Monty Hall paradox: formulation and explanation. Monty Hall's Paradox - a logical puzzle not for weaklings How to choose one of the three

The Monty Hall Mystery

In search of a car, the player chooses door 1. Then the host opens the 3rd door, behind which there is a goat, and invites the player to change his choice to door 2. Should he do this?

Monty Hall's paradox- 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 last year in the journal Parade Magazine, sounds like this:

Imagine that you are a participant in a game in which you need to choose one of three doors. Behind one of the doors there is a car, behind the other two doors there 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 task is called Monty Hall paradox.

Problem and solution

More precise formulation of the problem

The most common problem formulation published in the journal Parade, unfortunately, is not entirely accurate, since it leaves several essential conditions uncertain. A more complete and accurate formulation of the problem looks something like this:

Imagine that you are a participant in a game in which you are in front of three doors. The presenter, who is known to be honest, placed a car behind one of the doors, and a goat each behind the other two doors. You have no information about what is behind which door. The presenter tells you: “ First you must select one of the doors. After that, I will open one of the remaining doors, behind which is a goat. I will then have you change your original choice and choose the remaining closed door instead of the one you initially chose. You can follow my advice and choose a different door, or confirm your original choice. After that, I will open the door you chose and you will win whatever is behind that door.”

You choose door number 3. The host opens door number 1 and shows that there is a goat behind it. The presenter then asks you to choose door number 2. Will your chances of winning a car increase if you follow his advice?

This problem also implicitly assumes that the leader opening the door with the goat does not carry any information about what is behind the door that the player first chose. The easiest way to achieve this is to require that when the car is behind the door chosen by the player, the leader opens one of the remaining doors with goats at random.

Initially, the probability that the participant will hit the car is 1/3. After the host opens the door, most people think it should be 1/2, but this is not the case. The presenter knows where the car is and therefore does not open the door with the car. And the probability would be 1/2 only if the presenter did not know the position of the prizes, and then opening the door would not change anything.

The most significant addition to the above formulation here is that the player knows before the start of the game that after his choice the leader Anyway will open the door with the goat and Anyway will offer the player to change his choice, that is, the performance of these actions by the leader does not carry any information about whether the player’s initial choice was correct or incorrect.

Solution

The correct answer to this problem is the following: yes, the chances of winning a car are doubled 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 leader 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 because 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 (sic!). 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 (sic!). One is opened by the presenter, and the second by the player himself.

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 it should be ignored opening the door by the host. The player essentially chooses between one the door he chose first, and the others two- opening one of them only serves to distract the player's attention. 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 two others (the probability of this is 2/3). Moreover, it is already known that in any case there is a goat behind one of the two remaining doors, and, opening this door, the leader does not give The player receives no additional information about what is behind the player's chosen door. 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.

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. That is why it is important to keep in mind the exact formulation of the Monty Hall problem. (Although, even with this option, the correct strategy would be to change the choice of door (provided that the player does not know the leader’s “trick”). Since in this case, losing will mean the realization of the probability 1/3.)

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 winning is equal to .

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 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. The moment the player chooses a card, already defined, 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.

Proof using a table

When conducting a large number of experiments, the car should be detected behind each of the doors the same number of times, that is, very close to 1/3 of the total.

According to the laws of probability distribution, you will choose wrong door in 2 cases out of 3. This means that in 2 out of 3 cases you will get the car simply by changing your decision. The table shows that you are likely to get the first choice wrong, in which case you end up in the other two rows of the table. And here they will show you which door you need to choose.

Three Prisoners Problem

Another formulation of the paradox was presented by Martin Gardner in the column Math games, which he wrote for Scientific American, in.

Three prisoners A, B And C sentenced to death, but it is known that one will be pardoned. The sentence prohibits the offender from being told whether he will be pardoned or not. A persuades the guard to say which of the other two prisoners will be executed. Since the question does not concern A, the guard decides to announce that they will execute B. How the likelihood of execution has changed A And C? Or, to use an analogy with the Monty Hall problem, should A switch places with WITH, if he has such an opportunity?

Answer

The table shows the probabilities of which prisoner will be pardoned, before and after the guard's message.

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 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.

The Monty Hall Paradox began to appear more and more often on bookmaker websites. What is it and can the player use it to his advantage?

What is Monty Hall's Paradox?

Monty Hall's paradox is a problem from probability theory. It gained its popularity thanks to an American television show where the player has to open one of three doors. Naturally, the prize is behind only one door (a car), and behind the other two is a goat (the show, after all). First the player selects a door. It doesn't open yet. There are two doors left. Of these two doors, the leader must open the one behind which there is a goat. As a result, there are two unopened doors left, one of which is the one the player chose. Behind one is a goat, behind the other is a car. The host offers the player to change his initial choice and open another door. What happens to a player's chances of winning a prize if he changes his mind, and is there any point in doing so?

If the player changes his choice, he wins with a probability of 66.6%. If you remain with your original opinion, the chance of seeing the car will be limited to 33.7%. This is the paradox. It seems that there are always two doors left, in which there is one prize, and therefore the probability of winning (change/don’t change) is 50%. But in reality everything is completely different. If the presenter immediately opened the door with the goat, and then asked the player to choose one of two doors, then the chance would really be 50%. But first the player makes his choice and the probability of winning the initially chosen door is 1/3.

If this choice is repeated many times, its probability will always remain at the level of 1/3, regardless of any further actions of the leader or the player himself. Accordingly, for the two remaining doors, there will always be a probability of 2/3. And because the leader of these two doors always leaves one, then it takes on the value of this probability of 2/3.

So it turns out The player's initial choice will lead to a win in a third of all cases, and a change of decision will lead to a win in two thirds. That is why this task is called a paradox, because it defies logic and common sense. The human brain is accustomed to working in a pattern, which is why optical illusions, illusions, and paradoxes occur. This is nothing more than a person’s ignorance of a specific issue. Even the logical explanation of the problem written above is not accepted by everyone, and one has to use a more accessible method of enlightenment.

Let's present this problem in a slightly different, more expanded format. There are not 3, but 10 doors, but the conditions are still the same - the player chooses one door, and the presenter opens all the doors and leaves one again. The host can only open doors with a goat. Those. the player again faces a choice - door with a goat / door with a car. Here the conditions are more understandable for the average person to understand.

It is clear that initially it is very difficult to choose a door with a prize, or rather the probability is 1/10. And it is logical that most likely the car will be behind the remaining of the 9 doors. And because the presenter opens only non-winning ones, then the door that remains open after the presenter and will be offered to the player will be the door with the prize. If such a formulation has caused difficulties for a person, then the conditions can be simplified even more until, as they say, it gets it. This is not a sign of a person’s great or small intelligence, rather it is an excellent test of the subject “are you a humanist or a techie.” Options with two, ten, thousand, etc. doors are identical in essence, but differ in difficulty of perception. The fewer doors, the easier it is to confuse a person.

The appearance of the Monty Hall paradox on sites dedicated to various strategies is more pleasing than sad, especially for bookmakers. The truth is that for now the significance of the Monty Hall paradox is given exclusively to practical purposes. This is more like a clear example that not everything you see is what really is. That the same bookmaker odds may contain not only the real distribution of forces based on statistics and current news from the teams. Players can also move the line without any objective reasons. Here the usual herd reflex () and agreements can take place. Yes, just one big bet on an uncongested event can move the line.

Although there are also unique people who claim that this paradox can easily be applied to sports betting. Unfortunately, these are statements without any evidence. Let's imagine the Monty Hall paradox in the context of sports betting. To start you need to find an event with equal three chances of success. There are such things, although they are rare. There is a line on football where the odds for one team to win, a draw and the other team are 2.7 - an even line to the point of impossibility. We need to choose our option. Then it is required that at a certain stage one event disappears, and two, the most probable, remain. Until the end of the match, not a single event can be dismissed, even if it is unlikely.

Over the long haul, it will definitely play out and give its skew to the statistics. But, even if you imagine that it won’t play, then at the stage when there are two options left, these options will already have values ​​commensurate with the original ones. And all because the bookmaker moves the odds during the match. Roughly speaking, when you have to choose between two doors, it will no longer be a goat and a car, but a goat and a bicycle. A goat is zero, a loss is not going anywhere. And the car will turn from a coefficient of 2.7 into a bicycle with a much lower coefficient.

As a result, although changing the initial decision may increase the winning percentage, the winning itself will have a completely different value. Those. in Monty Hall's paradox the initial conditions do not change, but in sports betting they do. Hence its inapplicability in the fight against bookmakers. On the other hand, who knows? Maybe there is some kind of paradox here, it’s just that no one sees it yet.

Conclusion

We continue to strongly recommend using. Leave high-risk financial strategies for casinos or practice gaming accounts. For stable earnings on bets you need the right one, not all kinds of variations. HOW place a bet without understanding FOR WHAT.

About lotteries

This game has long become widespread and has become an integral part of modern life. And although the lottery is increasingly expanding its capabilities, many people still see it only as a way to get rich. It may not be free or reliable. On the other hand, as one of Jack London's heroes noted, in a game of chance one cannot ignore the facts - people sometimes get lucky.

Mathematics of chance. History of probability theory

Alexander Bufetov

Transcript and video recording of a lecture by Alexander Bufetov, Doctor of Physics and Mathematics, leading researcher at the Steklov Institute of Mathematics, leading researcher at the IITP RAS, professor at the Faculty of Mathematics at the Higher School of Economics, director of research at the National Center for Scientific Research in France (CNRS), given as part of the series “ Public lectures "Polit.ru" February 6, 2014

The illusion of regularity: why randomness seems unnatural

Our ideas about the random, the natural and the impossible often disagree with the data of statistics and probability theory. In the book “Imperfect Chance. How chance rules our lives,” American physicist and science popularizer Leonard Mlodinow talks about why random algorithms look so strange, what the catch is in “randomly” shuffling songs on an iPod, and what a stock analyst’s luck depends on. “Theories and Practices” publishes an excerpt from the book.

Determinism

Determinism is a general scientific concept and philosophical doctrine about causality, patterns, genetic connections, interaction and conditionality of all phenomena and processes occurring in the world.

God is a statistic

Deborah Nolan, a professor of statistics at the University of California at Berkeley, asks her students to complete a very strange task at first glance. The first group must toss a coin a hundred times and write down the result: heads or tails. The second must imagine that she is tossing a coin - and also make a list of hundreds of “imaginary” results.

What is determinism

If the initial conditions of a system are known, it is possible, using the laws of nature, to predict its final state.

The Picky Bride Problem

Huseyn-Zade S. M.

Zeno's paradox

Is it possible to get from one point in space to another? The ancient Greek philosopher Zeno of Elea believed that movement could not be accomplished at all, but how did he argue for this? Colm Keller will talk about how to resolve the famous Zeno's paradox.

Paradoxes of infinite sets

Imagine a hotel with an infinite number of rooms. A bus arrives with an endless number of future guests. But placing them all is not so easy. This is an endless hassle, and the guests are endlessly tired. And if you fail to cope with the task, then you can lose an infinite amount of money! What to do?

Dependence of child growth on parents' height

Young parents, of course, want to know how tall their child will be as an adult. Mathematical statistics can offer a simple linear relationship to approximate the height of children based only on the height of the father and mother, and also indicate the accuracy of such an estimate.

Monty Hall's paradox is probably the most famous paradox in probability theory. There are many variations of it, for example, the paradox of three prisoners. And there are many interpretations and explanations of this paradox. But here, I would like to give not only a formal explanation, but show the “physical” basis of what happens in the Monty Hall paradox and others like it.

The classic formulation is:

“You are a participant in the game. There are three doors in front of you. There's a prize for one of them. The host invites you to try to guess where the prize is. You point to one of the doors (at random).

Formulation of the Monty Hall Paradox

The host knows where the prize actually is. He doesn’t yet open the door you pointed to. But it opens one more of the remaining doors for you, behind which there is no prize. The question is, should you change your choice or stay with your previous decision?

It turns out that if you simply change your choices, your chances of winning will increase!

The paradox of the situation is obvious. It seems that everything that happens is random. It makes no difference whether you change your mind or not. But that's not true.

"Physical" explanation of the nature of this paradox

Let's first not go into mathematical subtleties, but simply look at the situation with an open mind.

In this game, you just make a random choice first. Then the presenter tells you Additional information, which allows you to increase your chances of winning.

How does the presenter give you additional information? Very simple. Note that it opens not any door.

Let's, for the sake of simplicity (although there is an element of deceit in this), consider a more likely situation: you pointed to a door behind which there is no prize. Then, behind one of the remaining doors is a prize There is. That is, the presenter has no choice. He opens a very specific door. (You pointed to one, there is a prize behind the other, there is only one door left that the leader can open.)

It is at this moment of meaningful choice that he gives you information that you can use.

In this case, the use of information is that you change your decision.

By the way, your second choice is already too not accidental(or rather, not as random as the first choice). After all, you choose from closed doors, but one is already open and it not arbitrary.

Actually, after these considerations, you may have the feeling that it is better to change your decision. This is true. Let's show this more formally.

A more formal explanation of the Monty Hall paradox

In fact, your first, random choice splits all the doors into two groups. Behind the door you chose there is a prize with a probability of 1/3, behind the other two - with a probability of 2/3. Now the leader makes a change: he opens one door in the second group. And now the entire 2/3 probability applies only to the closed door from the group of two doors.

It is clear that now it is more profitable for you to change your decision.

Although, of course, you still have a chance to lose.

However, changing your selection increases your chances of winning.

Monty Hall Paradox

The Monty Hall paradox is a probabilistic problem, the solution of which (according to some) is contrary to common sense. Problem formulation:

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.

Monty Hall paradox. The most inaccurate mathematics

He then asks you if you would like to change your choice and choose door number 2.
Will your chances of winning a car increase if you accept the presenter's offer and change your choice?

When solving a problem, it is often mistakenly assumed that the two choices are independent and, therefore, the probability will not change if the choice is changed. In fact, this is not the case, as you can see by remembering Bayes' formula or looking at the simulation results below:

Here: “strategy 1” - do not change the choice, “strategy 2” - change the choice. Theoretically, for the case with 3 doors, the probability distribution is 33.(3)% and 66.(6)%. Numerical simulations should yield similar results.

Links

Monty Hall Paradox– a problem from the section of probability theory, the solution of which contradicts common sense.

History[edit | edit wiki text]

At the end of 1963, a new talk show called “Let’s Make a Deal” aired. According to the quiz scenario, viewers from the audience received prizes for correct answers, having a chance to increase them by making new bets, but risking their existing winnings. The show's founders were Stefan Hatosu and Monty Hall, the latter of whom became its constant host for many years.

One of the tasks for the participants was the drawing of the Main Prize, which was located behind one of three doors. Behind the remaining two were incentive prizes, and the presenter, in turn, knew the order of their arrangement. The contestant had to determine the winning door by betting their entire winnings for the show.

When the guesser decided on the number, the presenter opened one of the remaining doors, behind which there was an incentive prize, and invited the player to change the initially chosen door.

Wording[edit | edit wiki text]

As a specific problem, the paradox was first formulated by Steve Selvin in 1975, when he sent The American Statistician magazine and host Monty Hall the question: would a contestant's chances of winning the Grand Prize change if, after opening the door with incentive will he change his choice? After this incident, the concept of the “Monty Hall Paradox” appeared.

In 1990, the most common version of the paradox was published in Parade Magazine with an example:

“Imagine yourself on a game show where you have to choose one of three doors: two of them are goats, and the third is a car. When you make a choice, assuming, for example, that the winning door is number one, the leader opens one of the remaining two doors, for example, number three, behind which is a goat. Then you are given a chance to change the selection to another door? Can you increase your chances of winning a car if you change your choice from door number one to door number two?

This formulation is a simplified version, because There remains the factor of influence of the presenter, who knows exactly where the car is and is interested in the participant’s loss.

For the task to become purely mathematical, it is necessary to eliminate the human factor by introducing the opening of a door with an incentive prize and the ability to change the initial choice as integral conditions.

Solution[edit | edit wiki text]

When comparing chances, at first glance, changing the door number will not give any advantages, because all three options have a 1/3 chance of winning (approx. 33.33% for each of the three doors). In this case, opening one of the doors will not in any way affect the chances of the remaining two, whose chances will become 1/2 to 1/2 (50% for each of the two remaining doors). This judgment is based on the assumption that the player’s choice of a door and the leader’s choice of a door are two independent events that do not affect one another. In reality, it is necessary to consider the entire sequence of events as a whole. In accordance with the theory of probability, the chances of the first selected door from the beginning to the end of the game are invariably 1/3 (approx. 33.33%), and the two remaining ones have a total of 1/3+1/3 = 2/3 (approx. 66.66%). When one of the two remaining doors opens, its chances become 0% (there is an incentive prize hidden behind it), and as a result, the chances of closing the unselected door will be 66.66%, i.e. twice as much as the one originally selected.

To make it easier to understand the results of a choice, you can consider an alternative situation in which the number of options will be greater, for example, a thousand. The probability of choosing the winning option is 1/1000 (0.1%). Given that nine hundred and ninety-eight incorrect ones are subsequently opened out of the remaining nine hundred and ninety-nine options, it becomes clear that the probability of the one remaining door out of the nine hundred and ninety-nine not chosen is higher than that of the only one chosen at the beginning.

Mentions[edit | edit wiki text]

You can find references to the Monty Hall Paradox in “Twenty-One” (a film by Robert Luketic), “The Klutz” (a novel by Sergei Lukyanenko), the television series “4isla” (TV series), “The Mysterious Murder of a Dog in the Night-Time” (a story by Mark Haddon), “XKCD” ( comic book), “MythBusters” (TV show).

See also[edit | edit wiki text]

The image shows the process of choosing between two buried doors from the three initially proposed

Examples of solutions to combinatorics problems

Combinatorics is a science that everyone encounters in everyday life: how many ways to choose 3 people on duty to clean the classroom or how many ways to form a word from given letters.

In general, combinatorics allows you to calculate how many different combinations, according to certain conditions, can be made from given objects (same or different).

As a science, combinatorics originated in the 16th century, and now every student (and often even schoolchildren) studies it. They begin studying with the concepts of permutations, placements, combinations (with or without repetitions); you will find problems on these topics below. The most well-known rules of combinatorics are the sum and product rules, which are most often used in typical combinatorial problems.

Below you will find several examples of problems with solutions using combinatorial concepts and rules that will help you understand typical tasks. If you have difficulties with the tasks, order a test on combinatorics.

Combinatorics problems with online solutions

Task 1. Mom has 2 apples and 3 pears. Every day for 5 days in a row she gives out one fruit. In how many ways can this be done?

Solution of combinatorics problem 1 (pdf, 35 Kb)

Task 2. An enterprise can provide work for 4 women in one specialty, 6 men for another, and 3 workers for a third, regardless of gender. In how many ways can vacancies be filled if there are 14 applicants: 6 women and 8 men?

Solution of problem in combinatorics 2 (pdf, 39 Kb)

Task 3. There are 9 carriages in a passenger train. In how many ways can 4 people be seated on a train, provided that they all travel in different carriages?

Solution of combinatorics problem 3 (pdf, 33 Kb)

Task 4. There are 9 people in the group. How many different subgroups can you form, provided that the subgroup includes at least 2 people?

Solution to combinatorics problem 4 (pdf, 34 Kb)

Task 5. A group of 20 students needs to be divided into 3 teams, and the first team should include 3 people, the second - 5 and the third - 12. In how many ways can this be done?

Solution of problem in combinatorics 5 (pdf, 37 Kb)

Task 6. The coach selects 5 boys out of 10 to be on the team. In how many ways can he form the team if 2 specific boys are to be on the team?

Combinatorics problem with solution 6 (pdf, 33 Kb)

Task 7. 15 chess players took part in the chess tournament, and each of them played only one game with each of the others. How many games were played in this tournament?

Combinatorics problem with solution 7 (pdf, 37 Kb)

Task 8. How many different fractions can be made from the numbers 3, 5, 7, 11, 13, 17 so that each fraction contains 2 different numbers? How many of them are proper fractions?

Combinatorics problem with solution 8 (pdf, 32 Kb)

Task 9. How many words can you get by rearranging the letters in the word Mountain and Institute?

Combinatorics problem with solution 9 (pdf, 32 Kb)

Problem 10. Which numbers from 1 to 1,000,000 are greater: those in which the unit occurs, or those in which it does not occur?

Combinatorics problem with solution 10 (pdf, 39 Kb)

Ready-made examples

Need solved combinatorics problems? Find in the workbook:

Other solutions to problems in probability theory

People are accustomed to considering what seems obvious to be correct. That’s why they often get into trouble by misjudging the situation, trusting their intuition, and not taking the time to critically think about their choices and their consequences.

Monty is a clear illustration of a person's inability to weigh their chances of success when choosing a favorable outcome in the presence of more than one unfavorable one.

Formulation of the Monty Hall Paradox

So, what kind of animal is this? What exactly are we talking about? The most famous example of the Monty Hall Paradox is a television show popular in America in the middle of the last century called Let's Make a Bet! By the way, it was thanks to the host of this quiz that the Monty Hall Paradox subsequently received its name.

The game consisted of the following: the participant was shown three doors that looked exactly the same. However, behind one of them the player was waiting for an expensive new car, but behind the other two he was impatiently pining for a goat. As is usually the case with game shows, whatever was behind the contestant's chosen door became his winnings.

What is the trick?

But it's not that simple. After the choice was made, the presenter, knowing where the main prize was hidden, opened one of the remaining two doors (of course, the one behind which the artiodactyl was hiding), and then asked the player if he would like to change his decision.

The Monty Hall paradox, formulated by scientists in 1990, is that, contrary to intuition that there is no difference in making a leading decision based on an issue, one must agree to change one's choice. If you want to get a great car, of course.

How it works?

There are several reasons why people will not want to give up their choice. Intuition and simple (but incorrect) logic say that nothing depends on this decision. Moreover, not everyone wants to follow the lead of another - this is real manipulation, isn’t it? No not like this. But if everything was immediately intuitive, they wouldn’t even name it. It's not strange to have doubts. When this puzzle was first published in one of the major magazines, thousands of readers, including recognized mathematicians, sent letters to the editor claiming that the answer printed in the issue was not true. If the existence of probability theory was not news to the person who got on the show, then perhaps he would be able to solve this problem. And thereby increase the chances of winning. In fact, the explanation of the Monty Hall paradox comes down to simple mathematics.

Explanation one, more complicated

The probability that the prize is behind the door that was originally chosen is one in three. The chance of finding it behind one of the remaining two is equal to two out of three. Logical, right? Now, after one of these doors is opened and a goat is found behind it, there is only one option left in the second set (the one that corresponds to the 2/3 chance of success). The value of this option remains the same, which is two out of three. Thus, it becomes obvious that by changing his decision, the player will double the probability of winning.

Explanation number two, simpler

After such an interpretation of the decision, many still insist that there is no point in this choice, because there are only two options and one of them is definitely winning, and the other definitely leads to defeat.

But probability theory has its own view on this problem. And this becomes even clearer if we imagine that there were initially not three doors, but, say, a hundred. In this case, it is possible to guess where the prize, the first time, is only one in ninety-nine. Now the participant makes his choice, and Monty eliminates ninety-eight doors with goats, leaving only two, one of which the player chose. Thus, the option chosen initially retains the odds of winning equal to 1/100, and the second option offered remains 99/100. The choice should be obvious.

Are there any refutations?

The answer is simple: no. There is not a single sufficiently substantiated refutation of the Monty Hall paradox. All the “revelations” that can be found on the Internet boil down to a misunderstanding of the principles of mathematics and logic.

For anyone who is well acquainted with mathematical principles, the non-randomness of probabilities is absolutely obvious. Only those who do not understand how logic works can disagree with them. If all of the above still sounds unconvincing, the rationale for the paradox was tested and confirmed on the famous program “MythBusters,” and who else to believe if not them?

Possibility to see clearly

Okay, let all this sound convincing. But this is only a theory, is it possible to somehow look at the work of this principle in action, and not just in words? Firstly, no one canceled living people. Find a partner who will take on the role of facilitator and help you play out the algorithm described above in reality. For convenience, you can take boxes, crates, or even draw on paper. After repeating the process several dozen times, compare the number of wins in case of changing the initial choice with how many wins brought by stubbornness, and everything will become clear. Or you can do it even simpler and use the Internet. There are many simulators of the Monty Hall paradox on the Internet, in which you can test everything yourself and without unnecessary props.

What is the use of this knowledge?

It may seem that this is just another puzzle designed to strain your brain, and it serves only entertainment purposes. However, the Monty Hall paradox finds its practical application primarily in gambling and various sweepstakes. Those with extensive experience are well aware of common strategies for increasing the chances of finding a value bet (from the English word value, which literally means “value” - a prediction that is more likely to come true than it was estimated by bookmakers). And one of these strategies directly involves the Monty Hall Paradox.

Example in working with betting

A sports example will differ little from a classic one. Let's say there are three teams from the first division. In the next three days, each of these teams must play one decisive match. The one that scores more points than the other two at the end of the match will remain in the first division, while the rest will be forced to leave it. The bookmaker's offer is simple: you need to bet on maintaining the position of one of these football clubs, while the betting odds are equal.

For convenience, conditions are accepted under which the rivals of the clubs participating in the selection are approximately equal in strength. Thus, it will not be possible to clearly determine the favorite before the start of the games.

Here you need to remember the story about goats and a car. Each team has a one in three chance of staying in its place. Any of them is selected and a bet is placed on it. Let it be Baltika. According to the results of the first day, one of the clubs loses, and two have yet to play. This is the same “Baltika” and, say, “Shinnik”.

The majority will retain their original bid - Baltika will remain in the first division. But it should be remembered that its chances remained the same, but Shinnik’s chances doubled. Therefore, it is logical to make another bet, a larger one, on Shinnik’s victory.

The next day comes, and the match involving Baltika ends in a draw. Shinnik plays next, and their game ends with a victory with a score of 3:0. It turns out that he will remain in the first division. Therefore, although the first bet on Baltika is lost, this loss is covered by the profit on the new bet on Shinnik.

One can assume, and most will do so, that Shinnik's win was just an accident. In fact, mistaking probability for chance is the biggest mistake for a person participating in sports betting. After all, a professional will always say that any probability is expressed primarily in clear mathematical patterns. If you know the basics of this approach and all the nuances associated with it, then the risks of losing money will be minimized.

Useful in forecasting economic processes

So, in sports betting, the Monty Hall paradox is simply necessary to know. But its scope of application is not limited to betting. Probability theory is always closely related to statistics, which is why understanding the principles of paradox is no less important in politics and economics.

In conditions of economic uncertainty, which analysts often deal with, one must remember the following conclusion arising from solving the problem: it is not necessary to know exactly the only correct solution. The chances of a successful forecast always increase if you know what will definitely not happen. Actually, this is the most useful conclusion from the Monty Hall paradox.

When the world is on the verge of economic turmoil, politicians always try to guess the right course of action in order to minimize the consequences of the crisis. Returning to previous examples, in the economic sphere the task can be described as follows: there are three doors before the leaders of countries. One leads to hyperinflation, the second to deflation, and the third to the cherished moderate economic growth. But how to find the right answer?

Politicians claim that their actions will lead to more jobs and economic growth. But leading economists, experienced people, including even Nobel Prize laureates, clearly demonstrate to them that one of these options will definitely not lead to the desired result. Will politicians change their choices after this? It is extremely unlikely, since in this respect they are not much different from the same participants in the television show. Therefore, the likelihood of error will only increase with an increase in the number of advisers.

Does this exhaust the information on the topic?

In fact, so far only the “classic” version of the paradox has been considered here, that is, the situation in which the presenter knows exactly which door is behind the prize, and only opens the door with the goat. But there are other mechanisms of the leader’s behavior, depending on which the principle of operation of the algorithm and the result of its execution will differ.

The influence of the leader's behavior on the paradox

So what can the presenter do to change the course of events? Let's allow different options.

The so-called "Devil Monty" is a situation in which the host will always offer the player to change his choice, provided that it was initially correct. In this case, changing the decision will always lead to defeat.

On the contrary, “Angel Monty” refers to a similar principle of behavior, but in the event that the player’s choice was initially incorrect. It is logical that in such a situation changing the decision will lead to victory.

If the presenter opens the doors at random, having no idea what is hidden behind each of them, then the chances of winning will always be fifty percent. In this case, there may be a car behind the open leading door.

The GM has a 100% chance of opening the door with a goat if the player chose a car, and with a 50% chance if the player chose a goat. With this algorithm of actions, if the player changes his choice, he will always win in one case out of two.

When the game is repeated over and over again, and the probability of a particular door winning is always arbitrary (as well as which door the presenter will open, while he knows where the car is hiding, and he always opens the door with a goat and offers to change the choice) - the chance of winning will always be equal to one in three. This is called a Nash equilibrium.

The same as in the same case, but provided that the leader is not obliged to open one of the doors at all - the probability of victory will still be equal to 1/3.

While the classic scheme is quite easy to test, experiments with other possible behavior algorithms for the presenter are much more difficult to carry out in practice. But with due meticulousness of the experimenter, this is also possible.

And yet, what is all this for?

Understanding the mechanisms of action of any logical paradoxes is very useful for a person, his brain and awareness of how the world can actually be structured, how much its structure can differ from the individual’s usual idea of ​​it.

The more a person knows about how things work that surround him in everyday life and about which he is not at all used to thinking, the better his consciousness works, and the more effective he can be in his actions and aspirations.