Monty Hall Problem Demonstration

This is a demonstration of a famous problem in probability.

Imagine you're a contestant on a game show. The host shows you three doors and tells you that behind one of the doors is a prize. If you can choose the door with the prize the prize is yours.

You choose one of the three doors. The host then opens a different door, showing you that there is no prize behind it. He then asks you if you'd like to switch and choose the other not yet opened door.

Should you switch? What are your odds of winning if you stay with your original choice, and what are your odds if you switch?

Counterintuitive though it may be, this demonstration, which chooses the first choice, the winning door, and the revealed door¹ all at random, shows that you double your chances of winning if you switch doors when given the chance (2/3 rather than 1/3).

A few simple explanations:

1) The probability of winning with the stay strategy is clearly 1/3. The switch strategy wins if and only if the stay strategy loses. So it must have probability 1 - 1/3 = 2/3.

2) The switch strategy is equivalent to choosing both of the doors not yet chosen, because you win if either of those hides the prize. If either does the one you switched to does, because clearly the door already open doesn't.

See the Wikipedia entry for more details.

¹ Actually, the script must know (as the host does) which is the winning door when choosing which door to reveal so that it can be sure to choose a losing door. It either chooses one of two at random or has only one choice.

This is another explanation of the probabilities - if the original choice isn't the winner, of the two doors left one is revealed and the other is the winner. Thus the switch strategy wins.