February 17, 2009 by esarsea
Monty asks you if you want what is behind Door #1, Door #2 or Door #3. Before you choose, Monty explains that there is a new car behind one of the doors, and a goat behind each of the other two doors.
You select Door #1. Before revealing your prize, Monty opens Door #3 and reveals a goat. The new car is behind Door #1 or Door #2! Then Monty asks you if you want to change your selection from Door #1 to Door #2.
Assuming you would prefer to win the new car instead of a goat, is it to your advantage to change your selection?
You might be surprised to learn that the answer is yes! While at first it might appear as a 50/50 proposition, it is not.
When you choose Door #1, you had a 1 in 3 chance of being correct, or inversely, you would pick a goat 2 out of 3 times. After Monty opens Door #3 to reveal one of the goats, you still have a 1 in 3 chance of winning a new car if you retain your original selection of Door #1. In other words, your original choice has a 2 out of 3 chance of being incorrect.
If you change your choice to Door #2, you have the inverse of keeping your original selection, or a 2 out of 3 chance of being correct! You will win the new car 2/3rds of the time by changing your choice in this example.
This is an example of what is commonly referred to as the Monty Hall Problem, which according to Wikipedia is a probability puzzle with a counterintuitive solution. It’s the mathematical equivalent to the Three Prisoners Problem and the Bertrand’s Box Paradox.
This is something that has been around for quite a while, but I just discoved it after reading an article in Card Player Magazine which I subscribe to (I enjoy playing poker, especially No Limit Texas Hold’em and Omaha).
Here is a webpage where you can play the Monty Hall game yourself, and track your own results.