With “Monty Hall problem” (or paradox of Monty Hall) refers to a famous and only apparent enigma which falls within the scope of probability calculation. This paradox takes its name from the conductor of game shows American Let’s Make a Deal in which participants must make agreements – deals – right with Monty Hall (pseudonym of Maurice Halprin).
The wording “paradox” which is actually a problem is due to the fact that the solution seems to contradict our basic intuitionsbut to be honest – contrary to what happens in the case of the barber’s paradox for example – here there is no no logical catch.
How the three-door game works: formulation of the problem
The game starts with a competitor placed in front of three closed doors. The competitor knows that behind only one of these doors is found a carwhile behind the other two there is one for each goat. Clearly the competitor has no idea which door hides the car. The player, as a first move, can choose one of the three doors and have as a prize what is behind the selected door, the hope is clearly to be able to win the car!
Once the selection has been made brings by the player it remains closed and takes over conductor. The latter you know Very well what’s behind every door and opens one of the two remaining doors, revealing one of the two goats.
At this point the handler gives the possibility to the player of change the chosen port initially, then selecting the only remaining port.
Given the situation, the mathematics always suggests change come to this step, because the choice of the remaining door makes increase the probability of winning the car carrying it by 1/3or approximately 33%, up to 2/3or approximately 66%. But how can this be so?
Solutions to the Monty Hall problem
To get to the solution of this problem let’s try to visualize the three possible scenariosas if we were external to the game and aware of everything that lies behind the doors:
- The player in the first choice select the door with a goat behind it. In this case we call it “goat 1” because there are a total of two goats.
Then the conductor chooses the other door with what we can call the “goat 2“.
At this point to the player it is better to change the door. Choosing the remaining door wins the car. - The player in the first choice select the door with the ” behind itgoat 2“.
The conductor choose the door that hides the “goat 1“.
Also in this case the player is better off changing doors in order to win the car. - The player in the first choice select the door with thecar.
The conductor choose the door that hides one of the two goatsit makes no difference which one.
Only in this case to the player it’s not worth changing brings.
The schematic solution to the problem
Looking carefully at these scenarios we see that at the moment of first choice the player always has 1/3 chance of selecting the door with the car And 2/3 probability of choosing one of the doors with the goat. After that there is the action of conductorwhich chooses e he opens a door in which there is definitely a goat (since he knows everything). What the host does obviously it doesn’t affect on the chance of choice initial of the player.
At this point the player is certain that one of the two doors remained closed (the one chosen by him or the other) hides the machine. So or it was particularly lucky with the initial choice (as we see in scenario 3) and it has no convenience to change (1 in 3 cases). Or, if the first choice had been unfortunate and had selected one of the two doors hiding a goat, then he would find himself in scenario 1 or 2 and therefore would always be Better for him change brings (2 out of 3 cases).
So in general, given the probability, it’s always better to change if you are not aware of the outcome of the first choice. All this can be understood schematically by visualizing it, but mathematical formulas also demonstrate it to us.
The rules of probability: Bayes’ Theorem
To help us in theoretical-mathematical solution we must rely on one of the most important theorems of probability calculus, known as Bayes’ theorem from the name of the scholar who developed it. This theorem relates the probability of an event like the event of finding the car behind the door, with the probability of this same event if the conditions of other events related to it change.
The formula is as follows:
In this formula TO And B there are two eventsP(A) is the probability of event A, P(B) is the probability of event B, P(A|B) is the probability of event A when B occurs, and P(B|A) is the probability of B when A also occurs.
Let’s go back to the Monty Hall paradox, we know they exist three doors and we call:
– E1 the event “there is a car behind door 1”;
– E2 the event “there is a car behind door 2”;
– E3 the event “there is a car behind door 3”.
And moreover, placing ourselves from the player’s point of view we also call:
– C1 the event “there is a goat behind door 1”;
– C2 the “there is a goat behind door 2” event;
– C3 the “there is a goat behind door 3” event.
Now let’s say that the competitor choose door 1 and that the conductor opens the door 3 behind which there is obviously a goat.
If we want to know what is the probability of finding the car behind door 2, (therefore the probability that by changing the choice you will find the machine) knowing that behind door 3 there is a goat, we apply Bayes’ theorem:
Now let’s calculate the values.
– P(E2) that is, the probability that there is a car behind door 2 is equal to 1/3 because initially there is only one car behind one of the 3 doors;
– P(C3) that is, the probability that there will be a goat behind door 3 before the handler opens it is equal to 1/2 because it is the probability that behind door 3 there is a goat knowing that the handler must choose between the two doors not chosen by the competitor.
– P(C3|E2) that is, the probability that there is a goat behind door 3 knowing that the car is behind door 2 (which the handler knows) is equal to 1.
Substituting these values into the formula we obtain that probability of finding the car behind door 2 And:
In this way it is proved mathematically that there is one probability of 2/3 or a little more than 66% Of win the car if you change one’s initial choice, which is clearly higher than the remaining 33% of probability of victory that one would have if it didn’t change the port selected initially.
