The original and most simple scenario of the Monty Hall problem is this: You are in a prize contest and in front of you there are three doors (A, B and C). Behind one of the doors is a prize (Car), while behind others is a loss (Goat). You first choose a door (let’s say door A). The contest host then opens another door behind which is a goat (let’s say door B), and then he ask you will you stay behind your original choice or will you switch the door. The question behind this is what is the better strategy?
The basis of the answer lies in related and unrelated events. The most common answer is that it doesn’t matter which strategy you choose because it is 50/50 chance – but it is not. The 50/50 assumption is based on the idea that the first choice (one of three doors) and the second choice (stay or switch door) are unrelated events, like flipping a coin two times. But in reality, those are related events, and the second event depends on the first event.
At the first step, when you choose one of three doors, the probability that you picked the right door is 33%, or in other words, there is 66,67% that you are on the wrong door. The fact that that in the second step you are given a choice between your door and the other one doesn’t change the fact that you are most likely starting with the wrong door. Therefore, it is better to switch door in the second step.
Read more at There’s Something About R
