Let's Make A Deal

While watching Numb3rs, a scene where the math wiz was explaining something caught my attention. He was explaining the 'Let's Make A Deal' phenomenon. Basically it goes like this:

The contestant is given 3 doors, behind one of which is a prize. The other two have donkeys or goats or whatever. The contestant picks one of the doors. The host then reveals a losing door from the remaining two. The contestant is now given the option to switch their pick to the remaining door.

Should they switch? Should they stick with their original pick? Most people's intuition says it doesn't matter if they switch, either way they will have a 1/2 chance of winning because there are two doors left and you pick one of them.

WRONG. According to the show anyway. According to Numb3rs, you should switch your pick.

This didn't sit right with me, so I decided to do an experiment. My girlfriend and I used three cards from a deck and used an Ace as the 'winner'. First, we did a series where she stayed with her pick, and somewhat surprisingly her winning ratio was approximately 1/3.

Then, we did a series where she switched her pick, and sure enough, she won 2/3 times (or so). So how did this happen? Why didn't she get 50/50 odds?

It's rather difficult for me to explain because I am still trying to get over the hump myself. It's something along these lines:

There are really two rounds of odds involved, the first with 3 cards, and the second with two cards. Picking from the first round nets you a 1/3 chance of winning. Sticking with the VERY same card during the second round guarantees that you retain a 1/3 chance of winning. On the other side of things, by removing a losing card for the second round, the remaining unpicked card now embodies a 2/3 chance of winning by combining the odds for both the unpicked cards into one.

Once you try the experiment that my girlfriend and I tried, it becomes obvious that sticking with your original pick is a bad idea. You essentially relegate yourself to a 1/3 chance of winning. While that remaining card in the second round embodies the remaining 2/3 chance of winning.

While browsing around the web, I was surprised to find all the talk surrounding this apparent 'paradox'. This was addressed in statistics journals as early as 1975, mostly due to the popularity of the show. One of the sites I found has this neat Java applet that will do the experiment for you:

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

Even more interesting to note is that Monty Hall himself responded to the question himself on the Let's Make A Deal homepage:

May 12, 1975

Mr. Steve Selvin
Asst. Professor of Biostatistics
University of California, Berkeley

Dear Steve:

Thank you for sending me the problem from "The American Statistician."

Although I am not a student of a statistics problems, I do know that these figures can always be used to one's advantage, if I wished to manipulate same. The big hole in your argument of problems is that once the first box is seen to be empty, the contestant cannot exchange his box. So the problems still remain the same, don't they. . . one out of three. Oh, and incidentally, after one is seen to be empty, his chances are no longer 50/50 but remain what they were in the first place, one out of three. It just seems to the contestant that one box having been eliminated, he stands a better chance. Not so. It was always two to one against him. And if you ever get on my show, the rules hold fast for you -- no trading boxes after the selection.

Next time let's play on my home grounds. I graduated in chemistry and zoology. You want to know your chances of surviving with our polluted air and water?

Sincerely,
Monty

So, according to Monty Hall himself, the contestant was never allowed to switch their pick after the first door was revealed. I don't remember the show well enough to remember if that is true, but it's funny how all this talk might be about a situation that never even occurred on the show!