Thanks to a problem that "pchatziko" raised in the infobeto forum, we searched for and found more about it and we list you an article written by a friend engineer.
Ο Monty Hall is a Canadian showman who presented the famous televison Let's make a deal in the ABC from 1963 to 1977 and in some other individual seasons up to 1991. This video game is one of the most historic ones that has gone through the TV, and it is characteristic that many of its elements have inspired and influenced many televised games to this day.
His name Monty Hall, however, is now known mainly to probability scientists, since it corresponds to one of the largest "veridical paradoxes" of this science.
It all started when 1975 o Steve Selvin he sent a letter to the magazine American Statistician, publishing a problem based on the particular video game, which he later named Monty Hall problem, although the problem was in fact equivalent to the one already known at the time Three Prisoners problemof Martin Gardner and other problems that have already been formulated by 1889.
The Monty Hall problem as follows.
There are three doors. One of them hides a car. Others hide from a goat. The player is asked to pick a door, according to the now known curtain play.
Let's say the player selects 1 the door. The presenter, of course, will not immediately open this door, but it will be a little delayed, letting go say the 3 door, which contains a goat, and increase the audience's distress.
At that moment, the presenter gives the player the chance to change if he wants to choose between the two remaining doors or, of course, if he wants to keep it.
If the player changes his choice and requests the other door, 2 times is more likely to find the car than if he persisted in his original choice. Does it apply or not?
Even mathematicians themselves, even Nobel laureates, as they say, do not find the right answer. Statistics show that as soon as 13% of people responds correctly to the above question.
The first approach says that if there are two doors left, one of which has the car and the other one the goat, the chances of winning the car are 50% on any door, the problem has now become a crown-type game , the chances are the same and the player has no particular reason to change his original choice.
This answer is considered, in this first approach, to be outrageous and obvious.
It is, however, the wrong one, and for that very reason it is Monty Hall problem is also called Monty Hall paradox...
In fact, changing his choice is twice as likely to hit the car.
The reason most are led to the wrong impression is that they underestimate the data. The situation we have found is not at all independent from its past, from the way it has emerged.
At first, when there were three closed doors, the probability of the player choosing the door by car was 1 / 3 or 33,3%, while the chance of choosing a goat door was 2 / 3 or 66,6%. Unless the presenter reveals the goat behind the door he opened, he did not change that.
Most of them overlook an element of the problem, that is, the fact that the presenter always a) knows what is behind each door and b) chooses which door to open first so that the agony is extended and to the next opening.
Without this, the opening of the first door would be a random experiment and its result, the revelation of the goat, would create new data, regardless of the original, and then the odds would actually be 50% 50%.
However, the above given tells us that the experiment of opening the first door is not a coincidence. Instead, it is not actually an experiment. The presenter is guided by our choice in choosing the first curtain to be opened, with the rationale, of course, for viewing purposes, in any case, the first door that opens to have a goat.
In the two following comparative tables made by wikipedia in the problem related to the problem, it is clear how the presenter selection for the door to be opened is formed and thus the way in which the data emphasized above influences the evolution of the game :
Considering the above-mentioned fact and the problem, we understand that with the opening of the first door the presenter does not change the chances at all, simply extends the agony by somehow transferring to a closed door that has a goat the whole 66,6 % that previously accounted for collectively in the two closed doors with goats.
The player's initial choice, therefore, still has a 33,3% chance of hiding a car and 66,6% likely to hide a goat.
Then the other door will have, conversely, 66,6% chance of having a car and 33,3% having a goat.
Changing the door, then, we are indeed twice as likely to find the car.
The following wikipedia plan explains to a certain extent all of the above.
What this method suggests, that is, basically, is to choose a door at the beginning randomly, knowing that the most likely (66,6%) is our choice not to correspond to a car, to expect the presenter to remove a door from the game goat, and then abandon the initial choice.
Finally, we end up on the right door if our initial choice was wrong, which is most likely, and if we were lucky enough at first to guess right at first with the 33,3% door, then change it, we lose it ...
This method has inspired several scenes in cinema.
It is not, however, the only time that an interesting probability or statistical problem arises through a televised game.
I also give you a video explaining it:
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