The problem of Monty Hall

The problem of Monty Hall

In a famous television contest the participant is required to choose a door between three (all closed) and his prize is to take away what is behind the chosen door. It is known that one of them hides a car and that after the other two there are goats. Once the contestant has chosen a door and communicates their choice, Monty, the presenter, opens one of the other doors and shows that behind it there is a goat. At this time the contestant is given the option to change doors if they wish.

Should the contestant keep their original choice or choose the other door?


The participant must change their initial choice to increase the chances of taking the car.

Let's see why. The probability that the contestant chooses the door that the car hides at its first opportunity is 1/3, so the probability of losing, that is, that the car is in one of the doors that it has not chosen is of 2/3.

What changes when the presenter shows a goat behind one of the other two doors? If the player has chosen in his first option the door that contains the car (with a probability of 1/3), then the presenter can open any of the two doors without prize. In this case, the player loses the car if it changes when the opportunity is offered.

But, if the player chooses a goat in his first option (with a probability of 2/3), the presenter has only one option to open a door that contains the goat. In that case, the door that the presenter did not open, must contain the car, so changing it wins.

In summary, if you keep your original choice, you win if you originally chose the car (with a probability of 1/3), while if you change, you win if you originally chose one of the two goats (with a probability of 2/3). Therefore, the contestant must always change their choice.

An erroneous assumption is that once there are only two doors left they both have the same probability (50%) of containing the car. It is wrong since the presenter opens the door after the choice of player. That is, the player's choice affects the door that the presenter opens as we have just seen.

A clearer way to see it is to rethink the problem. If instead of having only three doors there were 100, and after the original election the presenter opened 98 of the remaining ones to show that behind them there are only goats. If the contestant did not change his choice, he would win the car only if he had originally chosen it (1 out of 100 times), while if he changed it, he would win if he had not originally chosen it (and therefore it is what remains after opening the 98 doors ), 99 out of 100 times!

You can find more information about this problem on Wikipedia