Now, lest you think that this is only difficult for non mathematical types to grasp, I used to frequent the sci.math newsgroups and with a bemused grin read the long rants of Ph.Ds in mathematics argue over this problem when I now knew a group of 13 and 14 year olds who could demonstrate the result to them.
I’d like to point out that in the real game show, Monty didn’t always give you the option of switching. Sometimes he’d just open the door you chose. When you add that unpredictable human element, all bets are off.