The Monty Hall Conundrum

I've seen some recent discussion on this issue. This is a logic & mathematics problem. Most people, including some professional mathematicians, get it wrong.

Here's the set-up:

You are on "Let's Make a Deal." Monty Hall shows you three doors and tells you that behind one of the doors, there is placed a brand new car (or a large sum of cash). The other two doors conceal either an empty space or some sort of "Booby Prize."

Let's number the doors, 1, 2, & 3. After you pick your door, Monty decides (out of the goodness of his heart) to show you what is behind one of the other two doors. Obviously, the door that Monty reveals is empty or has a "Booby Prize." Monty then offers you the opportunity to switch to the remaining unchosen and unrevealed door.

Should you?

Does it matter?

Many folks come to the conclusion that it makes no difference because once Monty reveals the one "wrong" door, the odds shift to 50/50. Well, that would be true if no choice had been made prior to the reveal, so why isn't it true after the reveal?

When you made your initial choice, you had a 1 in 3 chance of being correct. You had a 2 in 3 chance of being wrong. After the reveal, that fact does not change. So, you should switch because based on statistics and odds, switching makes it twice as likely that you will win the good prize.

Still doubting?

Try this Simulation. NOTE: It is Java.