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19
May 08
Mon

21 and the Monty Hall problem

After initially appearing in Wired, the tale of the MIT Blackjack Club finally makes it to the big screen. It’s a decent watch, but nothing fantastic. Ironically, Mezrich probably would have made more money from his book and selling the rights to make the movie from it than the club members made pulling their long hours at the tables.

There is a scene from the movie where the character played by Kevin Spacey poses a question to our protagonist. It goes like this. You are on a game show. Before you are three closed doors. Behind one of them is a car, and behind the other two are goats. If you open the right door, you win what’s behind it (obviously, you want the car). The host, who knows what’s behind each door, gives you an opportunity to pick a door. After picking your door, the host does not open it, but instead opens another door – one which the host knows does not have the car. So now there are only two closed doors. The host then gives you a choice: you can stick with your original door, or you can switch to the other door. What do you do?

If you haven’t seen the problem before, pause here and have a think.

In the movie, our protagonist flips off an answer worthy of a 4.0 GPA mathematical whiz and opts to switch doors, saying that doing so will effectively double his chances of winning the car.

This is extremely counter-intuitive, but it is correct. The urge is of course to say, it doesn’t matter – you now have a 50/50 shot at the car since there are two doors. This would be true if the two occasions on which you get to pick doors are independent, but they’re not.

This problem is called the Monty Hall problem, after a game show host by that name. I’m not going to explain why it is the way it is (I find it easiest to explain to someone by drawing a probability tree, but Wikipedia explains it for you in several different ways). However, it is something which is conceptually difficult to explain someone purely orally.

I tried doing it with my flatmate (who is studying a double engineering degree) without much headway… so I tried simulating the game show. 10 runs later, he had only “won” the car 3 times, but he was unconvinced and said the sample size was too low. He remained unconvinced until I wrote a quick computer program to run through 10,000 samples (the car was won 3,331 times).

One way I like to explain it to remove the conceptual roadblock (ie, “It’s 50/50 because it’s an independent event!”) is that if it’s 50/50, then it shouldn’t matter that you choose not to switch doors when given the chance. So, if you stick with your original choice, you should have a 50/50 shot at winning. But, when you make that first choice you only have a 33% chance of winning. So why, if you stick with your first choice, would your chances of winning suddenly increase to 50% just because a door is removed? Well, the answer is, it doesn’t. Your chances of winning stay at 33%.

Here’s a variation. Say you have 4 doors. You get to pick one. After you do so, the game show host reveals one door which does not have the car and then gives you the choice to stay with your original door, or swap to one of the two other closed doors. What do you do?

Answer (highlight to reveal – sorry RSS readers): It will still increase your chances to swap. Not as much as in the 3-door scenario, but still enough to make a difference. If you stay with your first pick, you have a 1 in 4 chance of winning (25%). If you swap to one of the other doors (at random), I calculate your odds of winning improving to 3 in 8 (37.5%).

  8:52pm (GMT +10.00)  •  Movies  •   •  Tweet This  •  Comments (5)