3 door theory. Refutation of the "Monty Hall Paradox" (an imaginary refutation, as it turns out)

“There are three kinds of lies: lies, damned lies and statistics.” This phrase, attributed by Mark Twain to British Prime Minister Benjamin Disraeli, fairly reflects the attitude of the majority towards mathematical laws. Indeed, the theory of probability sometimes throws up surprising facts that are difficult to believe at first glance - and which, nevertheless, are confirmed by science. “Theories and Practices” recalled the most famous paradoxes.

Monty Hall Problem

This is exactly the problem that a cunning MIT professor presented to students in the movie Twenty-One. Having given the correct answer, the main character ends up on a team of brilliant young mathematicians who beat casinos in Las Vegas.

The classic formulation goes like this: “Let’s say a certain player is offered to take part in the famous American TV show Let’s Make a Deal, hosted by Monty Hall, and he needs to choose one of three doors. Behind two doors are goats, behind one is the main prize, a car, the presenter knows the location of the prizes. After the player makes his choice, the host opens one of the remaining doors, behind which there is a goat, and invites the player to change his decision. Should the player agree or is it better to keep his original choice?”

Here's a typical line of reasoning: after the host has opened one of the doors and shown the goat, the player has to choose between two doors. The car is located behind one of them, which means that the probability of guessing it is ½. So it makes no difference whether to change your choice or not. And yet, probability theory says that you can increase your chances of winning by changing your decision. Let's figure out why this is so.

To do this, let's take a step back. The moment we made our initial choice, we divided the doors into two parts: the one we chose and the other two. Obviously, the probability that the car is hiding behind “our” door is ⅓ - accordingly, the car is behind one of the two remaining doors with a probability of ⅔. When the presenter shows that there is a goat behind one of these doors, it turns out that this ⅔ chance falls on the second door. And this reduces the player’s choice to two doors, behind one of which (initially selected) the car is located with a probability of ⅓, and behind the other - with a probability of ⅔. The choice becomes obvious. Which, of course, does not change the fact that from the very beginning the player could choose the door with the car.

Three Prisoners Problem

The Three Prisoners Paradox is similar to the Monty Hall problem, although it takes place in a more dramatic setting. Three prisoners (A, B and C) are sentenced to death and placed in solitary confinement. The governor randomly selects one of them and gives him a pardon. The warden knows which of the three has been pardoned, but he is ordered to keep it a secret. Prisoner A asks the guard to tell him the name of the second prisoner (besides himself) who will definitely be executed: “if B is pardoned, tell me that C will be executed. If B is pardoned, tell me that B will be executed. If they are both executed , and I have been pardoned, toss a coin and say any of these two names.” The warden says that prisoner B will be executed. Should prisoner A be happy?

It would seem so. After all, before receiving this information, the probability of prisoner A’s death was ⅔, and now he knows that one of the other two prisoners will be executed - which means that the probability of his execution has decreased to ½. But in fact, prisoner A did not learn anything new: if he was not pardoned, he would be told the name of another prisoner, and he already knew that one of the two remaining would be executed. If he is lucky and the execution is canceled, he will hear a random name B or C. Therefore, his chances of salvation have not changed in any way.

Now imagine that one of the remaining prisoners finds out about prisoner A’s question and the answer received. This will change his views on the likelihood of a pardon.

If prisoner B overheard the conversation, he will know that he will definitely be executed. And if prisoner B, then the probability of his pardon will be ⅔. Why did it happen? Prisoner A has not received any information and still has a ⅓ chance of being pardoned. Prisoner B will definitely not be pardoned, and his chances are zero. This means that the probability that the third prisoner will be released is ⅔.

Paradox of two envelopes

This paradox became known thanks to the mathematician Martin Gardner, and is formulated as follows: “Suppose you and a friend were offered two envelopes, one of which contains a certain amount of money X, and the other contains an amount twice as much. You independently open the envelopes, count the money, and then you can exchange them. The envelopes are the same, so the probability that you will receive an envelope with a lower amount is ½. Let's say you open an envelope and find $10 in it. Therefore, it is equally likely that your friend's envelope contains $5 or $20. If you decide to exchange, then you can calculate the mathematical expectation of the final amount - that is, its average value. It is 1/2x$5+1/2x20=$12.5. Thus, the exchange is beneficial to you. And, most likely, your friend will think the same way. But it is obvious that the exchange cannot be beneficial for both of you. What is the mistake?

The paradox is that until you open your envelope, the probabilities behave well: you actually have a 50% chance of finding the amount X in your envelope and a 50% chance of finding the amount 2X. And common sense dictates that information about the amount you have cannot affect the contents of the second envelope.

However, as soon as you open the envelope, the situation changes dramatically (this paradox is somewhat similar to the story of Schrödinger's cat, where the very presence of an observer affects the state of affairs). The fact is that in order to comply with the conditions of the paradox, the probability of finding in the second envelope a larger or smaller amount than yours must be the same. But then any value of this sum from zero to infinity is equally probable. And if there is an equally probable infinite number of possibilities, they add up to infinity. And this is impossible.

For clarity, you can imagine that you find one cent in your envelope. Obviously, the second envelope cannot contain half the amount.

It is curious that discussions regarding the resolution of the paradox continue to this day. At the same time, attempts are being made both to explain the paradox from the inside and to develop the best strategy for behavior in such a situation. In particular, Professor Thomas Cover proposed an original approach to strategy formation - to change or not to change the envelope, guided by some intuitive expectation. Let's say, if you open an envelope and find $10 in it - a small amount in your estimation - it's worth exchanging it. And if there is, say, $1,000 in the envelope, which exceeds your wildest expectations, then there is no need to change. This intuitive strategy, if you are regularly asked to choose two envelopes, allows you to increase your total winnings more than the strategy of constantly changing envelopes.

Boy and girl paradox

This paradox was also proposed by Martin Gardner and is formulated as follows: “Mr. Smith has two children. At least one child is a boy. What is the probability that the second one is also a boy?

It would seem that the task is simple. However, if you start to look into it, a curious circumstance emerges: the correct answer will differ depending on how we calculate the probability of the gender of the other child.

Option 1

Let's consider all possible combinations in families with two children:

Girl/Girl

Girl boy

Boy/Girl

Boy/Boy

The girl/girl option does not suit us according to the conditions of the task. Therefore, for Mr. Smith's family, there are three equally probable options - which means that the probability that the other child will also be a boy is ⅓. This is exactly the answer that Gardner himself initially gave.

Option 2

Let's imagine that we meet Mr. Smith on the street when he is walking with his son. What is the probability that the second child is also a boy? Since the gender of the second child has no bearing on the gender of the first, the obvious (and correct) answer is ½.

Why is this happening, since it would seem that nothing has changed?

It all depends on how we approach the issue of calculating probability. In the first case, we considered all possible options for the Smith family. In the second, we considered all families that fall under the mandatory condition “there must be one boy.” The calculation of the probability of the sex of the second child was carried out with this condition (in probability theory this is called “conditional probability”), which led to a result different from the first.

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:
the car is equally likely placed behind any of the 3 doors;

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;

if the leader has a choice of which of 2 doors to open, he chooses any of them with equal probability)

At first glance, the odds should not change (sorry, this is no longer a paradox for me, and I can no longer come up with an incorrect explanation for why the odds will not change, which at first glance would look logical).

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!

Do you understand me now why I don’t see the paradox here? If there are two and only two mutually exclusive strategies, and one gives a win with probability p, then the other should give a win with probability 1-p, what kind of paradox is this?

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?

Greetings from sunny Vietnam! :) Come work with us! :)

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 “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

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 (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

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.

Problem and solution

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.

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

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.

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

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.

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 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. 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.

door 1	door 2	door 3
Selection Machine	Goat	Goat open
Choice	Car	Goat open
Choice	Goat open	Car

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.

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).

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 start 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

3 door theory. Refutation of the "Monty Hall Paradox" (an imaginary refutation, as it turns out)

Monty Hall Problem

Three Prisoners Problem

Paradox of two envelopes

Boy and girl paradox

Problem and solution

More precise formulation of the problem

Solution

Keys to Understanding

Increasing the number of doors

Decision Tree

Proof using a table

Three Prisoners Problem

Answer

Formulation of the Monty Hall Paradox

"Physical" explanation of the nature of this paradox

A more formal explanation of the Monty Hall paradox

Monty Hall Paradox

Monty Hall paradox. The most inaccurate mathematics

Links

History[edit | edit wiki text]

Wording[edit | edit wiki text]

Solution[edit | edit wiki text]

Mentions[edit | edit wiki text]

See also[edit | edit wiki text]

Examples of solutions to combinatorics problems

Combinatorics problems with online solutions

Ready-made examples