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?
Best Answer
The big difference between the two scenarios is that Monty deliberately avoids choosing the door the contestant chose, but he isn't even aware of what the man in the audience chose so that doesn't factor into his decision.
Let's do the math explicitly. Suppose (as you say) the contestant chooses $1$, the audience member chooses $2$, and Monty opens $3$. The audience member knows that Monty was choosing between doors $2$ and $3$, and could not open door $1$. Thus the audience member knows, as does the contestant, that Monty was forced to avoid door $1$ but elected to avoid $2$ as well. That deliberate selection is evidence that the prize is behind door $2$, so the audience member reasons in exactly the same way the contestant does, and concludes that he should stick with his original mental vote.
As another, similar, variant of the problem, suppose that Monty himself has no idea where the prize is, so just chooses between the unselected doors uniformly at random (possibly exposing the prize, in which case the game ends). In that case, convince yourself that there is no value to switching.