Monty Hall Problem: Reasoning Behind the Logic

Question

I watched Lecture 18: Probability Introduction from the MIT OpenCourseWare where the lecturer talks about the Monty Hall problem. He draws the decision tree and we find that actually you have a 2/3 chance of winning if you switch, and a 1/3 if you stick.

I understand the intuition, and the proof behind this (i.e. the decision tree).

However, what I dont understand is when a door is revealed, why doesn't the probability change from 1/3 to 1/2? Why is picking a door, then having one revealed, different from just having two doors and picking one at random?

Answer 1

Actually the Monty Hall problem becomes easier if you consider $100$ doors, $99$ goats and one car.

So obviously you want to win the car and you can pick a door. Do so, the chance of picking the right door is $\frac{1}{100}$. At this point the host opens 98 other doors revealing goats (which he can do, because he knows what's behind the doors). He then asks you whether you want to change doors, that is, if you want to swap to the only remaining door. Now, the chance that you picked the right door was $\frac{1}{100}$, so if you switch, you have a probability of $\frac{99}{100}$ of winning the car.

I'd go for the latter one.

Answer 2

Think of it this way. If your strategy is to stick to the door you originally chose then you will win if you chose the right door to begin with. The chance of that is $1/3$.

Now suppose your strategy is to switch. Then you will win if you chose either of the wrong doors to begin with. The chance of that is $2/3$.

It doesn't become $1/2$ because you are looking at a conditional probability.

Answer 3

Consider a different game.

The host picks a door among 3, opens it, and shows you a goat. Then you pick a door. You only have a 50/50 chance of winning this game.

Now a different game.

The host picks a door among 3, but he is only allowed to open one of the first 2. He opens it, shows you a goat. Then you get to pick one of the other doors.

Is the second game the same as the first game? No. In the first game, you may as well be playing with 2 doors and no host. In the second game, the unopened door among the first two has a much higher chance of having a car behind it.

Intuitively, most people think they are playing the first game when they play Monty Hall (giving the 50% intuitive result), but they are actually playing the second game. That's what confuses people.