A refutation of the Monty Hall Paradox (an imaginary refutation, as it turns out). Monty Hall's Paradox - a logical puzzle not for weaklings Monty Hall's Paradox explained using a formula

Probability theory is a branch of mathematics that is ready to confuse mathematicians themselves. Unlike the other, precise and unshakable dogmas of this science, this area is teeming with oddities and inaccuracies. A new paragraph, so to speak, was recently added to this section - the Monty Hall paradox. This is, in general, a task, but it is solved in a completely different way from the usual school or university ones.

Origin story

People have been racking their brains over the Monty Hall paradox since back in 1975. But it’s worth starting in 1963. It was then that a TV show called Let’s make a deal was released on the screens, which translates as “Let’s make a deal.” Its host was none other than Monty Hall, who presented viewers with sometimes unsolvable problems. One of the most striking became the one he presented in 1975. The problem became part of the mathematical theory of probability and the paradoxes that fit within its framework. It is also worth noting that this phenomenon has caused much debate and harsh criticism from scientists. Monty Hall's Paradox was published in the journal Parade in 1990, and since then has become an even more discussed and controversial issue of all times and peoples... Well, now we move directly to its formulation and interpretation.

Problem Statement

There are many interpretations of this paradox, but we decided to present you with the classic one, which was shown in the program itself. So, there are three doors in front of you. Behind one of them there is a car, behind the other two there is one goat each. The presenter invites you to choose one of the doors, and let’s say you stop at number 1. So far, you don’t know what’s behind this very first door, since they open the third one and show you that there’s a goat behind it. Therefore, you have not lost yet, because you have not chosen the door that hides the losing option. Therefore, your chances of getting a car increase.

But then the presenter invites you to change your decision. There are already two doors in front of you, behind one is a goat, behind the other is the desired prize. This is precisely the crux of the problem. It seems that whichever door you choose, the chances are 50/50. But in fact, if you change your mind, you are more likely to win. How so?

The first choice you make in this game is random. You can’t even remotely guess which of the three doors the prize is hidden behind, so you randomly point to the first one you come across. The presenter, in turn, knows where everything is. He has a door with a prize, a door that you pointed to, and a third without a prize, which he opens for you as the first clue. The second clue lies in his very proposal to change the choice.

Now you will no longer choose one of the three at random, and you can even change your decision to get the desired prize. It is the presenter’s proposal that gives the person the belief that the car is really not behind the door he chose, but behind another one. This is the whole essence of the paradox, since, in fact, you still have to choose (albeit from two, and not from three) at random, but the chances of winning increase. As statistics show, out of 30 players who changed their decision, 18 won the car. And this is 60%. And of the same 30 people who did not change their decision - only 11, that is, 36%.

Interpretation in numbers

Now let's give the Monty Hall paradox a more precise definition. The player's first choice splits the doors into two groups. The probability that the prize is located behind the door you chose is 1/3, and 2/3 behind the doors that remain. The leader then opens one of the doors of the second group. Thus, he transfers the entire remaining probability, 2/3, to the one door that you did not choose and that he did not open. It is logical that after such calculations it will be more profitable to change your decision. But it is important to remember that there is still a chance to lose. Sometimes the presenters are disingenuous, since you can initially point at the right prize door, and then voluntarily refuse it.

We are all accustomed to the fact that mathematics, as an exact science, goes hand in hand with common sense. What matters here are numbers, not words, precise formulas, not vague reflections, coordinates, not relative data. But its new section, called probability theory, blew up the entire familiar pattern. Problems in this area, it seems to us, do not fall within the framework of common sense and completely contradict all formulas and calculations. We suggest below that you familiarize yourself with other paradoxes of probability theory that have something in common with the one described above.

Boy and girl paradox

The problem, at first glance, is absurd, but it strictly obeys the mathematical formula and has two solutions. So, a certain man has two children. One of them is probably a boy. What is the probability that the second one will be a boy?

Option 1. We consider all combinations of two children in a family:

• Girl/girl.
• Girl boy.
• Boy/girl.
• Boy/boy.

The first combination obviously does not suit us, therefore, based on the last three, we get a 1/3 probability that the second child will be a small man.

Option 2. If we imagine such a case in practice, discarding fractions and formulas, then, based on the fact that there are only two sexes on Earth, the probability that the second child will be a boy is 1/2.

This experience shows us how cleverly statistics can be manipulated. So, the “sleeping beauty” is injected with sleeping pills and given a coin. If it lands on heads, she is woken up and the experiment ends. If it lands on heads, they wake her up, immediately giving her a second injection, and she forgets that she woke up, and after that they wake her up again only on the second day. After fully awakening, the “beauty” does not know on what day she opened her eyes, or what the probability is that the coin landed on heads. According to the first solution, the probability of landing heads (or tails) is 1/2. The essence of the second option is that if the experiment is carried out 1000 times, then in the case of heads the “beauty” will be awakened 500 times, and with rare ones - 1000. Now the probability of getting tails is 2/3.

The Monty Hall paradox is one of the well-known problems in probability theory, the solution of which, at first glance, contradicts common sense. The problem is formulated as a description of a hypothetical game based on the American television show “Let’s Make a Deal”, and is named after the host of this program. The most common formulation of this problem, published in 1990 in Parade Magazine, is as follows:
Imagine that you are a participant in a game in which you need to choose one of three doors. Behind one of the doors is a car, behind the other two doors are goats. You choose one of the doors, for example, number 1, after which the leader, who knows where the car is and where the goats are, opens one of the remaining doors, for example, number 3, behind which there is a goat. He then asks you if you would like to change your choice and choose door number 2. Will your chances of winning the car increase if you accept the host's offer and change your choice? Although this formulation of the problem is the best known, it is somewhat problematic because it leaves some important conditions of the problem undefined. Below is a more complete formulation.
When solving this problem, they usually reason something like this: after the leader has opened the door behind which the goat is, the car can only be behind one of the two remaining doors. Since the player cannot obtain any additional information about which door the car is behind, the probability of finding a car behind each door is the same, and changing the player's original door choice does not give the player any advantage. However, this line of reasoning is incorrect. If the host always knows which door is behind what is, always opens the one of the remaining doors behind which the goat is, and always invites the player to change his choice, then the probability that the car is behind the door chosen by the player is 1/3, and, accordingly, the probability that the car is behind the remaining door is 2/3. Thus, changing the initial choice increases the player's chances of winning the car by 2 times. This conclusion contradicts the intuitive perception of the situation by most people, which is why the described problem is called the Monty Hall paradox.

Verbal solution
The correct answer to this problem is the following: yes, the chances of winning a car increase by 2 times if the player follows the advice of the presenter and changes his original choice.
The simplest explanation for this answer is the following consideration. In order to win a car without changing the choice, the player must immediately guess the door behind which the car is located. The probability of this is 1/3. If the player initially lands on a door behind which there is a goat (and the probability of this event is 2/3, since there are two goats and only one car), then he can definitely win the car by changing his decision, since the car and one goat remain, and the presenter had already opened the door with the goat.
Thus, without changing the choice, the player remains with his initial probability of winning 1/3, and when changing the initial choice, the player benefits from twice the remaining probability that he guessed wrong at the beginning.
An intuitive explanation can also be made by swapping the two events. The first event is the player making a decision to change the door, the second event is the opening of an extra door. This is acceptable, since opening an extra door does not give the player any new information (see this article for documentation).
Then the problem can be reduced to the following formulation. At the first moment of time, the player divides the doors into two groups: in the first group there is one door (the one he chose), in the second group there are two remaining doors. At the next moment in time, the player makes a choice between groups. Obviously, for the first group the probability of winning is 1/3, for the second group it is 2/3. The player chooses the second group. In the second group, he can open both doors. One is opened by the presenter, and the second by the player himself.
Let's try to give the “most understandable” explanation. Let's reformulate the task: An honest presenter announces to the player that there is a car behind one of the three doors, and invites him to first point to one of the doors, and then choose one of two actions: open the indicated door (in the old formulation this is called “do not change your choice ") or open the other two (in the old formulation this would be just “change the choice”. Think, here lies the key to understanding!). It is clear that the player will choose the second of the two actions, since the probability of receiving a car in this case is twice as high. And the little thing that the presenter “showed the goat” even before choosing an action does not help or hinder the choice, because behind one of the two doors there is always a goat and the presenter will definitely show it at any turn of the game, so the player can use this goat don't look. It’s up to the player, if he chose the second action, to say “thank you” to the leader for saving him the trouble of opening one of the two doors himself, and opening the other. Well, or even simpler. Let's imagine this situation from the point of view of a presenter who performs a similar procedure with dozens of players. Since he knows perfectly well what is behind the doors, then, on average, in two cases out of three, he sees in advance that the player has chosen the “wrong” door. Therefore, for him there is definitely no paradox in the fact that the correct strategy is to change the choice after opening the first door: after all, then in the same two cases out of three the player will leave the studio in a new car.
Finally, the most “naive” proof. Let the one who stands by his choice be called “Stubborn,” and the one who follows the instructions of the leader be called “Attentive.” Then Stubborn wins if he initially guessed the car (1/3), and Attentive wins if he initially missed and hit the goat (2/3). After all, only in this case will he then point to the door with the car.
Keys to Understanding
Despite the simplicity of the explanation for this phenomenon, many people intuitively believe that the probability of winning does not change when the player changes his choice. Typically, the impossibility of changing the probability of winning is motivated by the fact that when calculating the probability, events that happened in the past do not matter, as happens, for example, when tossing a coin - the probability of heads or tails falling does not depend on how many times heads or tails have fallen before. Therefore, many believe that at the moment the player chooses one door out of two, it no longer matters that in the past there was a choice of one door out of three, and the probability of winning a car is the same both when changing the choice and when leaving the original choice.
However, although such considerations are true in the case of coin tosses, they are not true for all games. In this case, the opening of the door by the host should be ignored. The player essentially chooses between the one door they first chose and the other two - opening one of them only serves to distract the player. It is known that there is one car and two goats. The player's initial choice of one of the doors divides the possible outcomes of the game into two groups: either the car is behind the door chosen by the player (the probability of this is 1/3), or behind one of the other two (the probability of this is 2/3). At the same time, it is already known that in any case there is a goat behind one of the two remaining doors, and when opening this door, the presenter does not give the player any additional information about what is behind the door chosen by the player. Thus, the leader opening the door with the goat does not change the probability (2/3) that the car is behind one of the remaining doors. And since the player will not choose the already open door, then all this probability turns out to be concentrated in the event that the car is behind the remaining closed door.
More intuitive reasoning: Let the player use the “change choice” strategy. Then he will lose only if he initially chooses the car. And the probability of this is one third. Therefore, the probability of winning: 1-1/3=2/3. If the player follows the “don’t change choice” strategy, then he will win if and only if he initially chose the car. And the probability of this is one third.
Let's imagine this situation from the point of view of a presenter who performs a similar procedure with dozens of players. Since he knows perfectly well what is behind the doors, then, on average, in two cases out of three, he sees in advance that the player has chosen the “wrong” door. Therefore, for him there is definitely no paradox in the fact that the correct strategy is to change the choice after opening the first door: after all, then in the same two cases out of three the player will leave the studio in a new car.
Another common reason for the difficulty in understanding the solution to this problem is that people often imagine a slightly different game - when it is not known in advance whether the presenter will open the door with a goat and invite the player to change his choice. In this case, the player does not know the leader’s tactics (that is, essentially, does not know all the rules of the game) and cannot make the optimal choice. For example, if the presenter offers a change of option only if the player initially chose the door with the car, then, obviously, the player should always leave the original decision unchanged. This is why it is important to keep in mind the exact formulation of the Monty Hall problem. (with this option, the leader with different strategies can achieve any probability between the doors, in the general (average) case it will be 1/2 to 1/2).
Increasing the number of doors
In order to more easily understand the essence of what is happening, we can consider the case when the player sees in front of him not three doors, but, for example, a hundred. Moreover, behind one of the doors there is a car, and behind the other 99 there are goats. The player chooses one of the doors, and in 99% of cases he will choose the door with a goat, and the chances of immediately choosing the door with a car are very small - they are 1%. After this, the presenter opens 98 doors with goats and invites the player to choose the remaining door. However, in 99% of cases the car will be behind this remaining door, since the chances that the player immediately chose the correct door are very small. It is clear that in this situation a rationally thinking player should always accept the leader's offer.
When considering an increased number of doors, the question often arises: if in the original problem the leader opens one door out of three (that is, 1/3 of the total number of doors), then why should we assume that in the case of 100 doors the leader will open 98 doors with goats, and not 33? This consideration is usually one of the significant reasons why the Monty Hall paradox conflicts with the intuitive perception of the situation. It would be correct to assume that 98 doors will be opened because an essential condition of the task is the presence of only one alternative choice for the player, which is proposed by the presenter. Therefore, in order for the tasks to be similar, in the case of 4 doors the leader must open 2 doors, in the case of 5 doors - 3, and so on, so that there is always one unopened door other than the one that the player initially chose. If the presenter opens fewer doors, the task will no longer be similar to the original Monty Hall task.
It should be noted that in the case of many doors, even if the presenter leaves not one door closed, but several, and invites the player to choose one of them, then when changing the initial choice, the player’s chances of winning a car will still increase, although not so significantly. For example, consider a situation where a player chooses one door out of a hundred, and then the host opens only one of the remaining doors, inviting the player to change his choice. At the same time, the chances that the car is behind the door initially chosen by the player remain the same - 1/100, and for the remaining doors the chances change: the total probability that the car is behind one of the remaining doors (99/100) is now distributed not over There are 99 doors, but 98. Therefore, the probability of finding a car behind each of these doors will be not 1/100, but 99/9800. The increase in probability will be approximately 0.01%.
Decision Tree


A tree of possible decisions of the player and the presenter, showing the probability of each outcome
More formally, the game scenario can be described using a decision tree.
In the first two cases, where the player first chose the door behind which the goat is located, changing the choice results in a win. In the last two cases, when the player first chose the door with the car, changing the choice results in a loss.
The total probability that a change in choice will lead to a win is equivalent to the sum of the probabilities of the first two outcomes, that is

Accordingly, the probability that refusing to change the choice will lead to a gain is equal to

Conducting a similar experiment
There is a simple way to verify that changing your initial choice results in a win two out of three times on average. To do this, you can simulate the game described in the Monty Hall problem using playing cards. One person (distributing the cards) plays the role of the host Monty Hall, and the second plays the role of the player. For the game, three cards are taken, of which one depicts a door with a car (for example, an ace of spades), and the other two, identical (for example, two red deuces) represent doors with goats.
The presenter lays out three cards face down, inviting the player to take one of the cards. After the player selects a card, the leader looks at the two remaining cards and reveals a red two. After this, the cards remaining with the player and the presenter are opened, and if the card chosen by the player is the ace of spades, then a point is recorded in favor of the option when the player does not change his choice, and if the player turns out to have a red two, and the leader remains with the ace of spades, then a point is recorded in favor of the option when the player changes his choice. If many such rounds of the game are played, then the ratio of points in favor of two options will fairly well reflect the ratio of the probabilities of these options. It turns out that the number of points in favor of changing the initial choice is approximately twice as large.
Such an experiment allows us not only to verify that the probability of winning when changing the choice is twice as large, but also illustrates well why this happens. At the moment when the player chooses a card, it is already determined whether the ace of spades is in his hand or not. Further opening by the leader of one of his cards does not change the situation - the player already holds the card in his hand, and it remains there regardless of the actions of the leader. The probability for a player to choose the ace of spades from three cards is obviously 1/3, and thus the probability of not choosing it (and then the player will win if he changes his original choice) is 2/3.
Mention
In the film Twenty-One, the teacher, Miki Rosa, asks the main character, Ben, to solve a puzzle: behind three doors there are two scooters and one car, you need to guess the door in order to win the car. After the first choice, Miki suggests changing the choice. Ben agrees and argues mathematically for his decision. So he involuntarily passes the test for Mika’s team.
In Sergei Lukyanenko’s novel “The Klutz,” the main characters use this technique to win a carriage and the opportunity to continue their journey.
In the television series "4isla" (episode 13 of season 1 "Man Hunt"), one of the main characters, Charlie Epps, explains the Monty Hall paradox at a popular lecture on mathematics, visually illustrating it using marker boards with goats and a car drawn on the reverse sides. Charlie actually finds the car after changing his choice. However, it should be noted that he is conducting only one experiment, while the advantage of the choice switching strategy is statistical, and a series of experiments should be conducted to properly illustrate it.

Unhappy are those people who do not know how to program at least at the level of Excel formulas! For example, it will always seem to them that the paradoxes of probability theory are the quirks of mathematicians who are incapable of understanding real life. Meanwhile, probability theory actually models real processes, while human thought often cannot fully comprehend what is happening.

Let’s take Monty Hall’s paradox; I’ll give here its formulation from Russian Wikipedia:

Imagine that you have become a participant in a game in which you need to choose one of three doors. Behind one of the doors is a car, behind the other two doors are goats. You choose one of the doors, for example, number 1, after which the leader, who knows where the car is and where the goats are, opens one of the remaining doors, for example, number 3, behind which there is a goat. He then asks you if you would like to change your choice and choose door number 2. Will your chances of winning the car increase if you accept the host's offer and change your choice?

(in this case, the participant in the game knows the following rules in advance:

1. the car is equally likely placed behind any of the 3 doors;
2. In any case, the presenter is obliged to open the door with the goat (but not the one the player chose) and invite the player to change the choice;
3. if the leader has a choice of which of 2 doors to open, he chooses any of them with equal probability)

Typically, the narrators of this paradox begin to engage in complex reasoning or overwhelm the reader with formulas. But if you know even a little programming, you don't need this. You can run simulation experiments and see how often you win or lose with a particular strategy.

Really, what is probability? When they say “with a given strategy, the probability of winning is 1/3” - this means that if you conduct 1000 experiments, you will win about 333 of them. That is, in other words, the chances are “1 in 3” - this is literally one out of three experiments. “Probability 2/3” is exactly the same in literally two out of three cases.

So, let's do the Monty Hall experiment. One experiment easily fits into one line of an Excel table: here it is (the file is worth downloading to see the formulas), I will give a description here by columns:

A. Experiment number (for convenience)

B. Generate a random integer from 1 to 3. This will be the door behind which the car is hidden

C-E. for clarity, I placed “goats” and “cars” in these cells

F. Now we select a random door (in fact, we can choose the same door all the time, since randomness in choosing a door for a car is already enough for the model - check!)

G. The host now chooses a door from the remaining two to open for you

H. And here is the most important thing: it does not open the door behind which the car is, but if you initially pointed to the door with the goat, it opens the only other possible door with the goat! This is his tip for you.

I. Well, now let's calculate the chances. For now we won’t change the door – i.e. Let's count the cases when column B is equal to column F. Let there be “1” - won, and “0” - lost. Then the sum of the cells (cell I1003) is the number of wins. The number should be close to 333 (we are doing 1000 experiments in total). Indeed, finding a car behind each of the three doors is an equally probable event, which means that when choosing one door, the chance of guessing is one out of three.

J. It won't be enough! Let's change our choice.

K. Similarly: “1” – win, “0” – loss. So what's the total? And the total is a number equal to 1000 minus the number from cell I1003, i.e. close to 667. Does this surprise you? Could anything else have happened? After all, there are no other closed doors! If the initially chosen door gives you a win 333 times out of 1000, then the other door should give you a win in all the remaining cases!

If you liked this post, now try constructing a similar file for the boy-girl paradox in the following formulation:

Mr. Smith is the father of two children. We met him walking down the street with a little boy, whom he proudly introduced to us as his son. What is the probability that Mr. Smith's other child is also a boy?

Ecology of knowledge. One of the problems of probability theory is the most interesting and seemingly counterintuitive Monty Hall paradox, named after the host of the American TV show “Let’s Make A Deal.”

Many of us have probably heard about probability theory - a special branch of mathematics that studies patterns in random phenomena, random events, as well as their properties. And just one of the problems of probability theory is the most interesting and seemingly counterintuitive Monty Hall paradox, named after the host of the American TV show “Let’s Make A Deal.” We want to introduce you to this paradox today.

Definition of Monty Hall Paradox

As a problem, the Monty Hall paradox is defined in the form of descriptions of the above game, the most common among which is the formulation that was published by Parade Magazine in 1990.

According to it, a person must imagine himself as a participant in a game where he needs to choose one door out of three.

Behind one door is a car, and behind the others are goats. The player must choose one door, for example, door No. 1.

And the leader, who knows what is behind each door, opens one of the two doors that remain, for example, door No. 3, behind which there is a goat.

After this, the host asks the player if he would like to change his original choice and choose door No. 2?

Question: Will a player's chances of winning increase if he changes his choice?

But after the publication of this definition, it turned out that the player’s task was formulated somewhat incorrectly, because All conditions have not been discussed.

For example, the game host may choose a “Monty from Hell” strategy, offering to change the choice only if the player initially guessed the door behind which the car is located.

And it becomes clear that changing the choice will lead to a 100% loss.

Therefore, the most popular formulation of the problem was with special condition No. 6 from a special table:

• A car is equally likely to be behind each door
• The host is always obliged to open the door with a goat other than the one the player has chosen, and offer the player the opportunity to change the choice
• The presenter, having the opportunity to open one of two doors, chooses either one with equal probability

The analysis of the Monty Hall paradox presented below is considered precisely with this condition in mind. So, analysis of the paradox.

Analysis of the Monty Hall Paradox

There are three options for the development of events:

When solving the presented problem, the following reasoning is usually given: in each case, the leader removes one door with a goat, therefore, the probability of finding a car behind one of the two closed doors is equal to ½, regardless of what choice was made initially. However, it is not.

The idea is that by making the first choice, the participant divides the doors into A (selected), B and C (remaining). The chances (P) that the car is behind door A are 1/3, and the chances (P) that it is behind doors B and C are 2/3. And the chances of success when choosing doors B and C are calculated as follows:

P(B) = 2/3 * ½ = 1/3

P(C) = 2/3 * ½ = 1/3

Where ½ is the conditional probability that the car is behind this door, given that the car is not behind the door the player chose.

The presenter, opening a deliberately losing door from the remaining two, informs the player 1 bit of information and thereby changes the conditional probabilities for doors B and C to values ​​1 and 0. Now the chances of success will be calculated as follows:

P(B) = 2/3*1 = 2/3

P(C) = 2/3*0 = 0

And it turns out that if the player changes his initial choice, then his chance of success will be equal to 2/3.

This is explained as follows: By changing his choice after the leader’s manipulations, the player will win if he initially chose the door with a goat, because the presenter opens the second door with the goat, and the player can only change the doors. There are two ways to initially choose a door with a goat (2/3), respectively, if the player replaces the doors, he will win with a probability of 2/3. It is precisely because this conclusion contradicts intuitive perception that the problem received the status of a paradox.

Intuitive perception suggests the following: when the presenter opens a losing door, the player is faced with a new task, which at first glance is not related to the initial choice, because the goat behind the door opened by the leader will be there in any case, regardless of whether the player initially chose the losing or winning door.

After the leader opens the door, the player must make a choice again - either stay on the previous door or choose a new one. This means that the player makes a new choice, and does not change the original one. And the mathematical solution considers two sequential and interconnected tasks of the presenter.

But you need to keep in mind that the presenter opens the door of exactly the two that remain, but not the one the player chose. This means that the chance that the car is behind the remaining door increases, because the presenter did not choose her. If the presenter knows that there is a goat behind the door chosen by the player, he still opens it, thereby obviously reducing the likelihood that the player will choose the right door, because the probability of success will be equal to ½. But this is a game by different rules.

Here's another explanation: Let's say the player plays according to the system presented above, i.e. from doors B or C, he always chooses the one that differs from the original choice. He will lose if he initially chose the door with the car, because subsequently he will choose the door with the goat. In any other case, the player will win if he initially chose the losing option. However, the probability that he will initially choose it is 2/3, which means that to succeed in the game he must first make a mistake, which is twice as likely as the probability of choosing correctly.

Third explanation: Let's imagine that there are not 3 doors, but 1000. After the player has made a choice, the presenter removes 998 unnecessary doors - only two doors remain: the one chosen by the player and one more. But the chance that there is a car behind each door is not at all ½. Most likely (0.999%) the car will be behind the door that the player did not initially choose, i.e. behind the door selected from the 999 others remaining after the first choice. You need to think about the same when choosing from three doors, even if the chances of success decrease and become 2/3.

And the last explanation is the replacement of conditions. Let's say that instead of making an initial choice of, say, door #1, and instead of having the host open door #2 or #3, the player must make the right choice the first time if he knows that the probability of success with door #1 is 33 %, but he knows nothing about the absence of a car behind doors No. 2 and No. 3. It follows from this that the chance of success with the last door will be 66%, i.e. the probability of winning doubles.

But what will the state of affairs be if the presenter behaves differently?

Analysis of the Monty Hall paradox with different behavior of the presenter

The classic version of the Monty Hall Paradox states that the show's host must always give the player a choice of door, regardless of whether the player guessed right or not. But the leader can also complicate his behavior. For example:

• The presenter invites the player to change his choice if it is initially correct - the player will always lose if he agrees to change the choice;
• The presenter invites the player to change his choice if it is initially incorrect - the player will always win if he agrees;
• The presenter opens the door at random, not knowing what is where - the player’s chances of winning when changing the door will always be ½;
• The presenter opens the door with a goat, if the player actually chose the door with a goat, the player’s chances of winning when changing the door will always be ½;
• The host always opens the door with a goat. If the player chose the door with the car, the left door with the goat will open with probability (q) equal to p, and the right door with probability q = 1-p. If the leader opened the door on the left, then the probability of winning is calculated as 1/(1+p). If the leader opened the door on the right, then: 1/(1+q).But the probability that the door on the right will be opened is: (1+q)/3;
• Conditions from the example above, but p=q=1/2 - the player’s chances of winning when changing the door will always be 2/3;
• Conditions from the example above, but p=1 and q=0. If the leader opens the door on the right, then the player’s change of choice will lead to victory, if the door on the left is opened, then the probability of victory will be equal to ½;
• If the host always opens the door with a goat when the player chooses the door with the car, and with probability ½ if the player chooses the door with the goat, then the player's chances of winning when changing the door will always be ½;
• If the game is repeated many times, and the car is always behind one or another door with the same probability, plus the leader opens the door with the same probability, but the leader knows where the car is and always puts the player before a choice, opening the door with a goat, then the probability of winning will be equal 1/3;
• The conditions are from the example above, but the presenter may not open the door at all - the player’s chances of winning will be 1/3.

This is the Motney Hall paradox. Testing its classic version in practice is quite simple, but it will be much more difficult to conduct experiments with changing the behavior of the presenter. Although for meticulous practitioners this is possible. But whether you test the Monty Hall Paradox for yourself or not, you now know some of the secrets of the games played on people on various shows and television programs, as well as interesting mathematical patterns.

By the way, this is interesting: The Monty Hall paradox is mentioned in Robert Luketic's film "Twenty-One", Sergei Lukyanenko's novel "The Klutz", the television series "4isla", the story by Mark Haddon "The Mysterious Murder of a Dog in the Night-Time", the comic book "XKCD", and was also the "hero" of one of the episodes of the television show "MythBusters" published

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The Monty Hall paradox is one of the well-known problems in probability theory, the solution of which, at first glance, contradicts common sense. The problem is formulated as a description of a hypothetical game based on the American television show “Let’s Make a Deal”, and is named after the host of this program. The most common formulation of this problem, published in 1990 in Parade Magazine, is as follows:

Imagine that you are a participant in a game in which you need to choose one of three doors. Behind one of the doors is a car, behind the other two doors are goats. You choose one of the doors, for example, number 1, after which the leader, who knows where the car is and where the goats are, opens one of the remaining doors, for example, number 3, behind which there is a goat. He then asks you if you would like to change your choice and choose door number 2. Will your chances of winning the car increase if you accept the host's offer and change your choice?

Although this formulation of the problem is the best known, it is somewhat problematic because it leaves some important conditions of the problem undefined. Below is a more complete formulation.

When solving this problem, they usually reason something like this: after the leader has opened the door behind which the goat is, the car can only be behind one of the two remaining doors. Since the player cannot obtain any additional information about which door the car is behind, the probability of finding a car behind each door is the same, and changing the player's original door choice does not give the player any advantage. However, this line of reasoning is incorrect. If the host always knows which door is behind what is, always opens the one of the remaining doors behind which the goat is, and always invites the player to change his choice, then the probability that the car is behind the door chosen by the player is 1/3, and, accordingly, the probability that the car is behind the remaining door is 2/3. Thus, changing the initial choice increases the player's chances of winning the car by 2 times. This conclusion contradicts the intuitive perception of the situation by most people, which is why the described problem is called the Monty Hall paradox.

Verbal solution

The correct answer to this problem is the following: yes, the chances of winning a car increase by 2 times if the player follows the advice of the presenter and changes his original choice.

The simplest explanation for this answer is the following consideration. In order to win a car without changing the choice, the player must immediately guess the door behind which the car is located. The probability of this is 1/3. If the player initially lands on a door behind which there is a goat (and the probability of this event is 2/3, since there are two goats and only one car), then he can definitely win the car by changing his decision, since the car and one goat remain, and the presenter had already opened the door with the goat.

Thus, without changing the choice, the player remains with his initial probability of winning 1/3, and when changing the initial choice, the player benefits from twice the remaining probability that he guessed wrong at the beginning.

An intuitive explanation can also be made by swapping the two events. The first event is the player making a decision to change the door, the second event is the opening of an extra door. This is acceptable, since opening an extra door does not give the player any new information (see this article for documentation).

Then the problem can be reduced to the following formulation. At the first moment of time, the player divides the doors into two groups: in the first group there is one door (the one he chose), in the second group there are two remaining doors. At the next moment in time, the player makes a choice between groups. Obviously, for the first group the probability of winning is 1/3, for the second group it is 2/3. The player chooses the second group. In the second group, he can open both doors. One is opened by the presenter, and the second by the player himself.

Let's try to give the "most understandable" explanation. Let's reformulate the problem: An honest presenter announces to the player that there is a car behind one of the three doors, and invites him to first point to one of the doors, and then choose one of two actions: open the indicated door (in the old formulation this is called “do not change your choice ") or open the other two (in the old formulation this would be “change the choice”. Think, here lies the key to understanding!). It is clear that the player will choose the second of the two actions, since the probability of receiving a car in this case is twice as high. And the little thing that the presenter “showed the goat” even before choosing an action does not help or hinder the choice, because behind one of the two doors there is always a goat and the presenter will definitely show it at any turn of the game, so the player can use this goat don't look. The player’s job, if he chose the second action, is to say “thank you” to the leader for saving him the trouble of opening one of the two doors himself, and open the other. Well, or even simpler. Let's imagine this situation from the point of view of a presenter who performs a similar procedure with dozens of players. Since he knows perfectly well what is behind the doors, then, on average, in two cases out of three, he sees in advance that the player has chosen the “wrong” door. Therefore, for him there is definitely no paradox in the fact that the correct strategy is to change the choice after opening the first door: after all, then in the same two cases out of three the player will leave the studio in a new car.

Finally, the most “naive” proof. Let the one who stands by his choice be called “Stubborn,” and the one who follows the instructions of the leader be called “Attentive.” Then Stubborn wins if he initially guessed the car (1/3), and Attentive wins if he initially missed and hit the goat (2/3). After all, only in this case will he then point to the door with the car.

Keys to Understanding

Despite the simplicity of the explanation for this phenomenon, many people intuitively believe that the probability of winning does not change when the player changes his choice. Typically, the impossibility of changing the probability of winning is motivated by the fact that when calculating the probability, events that happened in the past do not matter, as happens, for example, when tossing a coin - the probability of heads or tails falling does not depend on how many times heads or tails have fallen before. Therefore, many believe that at the moment the player chooses one door out of two, it no longer matters that in the past there was a choice of one door out of three, and the probability of winning a car is the same both when changing the choice and when leaving the original choice.

However, although such considerations are true in the case of coin tosses, they are not true for all games. In this case, the opening of the door by the host should be ignored. The player essentially chooses between the one door they first chose and the other two - opening one of them only serves to distract the player. It is known that there is one car and two goats. The player's initial choice of one of the doors divides the possible outcomes of the game into two groups: either the car is behind the door chosen by the player (the probability of this is 1/3), or behind one of the other two (the probability of this is 2/3). At the same time, it is already known that in any case there is a goat behind one of the two remaining doors, and when opening this door, the presenter does not give the player any additional information about what is behind the door chosen by the player. Thus, the leader opening the door with the goat does not change the probability (2/3) that the car is behind one of the remaining doors. And since the player will not choose the already open door, then all this probability turns out to be concentrated in the event that the car is behind the remaining closed door.

More intuitive reasoning: Let the player use the “change choice” strategy. Then he will lose only if he initially chooses the car. And the probability of this is one third. Therefore, the probability of winning: 1-1/3=2/3. If the player follows the “don’t change choice” strategy, then he will win if and only if he initially chose the car. And the probability of this is one third.

Let's imagine this situation from the point of view of a presenter who performs a similar procedure with dozens of players. Since he knows perfectly well what is behind the doors, then, on average, in two cases out of three, he sees in advance that the player has chosen the “wrong” door. Therefore, for him there is definitely no paradox in the fact that the correct strategy is to change the choice after opening the first door: after all, then in the same two cases out of three the player will leave the studio in a new car.

Another common reason for the difficulty in understanding the solution to this problem is that people often imagine a slightly different game - when it is not known in advance whether the presenter will open the door with a goat and invite the player to change his choice. In this case, the player does not know the leader’s tactics (that is, essentially, does not know all the rules of the game) and cannot make the optimal choice. For example, if the presenter offers a change of option only if the player initially chose the door with the car, then, obviously, the player should always leave the original decision unchanged. This is why it is important to keep in mind the exact formulation of the Monty Hall problem. (with this option, the leader with different strategies can achieve any probability between the doors, in the general (average) case it will be 1/2 to 1/2).

Increasing the number of doors

In order to more easily understand the essence of what is happening, we can consider the case when the player sees in front of him not three doors, but, for example, a hundred. Moreover, behind one of the doors there is a car, and behind the other 99 there are goats. The player chooses one of the doors, and in 99% of cases he will choose the door with a goat, and the chances of immediately choosing the door with a car are very small - they are 1%. After this, the presenter opens 98 doors with goats and invites the player to choose the remaining door. However, in 99% of cases the car will be behind this remaining door, since the chances that the player immediately chose the correct door are very small. It is clear that in this situation a rationally thinking player should always accept the leader's offer.

When considering an increased number of doors, the question often arises: if in the original problem the leader opens one door out of three (that is, 1/3 of the total number of doors), then why should we assume that in the case of 100 doors the leader will open 98 doors with goats, and not 33? This consideration is usually one of the significant reasons why the Monty Hall paradox conflicts with the intuitive perception of the situation. It would be correct to assume that 98 doors will be opened because an essential condition of the task is the presence of only one alternative choice for the player, which is proposed by the presenter. Therefore, in order for the tasks to be similar, in the case of 4 doors the leader must open 2 doors, in the case of 5 doors - 3, and so on, so that there is always one unopened door other than the one that the player initially chose. If the presenter opens fewer doors, the task will no longer be similar to the original Monty Hall task.

It should be noted that in the case of many doors, even if the presenter leaves not one door closed, but several, and invites the player to choose one of them, then when changing the initial choice, the player’s chances of winning a car will still increase, although not so significantly. For example, consider a situation where a player chooses one door out of a hundred, and then the host opens only one of the remaining doors, inviting the player to change his choice. At the same time, the chances that the car is behind the door initially chosen by the player remain the same - 1/100, and for the remaining doors the chances change: the total probability that the car is behind one of the remaining doors (99/100) is now distributed not over There are 99 doors, but 98. Therefore, the probability of finding a car behind each of these doors will be not 1/100, but 99/9800. The increase in probability will be approximately 0.01%.

Decision Tree

A tree of possible decisions of the player and the presenter, showing the probability of each outcome

More formally, the game scenario can be described using a decision tree.

In the first two cases, where the player first chose the door behind which the goat is located, changing the choice results in a win. In the last two cases, when the player first chose the door with the car, changing the choice results in a loss.

The total probability that a change in choice will lead to a win is equivalent to the sum of the probabilities of the first two outcomes, that is

Conducting a similar experiment

There is a simple way to verify that changing your initial choice results in a win two out of three times on average. To do this, you can simulate the game described in the Monty Hall problem using playing cards. One person (dealing the cards) plays the role of host Monty Hall, and the second plays the role of the player. For the game, three cards are taken, of which one depicts a door with a car (for example, an ace of spades), and the other two, identical (for example, two red deuces) represent doors with goats.

The presenter lays out three cards face down, inviting the player to take one of the cards. After the player selects a card, the leader looks at the two remaining cards and reveals a red two. After this, the cards remaining for the player and the presenter are opened, and if the card chosen by the player is the ace of spades, then a point is recorded in favor of the option when the player does not change his choice, and if the player turns out to have a red two, and the leader remains with the ace of spades, then a point is recorded in favor of the option when the player changes his choice. If many such rounds of the game are played, then the ratio of points in favor of two options will fairly well reflect the ratio of the probabilities of these options. It turns out that the number of points in favor of changing the initial choice is approximately twice as large.

Such an experiment allows us not only to verify that the probability of winning when changing the choice is twice as large, but also illustrates well why this happens. At the moment when the player chooses a card, it is already determined whether the ace of spades is in his hand or not. Further opening by the leader of one of his cards does not change the situation - the player already holds the card in his hand, and it remains there regardless of the actions of the leader. The probability for a player to choose the ace of spades from three cards is obviously 1/3, and thus the probability of not choosing it (and then the player will win if he changes his original choice) is 2/3.

Mention

In the film Twenty-One, the teacher, Miki Rosa, asks the main character, Ben, to solve a puzzle: behind three doors there are two scooters and one car, you need to guess the door in order to win the car. After the first choice, Miki suggests changing the choice. Ben agrees and argues mathematically for his decision. So he involuntarily passes the test for Mika’s team.

In Sergei Lukyanenko’s novel “The Klutz,” the main characters use this technique to win a carriage and the opportunity to continue their journey.

In the television series "4isla" (episode 13 of season 1 "Man Hunt"), one of the main characters, Charlie Epps, explains the Monty Hall paradox at a popular lecture on mathematics, visually illustrating it using marker boards with goats and a car drawn on the reverse sides. Charlie actually finds the car after changing his choice. However, it should be noted that he is conducting only one experiment, while the advantage of the choice switching strategy is statistical, and a series of experiments should be conducted to properly illustrate it.

http://dic.academic.ru/dic.nsf/ruwiki/36146