Monty Hall

Imagine a game show where you’re the participant. You see in front of you three doors: One of them contains a car and the other two, a goat (I think is obvious what the prize is…). The presenter knows which door leads to the car and because he’s a generous guy, he offers you to choose a door. So, you choose one of the doors, but wait, don’t open it yet! The presenter seems to be more generous and opens one of the remaining doors, revealing a goat. Furthermore, you have the possibility of keeping your door, or choose the other one. At this point, is obvious to think the probability of having the car is 50 %, isn’t it? Well…. What if I told you… That statement is wrong?

This problem is known as the “Monty Hall” problem, asked in Parade Magazine in 1990, where Craig F. Whitaker sent a letter to Marilyn vos Savant’s column. The TV program problem’s is a reference to Let’s Make a Deal, a game show where they can’t change their boxes.

So here’s the thing: The presenter knows exactly where the car and the goats are, so he can’t eliminate the door which leads to the car. If the presenter wouldn’t know anything, then we have another probabilistic problem, but that’s not what we’re talking about. We’re talking about the presenter making his decision AFTER the player chooses a door.

When you have the three doors and choose one of them, the probabilities are 1/3 for the car and 2/3 for the goat. So it’s more probable that you first choose a goat and not the car. But either choosing the car or a goat the first time, there will be always a goat that can be eliminated. In that case, the probability of eliminating the car is 0.

So now, we have two doors, one leading to the prize and the one to the failure. Considering the fact that it’s more possible to choose a goat in the first choice, what happens when you change it? That your chance of obtaining the car is higher!

You have a simulator here in this link to contribute the Mathemathics. Keep calm and use logic!