How to choose one of the three. Monty Hall's paradox is a logic puzzle not for the faint of heart.
The solution of which, at first glance, contradicts common sense.
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The problem is formulated as a description of a game based on the American game show Let's Make a Deal, and is named after the host of that show. 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 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. After that, he 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? After publication, it immediately became clear that the task was formulated incorrectly: not all conditions were specified. For example, the presenter may follow the “Hell Monty” strategy: offer a change of choice if and only if the player chose a car as their first move. Obviously, changing the initial choice will lead to a guaranteed loss in such a situation (see below).
The most popular is a task with an additional condition - the participant in the game knows the following rules in advance:
- the car is equally likely placed behind any of the three 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 two doors to open, he chooses either of them with equal probability.
The following text discusses the Monty Hall problem in precisely this formulation.
Analysis
For the winning strategy, the following is important: if you change the choice of door after the actions of the leader, then you win if you initially chose the losing door. This is likely to happen 2 ⁄ 3 , since initially you can choose a losing door in 2 out of 3 ways.
But often when solving this problem, they reason something like this: the leader always ends up removing one losing door, and then the probability of a car appearing behind two not open ones becomes equal to ½, regardless of the initial choice. But this is not true: although there are indeed two possibilities for choice, these possibilities (taking into account the background) are not equally probable! This is true because all doors initially 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 problem is called Monty Hall paradox.
The situation with doors becomes even more clear if you imagine that there are not 3 doors, but, say, 1000, and after the player’s choice, the presenter removes 998 extra ones, leaving 2 doors: the one 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, 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 remains, 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 No. 1) and then the leader opening the door with the goat among the remaining ones (that is, always among No. 2 and No. 3), imagine that the player needs to guess the door on the first try, but he is previously informed that there may be a car behind door No. 1 with the initial probability (33%), and among the remaining doors it is indicated which of the doors there is definitely no car behind (0%). Accordingly, the last door will always account for 67%, and the strategy for choosing it is preferable.
Other presenter behavior
The classic version of the Monty Hall paradox states that the host will definitely offer the player to change the door, regardless of whether he chose the car or not. But more complex behavior of the leader is also possible. This table briefly describes several behaviors.
Possible behavior of the presenter Presenter behavior Result "Hell Monty": The host suggests changing if the door is right. A change will always produce a goat. "Angel Monty": the host suggests changing if the door is wrong. A change will always give you a car. “Ignorant Monty” or “Monty Buh”: the presenter accidentally falls, the door opens, and it turns out that there is no car behind it. In other words, the presenter himself does not know what is behind the doors, he opens the door completely at random, and only by chance there was no car behind it. The change gives a gain in ½ of the cases.
This is exactly how the American show “Deal or No Deal” works - however, a random door is opened by the player himself, and if there is no car behind it, the host offers to change it.The host chooses one of the goats and opens it if the player chose another door. The change gives a gain in ½ of the cases. The leader always opens the goat. If a car is selected, the left goat opens with the probability p and right with probability q=1−p. If the leader opened the left door, the shift gives a win with the probability 1 1 + p (\displaystyle (\frac (1)(1+p))). If right - 1 1 + q (\displaystyle (\frac (1)(1+q))). However, the subject cannot in any way influence the probability that the right door will be opened - regardless of his choice, this will happen with probability 1 + q 3 (\displaystyle (\frac (1+q)(3))). The same, p=q= ½ (classical case). The change gives a win with 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 leader opens the right door, the change gives a guaranteed win. If left - probability ½. The presenter always opens the goat if a car is chosen, and with a probability of ½ otherwise. The change gives a win with a probability of ½. General case: the game is repeated many times, the probability of hiding a car behind one or another door, as well as opening one or another door is arbitrary, but the leader knows where the car is and always offers a change, opening one of the goats. Nash equilibrium: the leader benefits most from the Monty Hall paradox in its classical form (probability of winning 2 ⁄ 3 ). The car hides behind any of the doors with probability ⅓; if there is a choice, we open any goat at random. The same thing, but the presenter may not open the door at all. Nash equilibrium: it is profitable for the leader not to open the door, the probability of winning is ⅓. See also
Notes
- Tierney, John (July 21, 1991), "Behind Monty's Hall"s Doors: Puzzle, Debate and Answer? ", The New York Times,
I met it under the name "The Monty Hall Paradox", and wow, solved it differently, namely: proved that this is a pseudo-paradox.
Friends, I will be glad to listen to 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 imagining myself as a thinker and think about changing my 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 are a participant in a game in which you are in front of three doors. The presenter, who is known to be honest, placed a car behind one of the doors, and a goat each behind the other two doors. You have no information about what is behind which door.
The host tells you: “First you must choose one of the doors. After that, I will open one of the remaining doors, behind which is a goat. I will then have you change your original choice and choose the remaining closed door instead of the one you initially chose. You can follow my advice and choose another door, or confirm your original choice. After that, I will open the door you chose and you will win whatever is behind that door.”
You choose door number 3. The host opens door number 1 and shows that there is a goat behind it. Then the host 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 in probability theory, the solution of which, at first glance, contradicts common sense.
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 initial 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.It seems to me that the chances will not change, i.e. there is no paradox.
And here's why: the first and second door choices are independent events. It’s like tossing a coin twice: what comes up the second time does not depend in any way on what comes up the first time.
So it is here: after opening the door with the 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 is 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 or an open one. To opening the doors was such a balance of probabilities, but after opening one door, all these probabilities become insignificant, 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 changing the choice.
Addition: 1) reasoning that:
a) the probability of finding a car behind the selected door is 1/3,
b) the probability that the car is behind two other unselected doors is 2/3,
c) because the leader opened the door with the goat, then the probability of 2/3 goes entirely to one unchosen (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 of 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 the 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 for each event the probability is calculated separately, 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 equal to 1 out of 2, those. 1/2.
3) Thus, this is not a paradox, but a delusion! (11/19/2009)
Appendix 2: Yesterday I came up with the simplest explanation that the re-election 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: you don’t have to mark the goats in the car, but cull the goats, and even the leader helps with this by opening the goat. And at the beginning of the game, with a probability of 2 out of 3, the player will succeed, so, having culled the goats, you need to change the choice. And this also suddenly became very obvious: 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.
“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 let’s imagine that one of the remaining prisoners finds out about prisoner A’s question and the answer he 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 this 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.
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.
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Young parents, of course, want to know how tall their child will be as an adult. Mathematical statistics can offer a simple linear relationship to approximate the height of children based only on the height of the father and mother, and also indicate the accuracy of such an estimate.
Monty Hall's paradox is probably the most famous paradox in probability theory. There are many variations of it, for example, the paradox of three prisoners. And there are many interpretations and explanations of this paradox. But here, I would like to give not only a formal explanation, but show the “physical” basis of what happens in the Monty Hall paradox and others like it.
The classic formulation is:
“You are a participant in the game. There are three doors in front of you. There's a prize for one of them. The host invites you to try to guess where the prize is. You point to one of the doors (at random).
Formulation of the Monty Hall Paradox
The host knows where the prize actually is. He doesn’t yet open the door you pointed to. But it opens one more of the remaining doors for you, behind which there is no prize. The question is, should you change your choice or stay with your previous decision?
It turns out that if you simply change your choices, your chances of winning will increase!
The paradox of the situation is obvious. It seems that everything that happens is random. It makes no difference whether you change your mind or not. But that's not true.
"Physical" explanation of the nature of this paradox
Let's first not go into mathematical subtleties, but simply look at the situation with an open mind.
In this game, you just make a random choice first. Then the presenter tells you additional information, which allows you to increase your chances of winning.
How does the presenter give you additional information? Very simple. Note that it opens not any door.
Let's, for the sake of simplicity (although there is an element of deceit in this), consider a more likely situation: you pointed to a door behind which there is no prize. Then, behind one of the remaining doors is a prize There is. That is, the presenter has no choice. He opens a very specific door. (You pointed to one, there is a prize behind the other, there is only one door left that the leader can open.)
It is at this moment of meaningful choice that he gives you information that you can use.
In this case, the use of information is that you change your decision.
By the way, your second choice is already too not accidental(or rather, not as random as the first choice). After all, you choose from closed doors, but one is already open and it not arbitrary.
Actually, after these considerations, you may have the feeling that it is better to change your decision. This is true. Let's show this more formally.
A more formal explanation of the Monty Hall paradox
In fact, your first, random choice splits all the doors into two groups. Behind the door you chose there is a prize with a probability of 1/3, behind the other two - with a probability of 2/3. Now the leader makes a change: he opens one door in the second group. And now the entire 2/3 probability applies only to the closed door from the group of two doors.
It is clear that now it is more profitable for you to change your decision.
Although, of course, you still have a chance to lose.
However, changing your selection increases your chances of winning.
Monty Hall Paradox
The Monty Hall paradox is a probabilistic problem, the solution of which (according to some) is contrary to common sense. Problem formulation:
Imagine that you are a participant in a game in which you need to choose one of three doors. Behind one of the doors is a car, behind the other two doors are goats.
You choose one of the doors, for example, number 1, after which the leader, who knows where the car is and where the goats are, opens one of the remaining doors, for example, number 3, behind which there is a goat.Monty Hall paradox. The most inaccurate mathematics
He then asks you if you would like to change your choice and choose door number 2.
Will your chances of winning a car increase if you accept the presenter's offer and change your choice?When solving a problem, it is often mistakenly assumed that the two choices are independent and, therefore, the probability will not change if the choice is changed. In fact, this is not the case, as you can see by remembering Bayes' formula or looking at the simulation results below:
Here: “strategy 1” - do not change the choice, “strategy 2” - change the choice. Theoretically, for the case with 3 doors, the probability distribution is 33.(3)% and 66.(6)%. Numerical simulations should yield similar results.
Links
Monty Hall Paradox– a problem from the section of probability theory, the solution of which contradicts common sense.
History[edit | edit wiki text]
At the end of 1963, a new talk show called “Let’s Make a Deal” aired. According to the quiz scenario, viewers from the audience received prizes for correct answers, having a chance to increase them by making new bets, but risking their existing winnings. The show's founders were Stefan Hatosu and Monty Hall, the latter of whom became its constant host for many years.
One of the tasks for the participants was the drawing of the Main Prize, which was located behind one of three doors. Behind the remaining two were incentive prizes, and the presenter, in turn, knew the order of their arrangement. The contestant had to determine the winning door by betting their entire winnings for the show.
When the guesser decided on the number, the presenter opened one of the remaining doors, behind which there was an incentive prize, and invited the player to change the initially chosen door.
Wording[edit | edit wiki text]
As a specific problem, the paradox was first formulated by Steve Selvin in 1975, when he sent The American Statistician magazine and host Monty Hall the question: would a contestant's chances of winning the Grand Prize change if, after opening the door with incentive will he change his choice? After this incident, the concept of the “Monty Hall Paradox” appeared.
In 1990, the most common version of the paradox was published in Parade Magazine with an example:
“Imagine yourself on a game show where you have to choose one of three doors: two of them are goats, and the third is a car. When you make a choice, assuming, for example, that the winning door is number one, the leader opens one of the remaining two doors, for example, number three, behind which is a goat. Then you are given a chance to change the selection to another door? Can you increase your chances of winning a car if you change your choice from door number one to door number two?
This formulation is a simplified version, because There remains the factor of influence of the presenter, who knows exactly where the car is and is interested in the participant’s loss.
For the task to become purely mathematical, it is necessary to eliminate the human factor by introducing the opening of a door with an incentive prize and the ability to change the initial choice as integral conditions.
Solution[edit | edit wiki text]
When comparing chances, at first glance, changing the door number will not give any advantages, because all three options have a 1/3 chance of winning (approx. 33.33% for each of the three doors). In this case, opening one of the doors will not in any way affect the chances of the remaining two, whose chances will become 1/2 to 1/2 (50% for each of the two remaining doors). This judgment is based on the assumption that the player’s choice of a door and the leader’s choice of a door are two independent events that do not affect one another. In reality, it is necessary to consider the entire sequence of events as a whole. In accordance with the theory of probability, the chances of the first selected door from the beginning to the end of the game are invariably 1/3 (approx. 33.33%), and the two remaining ones have a total of 1/3+1/3 = 2/3 (approx. 66.66%). When one of the two remaining doors opens, its chances become 0% (there is an incentive prize hidden behind it), and as a result, the chances of closing the unselected door will be 66.66%, i.e. twice as much as the one originally selected.
To make it easier to understand the results of a choice, you can consider an alternative situation in which the number of options will be greater, for example, a thousand. The probability of choosing the winning option is 1/1000 (0.1%). Given that nine hundred and ninety-eight incorrect ones are subsequently opened out of the remaining nine hundred and ninety-nine options, it becomes clear that the probability of the one remaining door out of the nine hundred and ninety-nine not chosen is higher than that of the only one chosen at the beginning.
Mentions[edit | edit wiki text]
You can find references to the Monty Hall Paradox in “Twenty-One” (a film by Robert Luketic), “The Klutz” (a novel by Sergei Lukyanenko), the television series “4isla” (TV series), “The Mysterious Murder of a Dog in the Night-Time” (a story by Mark Haddon), “XKCD” ( comic book), “MythBusters” (TV show).
See also[edit | edit wiki text]
The image shows the process of choosing between two buried doors from the three initially proposed
Examples of solutions to combinatorics problems
Combinatorics is a science that everyone encounters in everyday life: how many ways to choose 3 people on duty to clean the classroom or how many ways to form a word from given letters.
In general, combinatorics allows you to calculate how many different combinations, according to certain conditions, can be made from given objects (same or different).
As a science, combinatorics originated in the 16th century, and now every student (and often even schoolchildren) studies it. They 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 jobs 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
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?
Solution. Let us immediately note that this problem does not contain any paradox. A common task (initial level) on Bayes' formula, which follows from the definition of conditional probability.
Bayes formula
Let's denote by A the event - you won a car.
We put forward two hypotheses: H 1 - you do not change the door, and H 2 - you change the door.
P(H 1) = 1/3 - a priori (a priori means before the experiment, the presenter had not yet opened the door) probability of the hypothesis that you are changing the door.
P H1 (A) - conditional probability that you will guess the door behind which the car is located if the first hypothesis H 1 occurs
P H2 (A) - conditional probability that you will guess the door behind which the car is located, if the second hypothesis H 2 occurs
We find the probability of event A if hypothesis H 1 occurs (the probability that you won the car if you did not change the door):
Find the probability of event A if hypothesis H 2 occurs (the probability that you won a car if you changed the door):
Thus, the participant should change his original choice - in this case, the probability of winning will be equal to 2 ⁄ 3.
Statistical test of Monty Hall paradox
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.
