The Monty Hall Problem

1. Your first choice had a 1/3 chance of being correct.
2. The presenter doesn't remove a door at random: he removes a goat by knowing where it is.
3. Cambiar gana exactly when your first choice era a goat.

Answer: Yes, switching improves the probability of winning. When you switch, you win with probability $2/3$. At first, your door has a $1/3$ probability of hiding the car. The other two doors, together, have probability $2/3$. When the presenter opens a goat, he is not removing just any door. He knows where the car is and always avoids opening it. Therefore, the probability of $2/3$ that was distributed between the two non-chosen doors does not disappear: it remains concentrated in the only non-chosen door that is still closed. Like this: - if you keep your initial choice, you win with probability $1/3$;

• if you change, you win with probability $2/3$. Another way to look at it is this: trading wins exactly when your first choice was a goat. Since at the beginning there were two goats and a car, this happens in two out of three cases. Therefore, it is advisable to change doors.