To change or not to change the door?
In probability theory, there is a famous paradigm related to a TV presenter known as Monty Hall. In the final part of the show, participants chose between 3 doors with very good prizes (like a new car) and satisfactory prizes (like a piano).
Based on this context (but different from what happened at the show), the following problem is posed: If we have two goats and a new car behind the doors, and after choosing the participant, the presenter reveals what was behind one of the two other unchecked ports, and asks if the participant wants to stay at the chosen port or if the participant wants to change the port. What is the best strategy?
This issue raises a lot of discussion, because apparently at the moment when we can change doors, we have only two options, one with the prize and one with the goat (assuming the participant does not consider the goat as a prize). That is, it may seem that we have a 50% chance of winning or losing.
However, the situation is different. The presenter did not randomly open one of the remaining two doors. The presenter knows very well what is behind each door and, with that, opened a door that did not have the prize. Thus, the entire decision of the participant revolves around a question: did I initially choose a door with the prize or a door with a goat? For that is exactly what it means to change doors. Changing the door is believing that you chose the door with the goat, staying at the same door is believing that you chose the door with the prize.
In this situation, the initial choice of the right port is 1/3 and the wrong port is 2/3. The main problem with this reasoning is not guaranteeing that the exchange of doors gives the participant the prize, it is just a matter of probability. For example, I did some sample tests without changing the door:
For 10 tests, 50% of the time they chose the door with the prize;
For 100 tests, 32% of the time chose the door with the prize;
For 200 tests, 34% of the time chose the door with the prize;
For 1,000 tests, 31.8% of the time chose the door with the prize;
For 10,000 tests, 33.1% of the time chose the door with the prize.
Following the same logic, we can propose a case a little easier to understand. If we had 4 doors, and after the first choice, the presenter opens 2 doors and asks if the participant wants to change or not. In this case, the chance of initially choosing the door with the prize is 1/4, while the chance of choosing the goat is 3/4.
With this pattern, increasing the number of doors and maintaining only one award we have:
For 3 doors, 1/3 to choose the door with the prize initially, 2/3 to choose the goat;
For 4 doors, 1/4 to choose the door with the prize initially, 3/4 to choose the goat;
For 5 doors, 1/5 to choose the door with the prize initially, 4/5 to choose the goat;
For 6 doors, 1/6 to initially choose the door with the prize, 5/6 to choose the goat;
For 7 doors, 1/7 to choose the door with the prize initially, 6/7 to choose the goat;
For 8 doors, 1/8 to choose the door with the prize initially, 7/8 to choose the goat;
For 9 doors, 1/9 to choose the door with the prize initially, 8/9 to choose the goat;
For 10 doors, 1/10 to choose the door with the prize initially, 9/10 to choose the goat;
For 100 doors, 1/100 to choose the door with the prize initially, 99/100 to choose the goat;
For 1000 doors, 1/1000 of initially chooses the door with the prize, 999/1000 to choose the goat.
On the other hand, if there were 2 prizes and 4 doors, we would have a 2/4 chance to choose the door with the prize. In this scenario, the options for the presenter are:
1. The presenter shows a door with the goat;
2. The presenter shows a door with a goat and a door with a prize;
3. The presenter shows an award door.
In all three cases, the participant has a 2/4 chance of having initially chosen the door with the prize.
In the first case, moving means a new option, as there is a door with the goat and two doors with the prize. Thus, we have 9 options (listed below). Not changing the door, means believing that you initially chose one of the prizes, so we have three situations in which this occurs, and of these in one we end with the goat. Likewise, changing a door means believing that in the new choice you will find a prize. In this case, we have 6 situations in which this occurs, of which in four we end up with prizes. In other words, in both cases, we have a 2/3 chance of receiving the prize, so it is irrelevant to change the door or not.
He chose the goat, he won the goat;
He chose the goat; changed to award 1;
He chose the goat; changed to award 2;
He chose the 1st prize; changed to the goat;
He chose the 1st prize; won prize 1;
He chose the 1st prize; changed to award 2;
He chose prize 2; won prize 2;
He chose prize 2; changed to the goat;
He chose prize 2; changed to award 1.
In the second case, we have 4 options, of these we end with the goat and in two we end with the prize. Changing the door or not, in this case it does not matter, as both of us will have a 50% chance of winning.
He chose the goat, he won the goat;
He chose the goat; changed to the award;
He chose the prize; he won the prize;
He chose the prize; changed to goat.
In the third case, moving means a new choice, as there are two doors with a goat and one with the prize at stake. Thus, we have 9 options (listed below). Do not change the door, it means believing that you chose the prize initially, so we have 3 situations in which this occurs, and out of these 2 we end up with the goat. Likewise, changing a door means believing that in the new choice you will find the prize. In this case, we have six situations in which this occurs, of which in four we end up with the goat. That is, in both cases, we have a 1/3 chance of receiving the prize, so it is irrelevant to change the door or not.
He chose goat 1, he won goat 1;
He chose goat 1; changed to goat 2;
He chose goat 1; changed to the award;
He chose the prize; he won the prize;
He chose the prize; changed to goat 1;
He chose the prize; changed to goat 2.
Similar problems can be built by proposing N doors and M prizes, so that M + 1 <N. In closing we will make a more fun case. Suppose that the presenter does not know what is behind each door, and that the participant does not see the contents of the doors after the presenter has opened them. Thus, the participant chooses one of the N doors, the presenter opens other N-2 doors and asks if the participant wishes to remain at his door, or prefers to change. Perhaps now the result is indifferent, does it matter whether it stays or changes?
In this case, it is really indifferent to change or not, as the prize may have already been revealed and the participant does not know. That is, he can exchange one goat for another goat. If we analyze the possibilities of this problem, the chance of the participant having chosen the door with the prize is 1 / N, therefore its complement is (N-1) / N. So the chance of the presenter not having opened the door with the prize since the participant did not choose it is:
(1 / N-1). (N-1 / N) = 1 / N.