Probability theory...
You are participating in a TV game. And you are asked to choose one door out of 3. There is a car behind one of the doors, scooters behind the other two.Which door did you choose?
Let's say the first one.
And then the presenter, who knows that behind each of the doors, opens another door - at number 3. And there ... A scooter.
The host asks you: “Does your choice remain outside door number 1?†Or not?
Riddle itself: Should you change your choice in this situation? And why.
Your choice is interesting ... It is not necessary to look in response)))
Answer.
Spoiler Answer to riddle: Yes - worth it. Initially, the probability that the car is behind door number 1 is 33.3%. At the same time, the probability that the car is behind one of the other two is 66.7%. Since the presenter knows that he is behind each of the doors, he naturally opens the door in which the scooter is in order to create intrigue. It does this regardless of which door you choose. Even if you initially chose a scooter, it will not immediately open the door you have chosen, as there is a tough scenario for the TV game.
If you keep your choice, nothing really changes. The probability of 33.3% will remain. However, if you take into account that the presenter opens the scooter in a tough game scenario, then the probability that you will win the car is 66.7% when you change your choice. Probability theory ... And nothing more ...
a very complex and incomprehensible topic. according to school logic, the following is
if I need a prize and the actual ratio is 1k3. thus, the probability is 1 divided by 3.
if the presenter opened the door for me, he reduced the ratio to 1d2 and the probability is corresponding to 1 divided by 2, which is 50%.
then it is possible to connect supra-cerebral operations and assume that I chose the correct answer and personally they hate to embarrass me by offering to close up the repeated choice already with less variance so that I do not receive a prize. On this one, I, as a very long time ago, came up with invariant mechanisms to take into account this factor and the confactor. Tama is a very complex psychodynamics, because if the oponet is not as simple as the drake face, it will act with the same mechanism as I and me nana conr-conr-agreement to be, as it were, higher than this poson. those to calculate it.
another debugging mechanism is this numerical theory. what numbers and directions an ordinary person should choose. and subconsciously, an ordinary person will avoid numbers of extreme high polarity te 3 at one of three. even numbers and in islands of medium differential polarity. that almost 80% that a person will not choose the number 2 if the choice is 1-2-3-4. and similar crap with direction angles. with small numbers I find it difficult to work with many fractional numbers.
from this, I would close up the conclusion that the car is actually behind door 1 at 45% and behind door 2 at 50 percent. those who find themselves in a situation with the presenter and he probably is not as simple as the drake face should have chosen door 2. but it is necessary to take into account the publicity and it may seem that the presenter has biased towards the right choice. therefore, the nanopotsan should always be left at 1.
Three doors. Let the chosen one be at number 1. Numbers 2 and 3 remain. Red with a prize. The black one will be the door opened by the presenter. We count, and we are surprised - but by changing our choice the chances are doubled.
When solving this problem, they usually reason something like this: the presenter always removes one losing door in the end, and then the probabilities of a car appearing behind two unopened ones become equal to ½, regardless of the initial choice. But this is not true.
This conclusion contradicts the intuitive perception of the situation by most people, therefore the described task is called the Monty Hall paradox, that is, a paradox in the everyday sense.
And the intuitive perception is as follows: opening the door with the goat, the presenter sets a new task for the player, as if in no way connected with the previous choice - after all, the goat will be behind the open door regardless of whether the player has chosen a goat or a car in front of it. After the third door is open, the player has to make a choice again - and choose either the same door that he chose earlier, or another. That is, at the same time he does not change his previous choice, but makes a new one. The mathematical solution considers two sequential tasks of the leader, as related to each other.
However, one should take into account the factor from the condition that the leader will open the door with the goat from the two remaining ones, and not the door chosen by the player. Consequently, the remaining door has a better chance of being a vehicle since it was not selected by the presenter. If we consider the case when the leader, knowing that there is a goat behind the door chosen by the player, nevertheless opens this door, thereby he deliberately reduces the player's chances of choosing the correct door, since the probability of the correct choice will already be ½. But this kind of game will be according to different rules.
Let's give one more explanation. Suppose that you play according to the system described above, that is, of the two remaining doors, you always choose a door that is different from your initial choice. In which case will you lose? A loss will come when, and only then, when from the very beginning you have chosen the door behind which the car is located, because later you will inevitably change your mind in favor of the door with the goat, in all other cases you will win, that is, if you made a mistake from the very beginning with by choosing a door. But the probability of choosing a door with a goat from the very beginning is 2â„3, so it turns out that to win you need a mistake, the probability of which is twice as much as the correct choice.
An even more visual explanation can be given if we imagine that there are not 3 doors, but, say, 1000, and after the player's choice, the leader removes 998 excess ones, leaving 2 doors: the one that the player has chosen and another one. Obviously, the probability of finding a prize for each of them is not ½ at all. A much higher probability of finding it, namely 0.999, will take place when the decision is changed and the door selected from 999 is selected. In the case of 3 doors, the logic remains, but the probability of winning when changing the decision is lower, namely 2â„3.
Another alternative explanation is to replace the condition with an equivalent one. Imagine that instead of making the initial choice by the player (let it be always door number 1) and then opening the door with the goat among the remaining ones (that is, always among the number 2 and number 3), let's imagine that the player needs to guess the door on the first try, but he is preliminarily informed that the car can be behind door No. 1 with the initial probability (33%), and among the remaining doors it is indicated for which of the car doors is definitely not (0%). Accordingly, the last door will always account for 66%, and the strategy of choosing it is preferable.
Damn, this nonsense even on Wikipedia is passed off as the truth. Now Wikipedia has fallen in my eyes below the level of the yellow rag.
By the way, can someone without Wikipedia explain in their own words why it is necessary to change the choice? Otherwise, any fool can quote unverified sources.
nikita2112 wrote:
to explain in your own words
Here you have chosen the door. In one case out of three (on average), you got it right. And in two cases out of three, you did not guess. That is, I chose a dummy. And the host opens the last dummy. Remains with the prize. In two cases out of three!
-DENIMUSS -
And why in the picture in the second column there are 2 squares and one of them is painted in black, in honor of which it is painted there?
In Figure 2, I will name the graphs, and in each cubes in the first case, nothing is painted over, in the second, in theory, the 3rd painted over means that this option was excluded, why now it has become painted over in the second, this drawing is a lie, like the author's theme, and indeed the author is just a troll.
Probability distribution. Of those who changed the door (lower left corner), two received a car and one a goat. Of those who did not change (lower right corner) - the opposite. And what is there incomprehensible?
nikita2112
nikita2112 wrote:
And, yes, I CLAIM that the solution under the spoiler is wrong
And this is ... the correct answer!
For this is
-DENIMUSS - wrote:
If you keep your choice, nothing essentially changes. The probability of 33.3% will remain.
- bullshit. A choice of two, and the probability is 33%. Terver is such a terver.
Vova126
And you can't see from the picture that the car can be located anywhere, this does not affect the result. In the picture, you can clearly see how changing the door at the end results in two winners out of 3.
And where does the fucking flash have to do with it?