# Monty Hall problem: If contestant knows the door with goat…

I need to know the probability of this variation of the problem.

Suppose that the contestant is a psychic and in some way he/she knows one and only one door with a goat.

*The psychic can't know the future, so he/she don't know the door with car.

There doors are A = goat , B = car , C = goat.

1 - The host start the game and ask to contestant for pick a door.

2 - The psychic knows just one door with the goat, suppose that he avoid the door A, and pick anyone of the remaining doors (he don't know where is the prize in the remaining doors).

3 - Host reveals A door with goat.

4 - Host ask to contestant stay o change door.

¿How change the probability in this case?

In the original problem, when choosing the door in step $$1$$, each door has probability $$\frac{1}{3}$$. Here, in step $$1$$, we ignore $$A$$ because we know it is wrong, so the problem reduces to a $$50:50$$ guess. Door $$A$$ has probability $$0$$, and we know what is behind it, so when it is opened by the host in step $$3$$, we gain no new information and the probability of the other two doors does not change.
If the psychic knows that door $$A$$ has a goat behind it, and the host opens door $$C$$, revealing a second goat, we now know with probability $$1$$ that door B has the car behind it.