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The Monty Hall paradox. The most inaccurate math ever

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 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. If we change the door, then we lose only if we chose the prize door first, the probability of which is 1:1000. We win if our initial choice was Not correct, and the probability of this is 999 out of 1000. In the case of 3 doors, the logic is preserved, but the probability of winning when changing the decision is correspondingly 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))).
The same, p=q= ½ (classical case).	A change gives a win with a probability 2 ⁄ 3 .
The same, p=1, q=0 ("powerless Monty" - the tired presenter stands at the left door and opens the goat that is closer).	If the presenter opened the right door, the change gives a 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.

Notes

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

In December 1963, the American television channel NBC first aired the program Let's Make a Deal (“Let's make a deal!”), In which participants, selected from the audience in the studio, bargained with each other and with the host, played small games or simply guessed the answer to question. At the end of the broadcast, the participants could play the “deal of the day”. There were three doors in front of them, about which it was known that behind one of them was the Grand Prize (for example, a car), and behind the other two were less valuable or completely absurd gifts (for example, live goats). After the player had made his choice, Monty Hall, the host of the program, opened one of the two remaining doors, showing that there was no Prize behind it and letting the participant be glad that he had a chance to win.

In 1975, UCLA scientist Steve Selvin asked what would happen if, at that moment, after opening the door without a Prize, the participant was asked to change their choice. Will the player's chances of getting the Prize change in this case, and if so, in what direction? He sent the corresponding question in the form of a problem to The American Statistician (“American Statistician”), as well as to Monty Hall himself, who gave a rather curious answer to it. In spite of this answer (or perhaps because of it), the problem became popular under the name "Monty Hall problem".

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?

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 3 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.

Clue

Try to consider people who chose different doors in the same case (that is, when the Prize is, for example, behind door number 1). Who will benefit from changing their choice, and who will not?

Solution

As suggested in the tooltip, consider people who made different choices. Suppose the Prize is behind door #1, and behind doors #2 and #3 are goats. Suppose we have six people, and each door was chosen by two people, and from each pair one subsequently changed the decision, and the other did not.

Note that the Host who chooses door No. 1 will open one of the two doors to his taste, while, regardless of this, the Car will be received by the one who does not change his choice, but the one who changed his initial choice will remain without the Prize. Now let's look at those who chose doors #2 and #3. Since there is a Car behind door No. 1, the Host cannot open it, which leaves him no choice - he opens doors No. 3 and No. 2 for them, respectively. At the same time, the one who changed the decision in each pair will choose the Prize as a result, and the one who did not change will be left with nothing. Thus, out of three people who change their minds, two will get the Prize, and one will get the goat, while out of the three who left their original choice unchanged, only one will get the Prize.

It should be noted that if the Car were behind door #2 or #3, the result would be the same, only the specific winners would change. Thus, assuming that initially each door is chosen with equal probability, we get that those who change their choice win the Prize twice as often, that is, the probability of winning in this case is greater.

Let's look at this problem from the point of view of the mathematical theory of probability. We will assume that the probability of the initial choice of each of the doors is the same, as well as the probability of being behind each of the doors of the Car. In addition, it is useful to make a reservation that the Leader, when he can open two doors, chooses each of them with equal probability. Then it turns out that after the first decision, the probability that the Prize is behind the selected door is 1/3, while the probability that it is behind one of the other two doors is 2/3. At the same time, after the Host opened one of the two "unselected" doors, the entire probability of 2/3 falls on only one of the remaining doors, thereby creating the basis for changing the decision, which will increase the probability of winning by 2 times. Which, of course, does not guarantee it in any way in one specific case, but will lead to more successful results in the case of repeated repetition of the experiment.

Afterword

The Monty Hall problem is not the first known formulation of this problem. In particular, in 1959, Martin Gardner published in Scientific American a similar problem “about three prisoners” (Three Prisoners problem) with the following wording: “Out of three prisoners, one should be pardoned, and two should be executed. Prisoner A persuades the guard to tell him the name of the one of the other two who will be executed (either if both are executed), after which, having received the name B, he considers that the probability of his own salvation has become not 1/3, but 1/2. At the same time, prisoner C claims that the probability of his escape has become 2/3, while nothing has changed for A. Which one is right?"

However, Gardner was not the first, since back in 1889, in his Calculus of Probabilities, the French mathematician Joseph Bertrand (not to be confused with the Englishman Bertrand Russell!) Offers a similar problem (see Bertrand's box paradox): “There are three boxes, each of which contains two coins: two gold ones in the first one, two silver coins in the second one, and two different ones in the third one.

If you understand the solutions to all three problems, it is easy to notice the similarity of their ideas; mathematically, all of them are united by the concept of conditional probability, that is, the probability of event A, if it is known that event B has occurred. The simplest example: the probability that a unit fell out on a regular dice is 1/6; however, if the rolled number is known to be odd, then the probability that it is one is already 1/3. The Monty Hall problem, like the other two problems cited, shows that conditional probabilities must be handled with care.

These problems are also often called paradoxes: Monty Hall's paradox, Bertrand's box paradox (the latter should not be confused with the real Bertrand's paradox given in the same book, which proved the ambiguity of the concept of probability that existed at that time) - which implies some contradiction (for example, in " paradox of the Liar" the phrase "this statement is false" contradicts the law of the excluded middle). In this case, however, there is no contradiction with rigorous assertions. But there is a clear contradiction with the "public opinion" or simply the "obvious solution" of the problem. Indeed, most people, looking at the problem, believe that after opening one of the doors, the probability of finding the Prize behind any of the two remaining closed ones is 1/2. By doing so, they assert that it makes no difference whether they agree or disagree to change their mind. Moreover, many people find it difficult to comprehend an answer other than this, even after being told the detailed solution.

Monty Hall's response to Steve Selwyn

Mr. Steve Selvin,
assistant professor of biostatistics,
University of California, Berkeley.

Dear Steve,

Thank you for sending me the problem from American Statistical.

Although I didn't study statistics at university, I know that numbers can always be used to my advantage if I wanted to manipulate them. Your reasoning does not take into account one essential circumstance: after the first box is empty, the participant can no longer change his choice. So the probabilities remain the same: one in three, right? And, of course, after one of the boxes is empty, the chances do not become 50/50, but remain the same - one out of three. It only seems to the participant that by getting rid of one box, he gets more chances. Not at all. Two to one against him, as it was, and remains. And if you suddenly come to my show, the rules will remain the same for you: no changing boxes after the selection.

Imagine that a certain banker offers you to choose one of three closed boxes. In one of them 50 cents, in the other - one dollar, in the third - 10 thousand dollars. Whichever one you choose, you will get it as a prize.

You choose at random, say box number 1. And then the banker (who, of course, knows where everything is) right in front of your eyes opens a box with one dollar (let's say this is No. 2), after which he offers you to change the initially selected box No. 1 to box No. 3.

Should you change your mind? Will this increase your chances of getting 10 thousand?

This is Monty Hall's paradox - a problem of probability theory, the solution of which, at first glance, contradicts common sense. People have been scratching their heads over this problem since 1975.

The paradox was named after the host of the popular American TV show Let's Make a Deal. This TV show had similar rules, only the participants chose doors, two of which were hiding goats, and the third was a Cadillac.

Most of the players reasoned that after there were two closed doors and there was a Cadillac behind one of them, then the chances of getting it were 50-50. Obviously, when the host opens one door and invites you to change your mind, he starts a new game. Whether you change your mind or not, your chances will still be 50 percent. So right?

It turns out that it doesn't. In fact, by changing your mind, you double your chances of success. Why?

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 (see this article for proof). 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. Reformulate the problem: An honest host announces to the player that there is a car behind one of the three doors, and suggests that he 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 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 host 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 host 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 open 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.

Monty Hall, producer and host of the show Let's Make a Deal from 1963 to 1991.

In 1990, this problem and its solution were published in the American magazine Parade. The publication caused a flurry of indignant reviews from readers, many of whom had scientific degrees.

The main complaint was that not all the conditions of the problem were specified, and any nuance could affect the result. For example, the host could offer to change the decision only if the player chose a car on the first move. Obviously, changing the initial choice in such a situation will lead to a guaranteed loss.

However, in the entire existence of the Monty Hall TV show, people who changed their minds did win twice as often:

Out of 30 players who changed their mind, Cadillac won 18 - i.e. 60%

Out of the 30 players who remained with their choice, Cadillac won 11 - that is, approximately 36%

So the reasoning given in the decision, no matter how illogical they may seem, is confirmed by practice.

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 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 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 1%.

Possible Decision Tree of the 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.

If you still don't understand, spit on the formulas and justcheck everything statistically. Another possible explanation:

A player whose strategy would be to change the selected door every time would only lose if he initially chooses the door behind which the car is located.
Since the chance of choosing a car on the first try is one in three (or 33%), the chance of not choosing a car if the player changes his choice is also one in three (or 33%).
This means that the player who used the strategy to change the door will win with a probability of 66% or two to three.
This will double the chances of winning a player whose strategy is not to change their choice every time.

Still don't believe? Let's say you choose door #1. Here are all the possible options for what can happen in this case.

Met her called the Monty Hall Paradox, and wow solved it differently, namely: proved that this is a pseudo-paradox.

Friends, I will be glad to hear criticism of my refutation of this paradox (pseudo-paradox, if I'm right). And then I will see with my own eyes that my logic is lame, I will stop thinking of myself as a thinker and think about changing the type of activity to a more lyrical one: o). So, here is the content of the task. The proposed solution and my rebuttal are below.

Imagine that you have become a participant in a game in which you are in front of three doors. The host, who is known to be honest, placed a car behind one of the doors, and a goat behind the other two doors. You have no information about what is behind which door.

The facilitator tells you: “First you have to choose one of the doors. After that, I will open one of the remaining doors, behind which is a goat. Then I will suggest that you change your original choice and choose the remaining closed door instead of the one you chose at the beginning. You can follow my advice and choose another door, or you can confirm your original choice. After that, I will open the door you have chosen and you will win what is behind that door.”

You choose door number 3. The facilitator opens door number 1 and shows that there is a goat behind it. The host then asks you to choose door number 2.

Will your chances of winning a car increase if you follow his advice?
The Monty Hall paradox is one of the well-known problems of probability theory, the solution of which, at first glance, contradicts common sense.
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 get any additional information about which door the car is behind, 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 the 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.

It seems to me that the chances will not change; there is no paradox.

And here's why: the first and second door choices are independent events. It's like tossing a coin 2 times: what falls out the 2nd time does not depend in any way on what fell out in the 1st.

So here: after opening the door with a goat, the player finds himself in new situation when it has 2 doors and the probability of choosing a car or a goat is 1/2.

Once again: after opening one door out of three, the probability that the car is behind the remaining door, not equal to 2/3, because 2/3 is the probability that the car is behind any 2 doors. It is incorrect to attribute this probability to an unopened door and an open one. Before the opening of the doors was such an alignment of probabilities, but after opening one door, all these probabilities become void, because the situation has changed, and therefore a new probability calculation is needed, which ordinary people correctly carry out, answering that nothing will change from a change of choice.

Addendum: 1) reasoning that:

a) the probability of finding a car behind the chosen door is 1/3,

b) the probability that the car is behind two other unselected doors, 2/3,

c) because the host opened the door with the goat, then the probability of 2/3 goes entirely to one unselected (and unopened) door,

and therefore it is necessary to change the choice to another door, so that the probability from 1/3 becomes 2/3, not true, but false, namely: in paragraph "c", because initially the probability 2/3 concerns any two doors, including the 2 remaining not open, and since one door was opened, then this probability will be divided equally between 2 not open, i.e. the probability will be equal, and choosing another door will not increase it.

2) conditional probabilities are calculated if there are 2 or more random events, and the probability is calculated separately for each event, and only then the probability of the joint occurrence of 2 or more events is calculated. Here, at first, the probability of guessing was 1/3, but in order to calculate the probability that the car is not behind the door that was chosen, but behind the other one that is not open, you do not need to calculate the conditional probability, but you need to calculate the simple probability, which is 1 out of 2, those. 1/2.

3) Thus, this is not a paradox, but a fallacy! (19.11.2009)

Appendix 2: Yesterday I came up with the simplest explanation that reselection strategy is still more advantageous(the paradox is true!): with the first choice, getting into a goat is 2 times more likely than into a car, because there are two goats, and therefore, with the second choice, you need to change the choice. It's so obvious :o)

Or in other words: it is necessary not to mark in the car, but to reject the goats, and even the presenter helps in this, opening the goat. And at the beginning of the game, with a probability of 2 out of 3, the player will also succeed, so, having rejected the goats, you need to change the choice. And it also became very obvious all of a sudden :o)

So everything I've written so far has been a pseudo-refutation. Well, here is another illustration of the fact that you need to be more modest, respect someone else's point of view and not trust the assurances of your logic that its decisions are crystal logical.

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 of the problem undefined. 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 get any additional information about which door the car is behind, 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.

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. 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 formulation, this would just be "change the choice". Think, 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 “showed a goat” even before choosing the action 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 open 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.

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. In 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 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 (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.

Another common reason for the difficulty in understanding the solution to this problem is that often people imagine a slightly different game - where it is not known in advance whether the host will open the door with a goat and suggest that the player 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

decision tree

Possible decision tree of the 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) 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, the ace of spades), and two others that are identical (for example, two red deuces) are 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. The further opening of one of the cards by the leader 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, challenges the main character, Ben, to solve a puzzle: there are two scooters and one car behind three doors; you must guess the door 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 of Man Hunt), one of the main characters, Charlie Epps, explains the Monty Hall paradox in a popular lecture on mathematics, clearly illustrating it using marker boards with goats and a car drawn on the reverse sides. 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.

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

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