>>11983402Just because there are two doors left in the end doesn't mean it is 50, 50.
First understand the problem:
The problem is that you pick one door from three doors, one door has a car and two doors have a goat.
You therefore have a 1/3 chance of picking the car and a 2/3 chance of picking the goat.
Assume the car is in door 1.
If you pick door 2 then you have a goat, if you pick door 3 then you have a goat.
Because there are 2 goats, whether you pick a car or a goat, there will always be another door that has a goat in it. And the host will eliminate one of the goat options.
There is only a goat and a car left now.
But what you forget at this point is that the goats are not identical, there were two goats, there was goat 1 and goat 2, meaning you have a 2/3 chance of being right if you switch your door.
Because you could have picked goat 1 or goat 2 and he would eliminate the other goat.
The way it can go down is like this:
1 car
2 goat
3 goat
If you pick option 1, you get the car and switching will not win.
If you pick option 2, he eliminates option 3 and switching to door 1 will win
If you pick option 3, he eliminates option 2 and switching to door 1 will win.
So 2 out of 3 times, switching wins, therefore you have 2/3 probability of winning.
I made a code that simulates switching versus non-switching and the results are 2/3 wins if you switch and 1/3 wins if you don't switch.
It's not 50-50. Try thinking about it.
