Considering we all know about Monty Hall Problem of $2$ Goats and $1$ Car. Also, the solution of the probability $2/3$ concentrating on a single door.
Now, an alternate approach can be as follows:
Consider a person $A$ and doors $1,2,3.$ Now, also consider another person let say $B$.
$B$ comes into consideration only after Monty has shown the door behind which there is a goat.
Let's say $A$ always switches its door. $B$ selects the door which was previously selected by $A$.
Now in this scenario, as we say $A$ has probability $2/3$, whereas $B$ has probability $1/3$ of selecting door with the car.
But the problem is person $B$ comes into consideration where he has only $2$ choices one with the car and one with the goat, his probability of getting a door with the car behind should be $1/2$ and not $1/3$.
What am I missing here??
Player $A$ and $B$ agree they they will pick the same door. As there are $3$ doors the probability of them picking the right door is $1$ in $3$. At that moment a pretty woman from the studio was hired to distract player $B$ and they step out to have a drink. Meanwhile Monty Hall and player $A$ do a bunch of stuff but Player $B$ doesn't pay any attention. He comes back and the game is over. He asks player $A$ "did we win?"
Nothing that happened while he was in the bar having a drink had anything to do with anything. So his odds of winning is still $1/3$. What difference would seeing a goat door do, if he was determined not to switch?
The entire point of the Monty Hall problem is that being shown a goat door does not change the probability of the door you picked. As he was going to show you a goat door anyway the probability was $1/3$ and stays one third. It just that the $2/3$ probability of being wrong is no longer represented by $2$ unknown doors with a $2/3$ probability of a car being behind one of them, but is represented by an unknown door with a $2/3$ probability of a car being behind it.