This is a probability problem which most people get wrong. Monty presents three doors. Behind two doors there are Goats. Behind the other there is a New Car.
You are instructed to choose one door. After you choose, Monty opens one of the remaining doors to show you a goat and then asks if you would like to trade the door you originally chose for the remaining door.
Most people believe that the probability of winning the car are the same whether they trade or not. Please give this some thought before you continue reading.
In actuality the odds are twice as good if you make the trade.
There have been lengthy discussions by many intelligent people. The problem finally being put to bed by running a computer simulation which show that trading is the correct choice.
My view on this is simple. This problem is a sleight of hand which confuses people. The sleight of hand is showing that there is a Goat behind one of the remaining doors. Here is why.
How many doors would you have to open to guarantee that you would find a Goat? There is only one car so if you open two doors there has to be a goat behind one of them. Therefor opening one of the two doors to show a goat is MEANINGLESS. It changes nothing in the problem. It is just the same as if Monty had said “behind one of these doors is a Goat”.
Since exposing the Goat is meaningless let’s dispense with it and revisit the problem.
This time you chose one door and Monty asks if you would like to trade your door for the remaining two doors. At this point nobody should have a problem reaching the correct decision. By choosing two doors out of three you are twice as likely to find the Car as you would with only one door.
Now you can all sleep at night.
