A clearer explanation of the solution to the Monty Hall problem
The Monty Hall problem is a famous probability puzzle named after the host of the television game show “Let’s Make a Deal,” Monty Hall. The problem is described as follows:
1. You are a contestant on a game show, and there are three doors in front of you.
2. Behind one of the doors is a car (the prize you want), and behind the other two doors are goats (which are undesired).
3. You choose one of the doors, but the door is not opened immediately.
4. The game show host, who knows what is behind each door, opens one of the other two doors to reveal a goat.
5. Now, you are left with your original choice and one other unopened door.
The crucial decision is whether to stick with your initial choice or switch to the remaining unopened door. The question is: What is the optimal strategy to maximize your chances of winning the car?
Your immediate intuition may be to say that changing your choice will make no difference since there are now two doors left and each door has a 50% chance of having the car behind it, so your original choice and the remaining door have an equal probability of winning. But actually, the correct answer is that it is better to switch to the other remaining door. This becomes more apparent when you are reminded that the host knows beforehand where the car is hidden. Most explanations to this problem go something like: “imagine we had 100 doors instead of 3, and the host reveals 98 of them, leaving only 2 doors like before. In this scenario it is clearer that the remaining unopened door is more likely to hold the car than the initially selected door”. The problem with this “explanation” is that it only makes the right answer more intuitive but doesn’t actually explain why.
Full explanation
Three crucial information are key to understanding the correct answer:
1. The host knows where the car is beforehand.
2. When the host reveals a door, he will be sure not to open a door with the car. Because doing so will ruin the game.
3. The host can’t open your door after you’ve already picked it, whether it holds a car or a goat.
At the start of the game each door has a probability of 1/3 of having the car. After you pick one door, the host will proceed to reveal one of the remaining doors. Here lies the crucial point; the host is avoiding the door with the car behind it when he reveals one of the two doors that you didn’t pick. So, this act of avoiding the door with the car behind it by selecting another door gives us the information that the winning door is one of the remaining ones. But this information is not shared equally between the remaining doors (which is what our first intuition is telling us when we think that the two remaining doors have an equal 50% probability of having the car). The host would never reveal the door you already selected, so this door will not gain any information and will stay at 1/3 probability. But the remaining other door might have been revealed by the host but wasn’t, this act of avoidance could be because it holds the door and hence, this door will gain information. When the host picks the door, he 100% knows that the door he is revealing has a goat behind it, so this door’s initial 1/3 probability of having the car is turned to 0. This probability has to shift to the doors he purposefully avoided, which is the single unrevealed and initially unselected door in this case. This remaining door will get all this 1/3 probability added to its original 1/3 probability, resulting in 2/3 probability. Again, because it is included in the information gain while the initially picked door will remain at 1/3 because it was excluded from the information gain by the rules of the game. So, the correct answer is to switch to the other remaining door because it is twice more likely to hold the car than the original door you selected.
