The Monty Hall Problem

Continuing the discussion about probabilities and their intuition, here is a classical problem: the Monty Hall problem.

The setting is as follows: you are presented with three doors. Behind each door there is either a goat or a car. There are two goats and only one car. You get to pick a door, and someone who knows where the car is located, opens one of the two remaining doors and reveals a goat. Now there are two doors left, the one which you pick, and another one. Behind those two doors there is either a goat or a car. 

Then you are given a choice: switch the door, or stay with the original one. What should you do? 

Now there are two schools of thought: 

• stay because it makes no difference, your new odds are 50/50.
• switch because it increases your odds

Before answering the question, to build up the intuition on the correct answer, let's consider a similar problem:

 

Instead of 3 doors, consider 1 million doors, 999,999 goats and one car. You pick one door at random, and the chances to get the car are obviously 1 in a million. Then the host of the game, knowing the car location, opens 999,998 doors revealing 999,998 goats. Sticking with your original choice, you still have 1/1,000,000 chances to get the car, switching increases your chances to 999,999/1,000,000. There is no such thing as 50/50 in this problem (or for the original problem). For the original problem switching increases your odds from 1/2 to 2/3. Still not convinced? Use 100 billion doors instead. You are more likely to be killed by lightening than finding the car on the first try. Switching the doors is a sure way of getting the car.

The incorrect solution of 50/50 comes from a naive and faulty application of Bayes' theorem of information update. Granted the 1/3-2/3 odds are not intuitive and there are a variety of ways to convince yourself this is the correct answer, including playing this game with a friend many times. 

One thing to keep in mind is that the game show host (Monty Hall) is biased because he does know where the car is and he is always avoiding it. If the game show host would be unbiased and by luck would not reveal the car, then the odds would be 50/50 in that case. An unbiased host would sometimes reveal the car accidentally. It is the bias of the game show host which tricks our intuition to make us believe in a fair 50/50 solution. The answer is not a fair 50/50 because the game show host bias spoils the fairness overall. 

The amazing thing is that despite all explanations, about half of the population strongly defends one position, and half strongly defends the other position. If you think the correct answer is 50/50, please argue your point of view here and I'll attempt to convince you otherwise. 

Next time we'll continue discussing probabilities with another problem: the sleeping beauty problem. Unlike the Monty Hall problem, the sleeping beauty problem lacks consensus even among experts.