The decision of which, at first glance, is contrary to common sense.

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  The problem is formulated as a description of a game based on the American television game "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 the journal Parade Magazine, sounds like this:

  Imagine that you have become a participant in a game in which you have 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 that the host, 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. After that, he asks you - would you like to change your choice and choose door number 2? Will your chances of winning a car increase if you accept the host's offer and change your choice?

  After the publication, it immediately became clear that the problem was formulated incorrectly: not all conditions were stipulated. For example, the facilitator may follow the “hellish Monty” strategy: offer to change the choice if and only if the player has chosen a car on the first move. Obviously, changing the initial choice will lead to a guaranteed loss in such a situation (see below).

  The most popular is the problem with an additional condition - the participant of the game knows the following rules in advance:

  • the car is equally likely to be placed behind any of the three doors;
  • in any case, the host is obliged to open the door with the goat (but not the one that the player has chosen) and offer the player to change the choice;
  • if the leader has a choice of which of the two doors to open, he chooses either of them with the same probability.

  The following text discusses the Monty Hall problem in this formulation.

  Parsing

  For the winning strategy, the following is important: if you change the choice of the door after the leader's actions, then you win if you initially chose the losing door. It's likely to happen 2 ⁄ 3 , since initially you can choose a losing door in 2 ways out of 3.

  But often, when solving this problem, they argue something like this: the host always removes one losing door in the end, and then the probabilities of a car appearing behind two not open ones become equal to ½, regardless of the initial choice. But this is not true: although there are indeed two possibilities of choice, these possibilities (taking into account the background) are not equally probable! This is true because initially all doors had an equal chance of winning, but then had different probabilities of being eliminated.

  For most people, this conclusion contradicts the intuitive perception of the situation, and due to the resulting discrepancy between the logical conclusion and the answer to which the intuitive opinion inclines, the task is called Monty Hall paradox.

  The situation with the doors becomes even more obvious if we imagine that there are not 3 doors, but, say, 1000, and after the choice of the player, the presenter removes 998 extra ones, leaving 2 doors: the one that the player chose and one more. It seems more obvious that the probabilities of finding a prize behind these doors are different, and not equal to ½. A much higher probability of finding it, namely 0.999, will occur when changing the decision and choosing a door selected from 999. In the case of 3 doors, the logic is preserved, but the probability of winning when changing the decision is lower, namely 2 ⁄ 3 .

  Another way of reasoning is to replace the condition with an equivalent one. Let's imagine that instead of the player making the initial choice (let it always be door #1) and then opening the door with the goat among the remaining ones (that is, always between #2 and #3), let's imagine that the player needs to guess the door on the first try, but he is informed in advance that there can be a car behind door No. 1 with an initial probability (33%), and among the remaining doors it is indicated for which of the doors the car is definitely not behind (0%). Accordingly, the last door will always account for 67%, and the strategy of choosing it is preferable.

  Other leader behavior

  The classic version of the Monty Hall paradox states that the host will prompt the player to change the door, regardless of whether he chose the car or not. But more complex behavior of the host is also possible. This table briefly describes several behaviors.

  Possible Leader Behavior
  Host behavior	Result
  "Infernal Monty": The host offers to change if the door is correct.	Change will always give a goat.
  "Angelic Monty": The host offers to change if the door is wrong.	Change will always give a car.
  "Ignorant Monty" or "Monty Buch": the host inadvertently falls, the door opens, and it turns out that there is not a car behind it. In other words, the host himself does not know what is behind the doors, opens the door completely at random, and only by chance there was no car behind it.	A change gives a win in ½ of the cases. This is how the American show “Deal or No Deal” is arranged - however, the player himself opens a random door, and if there is no car behind it, the presenter offers to change it.
  The host chooses one of the goats and opens it if the player has chosen a different door.	A change gives a win in ½ of the cases.
  The host always opens the goat. If a car is selected, the left goat is opened with probability p and right with probability q=1−p.	If the leader opened the left door, the shift gives a win with probability 1 1 + p (\displaystyle (\frac (1)(1+p))). If the right 1 1 + q (\displaystyle (\frac (1)(1+q))). However, the subject cannot influence the probability that the right door will be opened - regardless of his choice, this will happen with a probability 1 + q 3 (\displaystyle (\frac (1+q)(3))).
  Same, p=q= ½ (classical case).	A change gives a win with a probability 2 ⁄ 3 .
  Same, p=1, q=0 ("powerless Monty" - the tired presenter stands at the left door and opens the goat that is closer).	If the leader opened the right door, the shift gives guaranteed win. If left - probability ½.
  The host always opens the goat if a car is chosen, and with probability ½ otherwise.	The change gives a win with a probability of ½.
  General case: the game is repeated many times, the probability of hiding the car behind one or another door, as well as opening this or that door is arbitrary, but the host knows where the car is and always offers a change by opening one of the goats.	Nash equilibrium: it is Monty Hall's paradox in its classical form that is most beneficial to the host (the probability of winning 2 ⁄ 3 ). The car hides behind any of the doors with probability ⅓; if there is a choice, open any goat at random.
  The same, but the host may not open the door at all.	Nash equilibrium: it is beneficial for the host not to open the door, the probability of winning is ⅓.

  see also

  Notes

  1. Tierney, John (July 21, 1991), "Behind Monty Hall"s Doors: Puzzle, Debate and Answer? ", The New York Times, . Retrieved 18 January 2008.

  The Monty Hall paradox is one of the well-known problems of 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 TV show Let's Make a Deal and is named after the host of this show. The most common formulation of this problem, published in 1990 in Parade Magazine, is as follows:
  Imagine that you have become a participant in a game in which you have 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 that the host, 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. After that, he 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 tasks are uncertain. The following is a more complete statement.
  When solving this problem, they usually reason something like this: after the host has opened the door behind which the goat is located, the car can only be behind one of the two remaining doors. Since the player cannot receive any additional information about which door the car is behind, then the probability of finding a car behind each of the doors is the same, and changing the initial choice of the door does not give the player any advantage. However, this line of reasoning is incorrect. If the host always knows what door is behind, always opens the remaining door that contains a goat, and always prompts 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 doubles the player's chances of winning the car. 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 decision
  The correct answer to this problem is the following: yes, the chances of winning a car are doubled if the player follows the host's advice and changes his initial 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 standing. The probability of this is 1/3. If the player initially hits the door with a goat behind it (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 mind, since the car and one goat remain, and the host has 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 turns to his advantage twice the remaining probability that he did not guess correctly at the beginning.
  Also, an intuitive explanation can be made by swapping the two events. The first event is the player's 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(document see in this article).
  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 of time, the player makes a choice between groups. It is obvious that for the first group the probability of winning is 1/3, for the second group 2/3. The player chooses the second group. In the second group, he can open both doors. One is opened by the host, and the second by the player himself.
  Let's try to give the "most understandable" explanation. Let's reformulate the problem: An honest host 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 specified door (in the old formulation, this is called "do not change your choice ”) or open the other two (in the old wording, this would just be “change the choice”. Think about it, this is the key to understanding!). It is clear that the player will choose the second of the two actions, since the probability of obtaining a car in this case is twice as high. And the little thing that the leader even before choosing the action “showed a goat” does not help and does not interfere with the choice, because behind one of the two doors there is always a goat and the leader will definitely show it at any time during the game, so the player can on this goat and don't watch. The player's business, if he chose the second action, is to say "thank you" to the host for saving him the trouble of opening one of the two doors himself, and opening the other. Well, or even easier. Let's imagine this situation from the point of view of the host, who is doing 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 that the correct strategy is to change the choice after opening the first door: after all, 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 the Stubborn one wins if he initially guessed the car (1/3), and the Attentive one - if he first missed and hit the goat (2/3). After all, only in this case he will then point to the door with the car.
  Keys to understanding
  Despite the simplicity of explaining this phenomenon, many people intuitively believe that the probability of winning does not change when the player changes his choice. Usually, the impossibility of changing the probability of winning is motivated by the fact that when calculating the probability, events that occurred in the past do not matter, as happens, for example, when tossing a coin - the probability of getting heads or tails does not depend on how many times heads or tails fell out 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 when changing the choice, and leaving the original choice.
  However, while such considerations are true in the case of a coin toss, they are not true for all games. AT this case the opening of the door by the master should be ignored. The player essentially chooses between the one door they chose first and the other two - opening one of them only serves to divert 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 (probability of this is 1/3), or behind one of the other two (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 by opening this door, the host does not give the player any additional information about what is behind the door chosen by the player. Thus, opening the door with the goat by the leader does not change the probability (2/3) that the car is behind one of the remaining doors. And since the player does not choose an already open door, then all this probability is concentrated in the event that the car is behind the remaining closed door.
  More intuitive reasoning: Let the player act on the "change choice" strategy. Then he will lose only if he initially chooses a car. And the probability of this is one third. Therefore, the probability of winning: 1-1/3=2/3. If the player acts according to the “do not 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 the host, who is doing 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 that the correct strategy is to change the choice after opening the first door: after all, in the same two cases out of three, the player will leave the studio in a new car.
  Other common cause The difficulty in understanding the solution to this problem lies in the fact that often people imagine a slightly different game - when it is not known in advance whether the host will open the door with a goat and offer the player to change his choice. In this case, the player does not know the leader's tactics (that is, in fact, does not know all the rules of the game) and cannot make the optimal choice. For example, if the facilitator will only offer a change of option 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. (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 by 1/2).
  Increase in the number of doors
  In order to make it easier to understand the essence of what is happening, we can consider the case when the player sees not three doors in front of him, but, for example, a hundred. At the same time, there is a car behind one of the doors, and goats behind the other 99. The player chooses one of the doors, while 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 that, the host opens 98 doors with goats and asks the player to choose the remaining door. In this case, 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 proposal.
  When considering the 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 Monty Hall's paradox conflicts with the intuitive perception of the situation. It would be correct to assume the opening of 98 doors, because the essential condition of the problem is that there is only one alternative choice for the player, which is offered by the host. 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 facilitator opens fewer doors, then 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 host leaves not one door closed, but several, and offers the player to choose one of them, then when changing the initial choice, the player's chances of winning the 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 facilitator 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 originally 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 on 99 doors, but 98. Therefore, the probability of finding a car behind each of these doors will not be 1/100, but 99/9800. The increase in probability will be approximately 0.01%.
  decision tree
  
  
  Wood possible solutions player and host, showing the probability of each outcome
  More formally, a game scenario can be described using a decision tree.
  In the first two cases, when the player first chose the door behind which the goat is, 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 win is equal to
  
  Conducting a similar experiment
  There is an easy way to make sure that changing the original 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 cards) in this case plays the role of the leading Monty Hall, and the second - the role of the player. Three cards are taken for the game, of which one depicts a door with a car (for example, an ace of spades), and two others, identical (for example, two red deuces) - doors with goats.
  The host lays out three cards face down, inviting the player to take one of the cards. After the player chooses a card, the leader looks at the two remaining cards and reveals the red deuce. After that, the cards left by the player and the leader 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 has a red deuce, and the leader has an ace of spades, then a point is scored in favor of the option when the player changes his choice. If we play many such rounds of the game, then the ratio between the points in favor of the two options reflects quite well the ratio of the probabilities of these options. In this case, it turns out that the number of points in favor of changing the initial choice is approximately twice as large.
  Such an experiment not only makes sure that the probability of winning when changing the choice is twice as high, but also well illustrates why this happens. At the moment when the player has chosen a card for himself, 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 the 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 the initial choice) is 2/3.
  Mention
  In the movie Twenty-one, the teacher, Miki Rosa, offers the protagonist, Ben, to solve a puzzle: there are two scooters and one car behind three doors, you must guess the door in order to win the car. After the first choice, Miki offers to change the choice. Ben agrees and mathematically justifies his decision. So he involuntarily passes the test for Miki's team.
  In Sergei Lukyanenko's novel "Nedotepa", the main characters, using this technique, 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, in a popular lecture on mathematics, explains the Monty Hall paradox, clearly illustrating it using marker boards, on reverse sides which are painted goats and a car. Charlie does find the car by changing the selection. However, it should be noted that he only runs one experiment, while the benefit of the changeover strategy is statistical, and a series of experiments should be run to illustrate correctly.

  We all know the situation when we relied on our intuition instead of sober calculation. After all, we must admit that it is far from always possible to calculate everything before making a choice. And no matter how cunning people are, who are used to making their choice only after a thorough analysis, they have not once had to do it on the principle of “probably so”. One of the reasons for such an action may be the banal lack of the necessary time to assess the situation.

  At the same time, the choice is waiting for the current situation right now, and does not allow you to get away from the answer or action. But even more tricky situations for us, which literally causes a spasm of the brain, is the destruction of confidence in the correctness of the choice or in its probable superiority over other options based on logical conclusions. All existing paradoxes are based on this.

  Paradox in the game of the TV show "Let's Make a Deal"

  One of the paradoxes that causes heated debate among puzzle lovers is called the Monty Hall paradox. It is named after the leading TV show in the US called "Let's Make a Deal". On the TV show, the host offers to open one of the three doors, where the prize is a car, while behind the other two there is one goat.

  The participant in the game makes his choice, but the host, knowing where the car is, opens not the door that the player indicated, but the other one, in which the goat is located and offers to change the player's initial choice. For further analysis, we accept this particular behavior of the host, although in fact it may change periodically. We will simply list other options for the development scenario below in the article.

  What is the essence of the paradox?

  Once again, point by point, we will designate the conditions and change the objects of the game for a change to our own.

  The participant of the game is in a room with three bank cells. In one of the three cells there is a gold ingot of gold, in the other two, one coin with a face value of 1 kopeck of the USSR.

  So, the participant before the choice and the conditions of the game are as follows:

  1. The participant can choose only one of the three cells.
  2. The banker knows initially the location of the ingot.
  3. The banker always opens a coin cell other than the player's choice and prompts the player to change the choice.
  4. The player can in turn change his choice or keep the original one.

  What does intuition say?

  The paradox is that for most people who are used to thinking logically, the chances of winning if they change their initial choice are 50 to 50. After all, after the banker opens another cell with a coin different from the player’s initial choice, 2 cells remain , in one of which there is an ingot of gold, and in the other a coin. The player wins an ingot if he accepts the banker's offer to change the cell, provided that there was no ingot in the cell originally selected by the player. And vice versa, under this condition, he loses if he refuses to accept the offer.

  

  As we suggest common sense, the probability of choosing an ingot and winning in this case is 1/2. But in fact the situation is different! “But how is it, everything is obvious here?” - you ask. Let's say you chose cell number 1. Intuitively, yes, no matter what choice you had initially, in the end you actually have a coin and an ingot before choosing. And if initially you had a probability of receiving a prize 1/3, then in the end, when you open one cell by a banker, you get a probability of 1/2. The probability seemed to have increased from 1/3 to 1/2. Upon careful analysis of the game, it turns out that when the decision is changed, the probability increases to 2/3 instead of the intuitive 1/2. Let's take a look at why this happens.

  Unlike the intuitive level, where our consciousness considers the event after the cell change as something separate and forgets about the initial choice, mathematics does not break these two events, but rather preserves the chain of events from beginning to end. So, as we said earlier, the chances of winning when hitting an ingot immediately are 1/3, and the probability that we will choose a cell with a coin is 2/3 (since we have one ingot and two coins).

  1. We initially select a bank cell with an ingot - the probability is 1/3.
     • If the player changes his choice by accepting the banker's offer, he loses.
     • If the player does not change his choice without accepting the banker's offer, he wins.
  2. We choose from the first time a bank cell with a coin - the probability is 2/3.
     • If the player changes his choice, he wins.
     • If the player does not change the choice - lost.

  So, in order for the player to leave the bank with a gold bar in his pocket, he must choose an initially losing position with a coin (probability 1/3), and then accept the banker's offer to change the cell.

  In order to understand this paradox and break out of the shackles of the initial selection pattern and the remaining cells, let's imagine the player's behavior in exactly the opposite way. Before the banker offers a cell for selection, the player is mentally precisely determined that he changes his choice, and only after that the event of opening an extra door follows for him. Why not? After all, the open door does not give him more information in such a logical sequence. At the first stage of time, the player divides the cells into two different areas: the first is a one-cell area with his original selection, the second is with the two remaining cells. Next, the player has to make a choice between two areas. The probability of getting a gold ingot from the cell from the first area is 1/3, from the second 2/3. The choice follows the second area in which he can open two cells, the first one will be opened by the banker, the second by himself.

  There is an even better explanation for the Monty Hall paradox. To do this, you need to change the wording of the task. The banker makes it clear that one of the three bank cells contains a gold bar. In the first case, he offers to open one of the three cells, and in the second - two at the same time. What will the player choose? Well, of course, two at once, by doubling the probability. And the moment when the banker opened the cell with a coin, this actually does not help the player in any way and does not interfere with the choice, because the banker will show this cell with a coin anyway, so the player can simply ignore this action. On the part of the player, one can only thank the banker for making life easier for him, and instead of two, he had to open one cell. Well, you can finally get rid of the paradox syndrome if you put yourself in the place of a banker who initially knows that the player points to the wrong door in two out of three cases. For the banker, there is no paradox as such, because he is sure in such an inversion of events that in the event of a change in events, the player takes the gold bar.

  

  Monty Hall's paradox clearly does not allow conservatives to win, who are ironclad in their original choice and lose their chance of increasing the probability. For conservatives, it will remain 1/3. For vigilant and reasonable people, it grows to the above 2/3.

  All the above statements are relevant only in compliance with the initially stipulated conditions.

  What if we increase the number of cells?

  What if we increase the number of cells? Let's say instead of three of them there will be 50. The gold ingot will lie in only one cell, and in the remaining 49 - coins. Accordingly, unlike the classical case, the probability of hitting the target on the move is 1/50 or 2% instead of 1/3, while the probability of choosing a cell with a coin is 98%. Further, the situation develops, as in the previous case. The banker offers to open any of the 50 cells, the participant chooses. Let's say the player opens a cell at serial number 49. The banker, in turn, as in the classical version, is in no hurry to fulfill the player's desire and opens other 48 cells with coins and offers to change his choice to the remaining one at number 50.

  It is important to understand here that the banker opens exactly 48 cells, not 30, and at the same time leaves 2, including the one chosen by the player. It is this choice that allows the paradox to run counter to intuition. As in the case with classic version, opening 48 cells by the banker leaves only one single alternative to choose from. The case of a variant of a smaller opening of cells does not allow one to put the problem on a par with the classics and feel the paradox.

  But since we have already touched on this option, then let's assume that the banker leaves not one, except for the one chosen by the player, but several cells. Presented, as before, 50 cells. The banker, after the choice of the player, opens only one cell, while leaving 48 cells closed, including the one chosen by the player. The probability of choosing an ingot on the first try is 1/50. In sum, the probability of finding an ingot in the remaining cells is 49/50, which, in turn, spreads not into 49, but into 48 cells. It is not difficult to calculate that the probability of finding an ingot in this case is (49/50)/48=49/2900 . The probability, although not by much, is still higher than 1/50 by approximately 1%.

  As we mentioned at the very beginning, the host Monty Hall in the classic scenario of the game with doors, goats and a prize car can change the conditions of the game and with it the probability of winning.

  

  The mathematics of paradox

  Can they mathematical formulas prove the increase in probability when changing choices?
  Let's imagine the chain of events as a set divided into two parts, the first part will be taken as X - this is the player's choice at the first stage of the safe cell; and the second set Y - the remaining two remaining cells. The probability (B) of winning for cells 2 and 3 can be expressed using formulas.

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

  Where 1/2 is the probability with which the banker will open cell 2 and 3, provided that the player initially selected a cell without an ingot.
  Further, the conditional probability 1/2 when the banker opens the cell with the coin changes to 1 and 0. Then the formulas take the following form:

  B(2) = 0 * 2/3 = 0
  B(3) = 1 * 2/3 = 1

  Here we clearly see that the probability of choosing an ingot in cell 3 is 2/3, which is just over 60 percent.
  A very beginner-level programmer can easily test this paradox by writing a program that calculates the probability when changing choices, or vice versa, and check the results.

  Explanation of the paradox in the film 21 (Twenty-one)

  A visual explanation of the Monty Paul paradox is given in the film "21" (Twenty-one), directed by Robert Luketic. Professor Mickey Rosa in a lecture gives an example from the Let's Make a Deal show and asks student Ben Campbell (actor and singer James Anthony) about the probability distribution, who gives the correct alignment and thus surprises the teacher.

  Independent study of the paradox

  For people who want to check the result on their own in practice, but do not have a mathematical basis, we suggest that you independently simulate a game in which you will be the leader and someone will be the player. You can involve children in this game, who will choose sweets or candy wrappers from them in pre-prepared cardboard boxes. With each choice, be sure to record the result for further calculation.

  Wording

  The most popular is the problem with additional condition No. 6 from the table - the participant of the game knows the following rules in advance:

  • the car is equally likely placed behind any of the 3 doors;
  • the host in any case is obliged to open the door with the goat and offer the player to change the choice, but not the door that the player has chosen;
  • if the leader has a choice which of the 2 doors to open, he chooses any of them with the same probability.

  The following text discusses the Monty Hall problem in this formulation.

  Parsing

  When solving this problem, one usually argues something like this: the host always removes one losing door in the end, and then the probabilities of a car appearing behind two unopened doors become 1/2, regardless of the initial choice.

  The whole point is that with his initial choice, the participant divides the doors: the chosen A and two others - B and C. The probability that the car is behind the selected door = 1/3, that behind the others = 2/3.

  For each of the remaining doors, the current situation is described as follows:

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

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

  Where 1/2 is the conditional probability that the car is behind the given door, provided that the car is not behind the door chosen by the player.

  The host, opening one of the remaining doors, which is always losing, thereby informs the player exactly 1 bit of information and changes conditional probabilities for B and C respectively to "1" and "0".

  As a result, the expressions take the form:

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

  Thus, the participant should change his initial choice - in this case, the probability of his winning will be equal to 2/3.

  One of the simplest explanations is this: if you change the door after the GM's actions, then you win if you originally chose the losing door (then the GM will open a second losing one and you will have to change your choice to win). And initially you can choose a losing door in 2 ways (probability 2/3), i.e. if you change the door, you win with a probability of 2/3.

  This conclusion contradicts the intuitive perception of the situation by most people, therefore the described task is called Monty Hall paradox, i.e. a paradox in the everyday sense.

  And the intuitive perception is as follows: opening the door with a goat, the host sets a new task for the player, which has nothing to do with the previous choice - after all, the goat is behind open door will appear regardless of whether the player has previously chosen a goat or a car. After the third door is opened, the player has to make a choice again - and choose either the same door that he chose before, or a different one. That is, while he does not change his previous choice, but makes a new one. The mathematical solution considers two successive tasks of the leader as related to each other.

  However, one should take into account the factor from the condition that the host will open the door with the goat from the remaining two, and not the door chosen by the player. Therefore, the remaining door has a better chance of a car, as it was not chosen as the master. If we consider the case when the leader, knowing that there is a goat behind the door chosen by the player, nevertheless opens this door, by doing so he deliberately reduces the chances of the player to choose the correct door, because. the probability of a correct choice will be already 1/2. But this kind of game will have different rules.

  Let's give one more explanation. Let's assume that you are playing according to the system described above, i.e. of the two remaining doors, you always choose a door different from your original choice. In which case will you lose? The loss will come when, and only then, when from the very beginning you have chosen the door behind which the car is located, because subsequently you will inevitably change your mind in favor of the door with a goat, in all other cases you will win, i.e., if from the very beginning Wrong choice of door. But the probability of choosing the door with the goat from the very beginning is 2/3, so it turns out that to win, you need a mistake, the probability of which is twice as large as the correct choice.

  Mentions

  • In the film Twenty-one, the teacher, Miki Rosa, offers the main character, Ben, to solve a problem: there are two scooters and one car behind three doors, you must guess the door with a car. After the first choice, Miki offers to change the choice. Ben agrees and mathematically justifies his decision. So he involuntarily passes the test for Miki's team.
  • In Sergei Lukyanenko's novel "Kluttyopa", the main characters, with the help of this technique, 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, at a popular lecture on mathematics, explains the Monty Hall paradox, clearly illustrating it with the help of marker boards, on the reverse sides of which goats and a car are drawn. Charlie does find the car by changing the selection. However, it should be noted that he only runs one experiment, while the benefit of the changeover strategy is statistical, and a series of experiments should be run to illustrate correctly.
  • The Monty Hall paradox is discussed in the diary of the hero of Mark Haddon's story The Curious Incident of the Dog in the Night.
  • Monty Hall's paradox tested by the MythBusters

  see also

  • Bertrand's paradox

  Links

  	The Monty Hall Paradox at Wikibooks
  	The Monty Hall Paradox at Wikimedia Commons

  • Interactive prototype: for those who want to fool around (generation occurs after the first choice)
  • Interactive prototype: real prototype of the game (cards are generated before selection, the work of the prototype is transparent)
  • Explainer video on Smart Videos.ru
  • Weisstein, Eric W. The Monty Hall Paradox (English) on the Wolfram MathWorld website.
  • The Monty Hall Paradox on the website of the TV show Let's Make a Deal
  • An excerpt from the book by S. Lukyanenko, which uses the Monty Hall paradox
  • Another Bayesian Solution Another Bayesian Solution at the Novosibirsk State University Forum

  Literature

  • Gmurman V.E. Probability theory and mathematical statistics, - M .: Higher education. 2005
  • Gnedin, Sasha "The Mondee Gills Game." magazine The Mathematical Intelligencer, 2011 http://www.springerlink.com/content/8402812734520774/fulltext.pdf
  • Parade Magazine dated 17 February.
  • vos Savant, Marilyn. Ask Marilyn column, magazine Parade Magazine dated 26 February.
  • Bapeswara Rao, V. V. and Rao, M. Bhaskara. "A three-door game show and some of its variants". Magazine The Mathematical Scientist, 1992, № 2.
  • Tijms, Henk. Understanding Probability, Chance Rules in Everyday Life. Cambridge University Press, New York, 2004. (ISBN 0-521-54036-4)

  Notes

  
  

  Wikimedia Foundation. 2010 .

  See what the "Monty Hall Paradox" is in other dictionaries:

  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? The Monty Hall paradox is one of the well-known problems of the theory ... ... Wikipedia

  - (The tie paradox) is a well-known paradox similar to the problem of two envelopes, which also demonstrates the features of the subjective perception of probability theory. The essence of the paradox: two men give each other for Christmas ties bought by them ... ... Wikipedia