# C3W1 Monty Hall problem intuition behind it? Cannot understand how it works

Dear All!
I am a music major, so math is something very new to me. I am now doing the first lab: “Monty Hall problem” and I cannot intuitively understand how it works. We have three doors, from which we choose one. This I understand - a 1/3 probability of having a car behind it. But once the host opens one door with the goat - aren’t we left with just two doors and a 50/50 chance of having a car opening one of them? One car, two doors… This is what I cannot understand! When I look into the analytical explanation - everything makes sense, but intuitively nothing makes sense. Three doors with the car behind one of them. One of the goats has been removed, so we are left with two doors. A car is behind one of the doors. Why isn’t it a 1/2 probability of opening the right door? Let’s imagine that after the host opened a door with the goat and we are left with just two doors, some new person comes, who knows nothing about the host, or about the fact that initially there were three doors. He sees just two doors and there is a car behind one of them… If he is to choose a door, won’t it be 1/2 probability for him to get a car? I am lost. If someone could please explain intuitively how come that the chances become 2/3 I would be immensely grateful.
Thank you!

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This is largely how probability works. It can be very counter-intuitive.

More later after I study the rest of your message.

The dilemma here is that as doors are opened, the data about the situation changes.

At the start, there are three doors, and a 1/3 chance that any door has the car.

Within the context of the original problem, opening one door does not change anything about what is behind the other two doors - but opening that door does provide new information for re-framing the situation.

If you ignore the door that has been opened, now you have a different scenario, where there are only two doors.

So the dilemma has to do with how the problem is framed.

Folks who teach probability love to confuse new student with this sort of mental gymnastics.

This is truly mind blowing! Thank you TMosh. But am I correct thinking, that if after opening the door with a goat the host and that door both disappear and some new person arrives and sees just two doors, he will have 1/2 probability of getting the car?

I am thinking in same way:
Mathematically, after host opens a door, there left two doors from which we have to choose. i.e. we are given an event and we have to choose from that event. looks like conditional probability type of problem.
Event A = The door with car
Event B = Remaining two doors after host opens a door with goat

P(A|B) = P(A intersection B)/P(B)

p(A|B) = (1/3) / (2/3)

p(A|B) = 1/2 = 50%

So the probability of winning the car after host open a door becomes 1/2 and not 2/3.

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Yes, because he’s been told there are two doors, and one of them hides a car.

He’s looking at a totally different situation than the original scenario.

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Thank you TMosh.

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