I just learned about the Monty Hall Problem and it seemed pretty much amazing to me.I am just a bit confused with it.
So,according to the problem we are on a game show, and we are given the choice of three doors: Behind one of them is a car and behind the others are goats. We start by picking one of the doors. After our selection, the host, who knows what's behind the doors reaveals one of the other two doors that has a goat. Now we are asked if we want to change our mind or stay with our initial pick.
According to Probabilities, if we don't swap and keep our first selection we get $(1/3)$ $33.3\% $ chance of wining the car since the elimination of the door that was opened by the host doesn't affect the probability of our door having the car which remains $33.3\%$ as it is at the initial problem.
On the other hand, if we switch the door with the other one left , then we get $(2/3)$ $66.6\%$ chances of wining the car since only two doors are remaining and the fact that the host revealed a goat in one of the unchosen doors changed nothing about the initial probability of our door having the car.
Till this point it makes pretty much sense to me.
But what about assuming we have a man in the audience that makes a different initial choice in his head. For instance, let's say that the contestant picked door number $1$ and he picked door number $2$. If door number $3$ is the door that is being revealed by the host (has a goat) then both doors $1$ and $2$ remain in the game. For the contestant door number $2$ has $66.6\%$ chances of having the car when for the man in the audience door number $1$ has $66.6\%$ chances of having the car. Isn't that weird? From two different perspectives we get two different probabilities about the same unopened doors. How's that possible?