The simulation of the Monty Hall Dilemma found below is pretty simple. You select a door by clicking on it. There is in fact no need to wait for Monty to open a door. All doors are open at once so that you can immediately pass your judgement as to what would happen if Monty had opened a door and made you decide whether to switch to another door or not. But the burden of having to make a decision has been removed.
Back from such a radical modification, the problem then can be reformulated as follows: there are three doors with a prize (a car) behind one of them (and a poster of an undesirable goat behind the other two.) You are asked to select a door. After which you are offered to either open that door or open the other two. You get the prize if it's behind one of the doors you open, you get nothing (or a poster of a goat or two) otherwise.
In the original problem, Monty does you a favor by opening a door with no prize behind. Just saves you an effort of opening the door, do not you agree? But do you care?
Thus you click on your original selection. The doors open. You can see right away what result your decision to switch the door or to stay with the initial selection might have had. The applet also keeps the scores. To repeat the experiment press the Reset button.
Note:About a year after I wrote this page, I've been cleaning my email box when I came across a message from Graeme Melham from Australia (August 13, 2004):
Well my explanation for this problem has come under a lot of criticism. I would like to share it.
It is my conjecture that if these were 3 boxes and one of them had lets say $1000 dollars in it while the other two were empty then we have the same problem at hand.
I say that if Monty did not open the empty box and asked the contestant whether they would like to swap their one box for his two boxes then the result would be no different.
The contestant would agree that Monty has twice the likelihood of having the prize, since he has two boxes. And hence it would be more attractive to swap.
I then get the person I am trying to explain the solution to to admit that indeed one of Monty's boxes will always be empty. So whether he leans over and opens the box after he handed the two boxes to the contestant (after the swap), or whether he opens it first and then asks you to swap the unopened box, makes no difference. Monty is twice as likely to have the prize because he has two boxes, the contestant has one box and one of Monty's boxes is always empty.