I Can’t Sleep - Monty Hall Problem | Calm Educational Reading for Sleep

Episode Date: January 26, 2025

Drift off with this calm bedtime reading about the Monty Hall problem, created to ease insomnia and bring restful focus. In this soothing episode, Benjamin explains the famous probability puzzle inspired by the game show Let’s Make a Deal, exploring its origins, reasoning, and surprising solution. His steady, reassuring narration transforms tricky mathematical concepts into peaceful storytelling, helping you release stress and quiet your thoughts. There is no whispering or hypnosis, only calm, fact-filled narration designed to guide you toward rest. Press play, close your eyes, and let the Monty Hall problem carry you into sleep. Want More? Request a Topic: https://www.icantsleeppodcast.com/request-a-topic Ad-Free Episodes: https://icantsleep.supportingcast.fm/ Shop Sleep-Friendly Products: https://www.icantsleeppodcast.com/sponsors Join the Discussion on Discord: https://discord.gg/myhGhVUhn7 This content is derived from the Wikipedia article on Monty Hall Problem, available under the Creative Commons Attribution-ShareAlike (CC BY-SA) license. Read the full article: Wikipedia – Monty Hall Problem. Happy sleeping! Learn more about your ad choices. Visit megaphone.fm/adchoices

You're listening to a Glassbox media podcast. What if I told you that most of the modern day self-help advice you've been hearing could actually make you worse? The key to a better life isn't about feel-good gimmicks that sound catchy. The Mentally Stronger Podcast gives you access to a licensed therapist who shares science-backed tools that will actually change your life. Hi, I'm Amy Morin, psychotherapist, mental strength trainer, and international best-selling author. In each episode, we cover research-back strategies, like how to stop relying on willpower and start creating habits for lasting change. And the five mental strength-building exercises you can do from your couch.

I also speak to world-class experts like Dr. Nicole Kane, who shares how to permanently heal anxiety by addressing the root cause. With over 200 episodes in our catalog, this podcast is for you if you're ready to crush self-doubt, conquer challenges, and become stronger than ever with therapist-approved strategies that can change your life. Listen to Mentally Stronger with Therapist Amy Morin, wherever you get your podcasts. Welcome to the I Can't Sleep podcast, where I read random articles from across the web to bore you to sleep with my soothing voice. I'm your host, Benjamin Boster, and today's episode is from a Wikipedia article titled Monty Hall Problem.

The Monty Hall problem is a brain teaser in the form of a probability puzzle. Based nominally on the American television game show, let's make a deal, and named after its original host, Monty Hall. The problem was originally posed and solved in a letter by Steve Selvin to the American statistician in 1975. It became famous as a question from reader Craig F. Whitaker's, letter quoted in Marilyn Vos Savants Ask Maryland column in Parade Magazine in 1990.

Suppose you're on a game show and you're given the choice of three doors. Behind one doors a car, behind the others, goats. You pick a door, say number one, and the host, who knows what's behind the doors, opens another door, say number three, which has a goat.

He then says, to you, do you want to pick door number two? Is it to your advantage to switch your choice? Savant's response was that the contestant should switch to the other door. By the standard assumptions, a switching strategy has a two-thirds probability of winning the car, while the strategy of keeping the initial choice has only a third probability. When the player first makes their choice, there is a two-thirds chance that the car is behind one of the doors not chosen. This probability does not change after the host reveals a goat behind one of the unchosen doors.

When the host provides information about the two unchosen doors, revealing that one of them does not have the car behind it, the two-thirds chance of the car being behind one of the unchosen doors, rests on the unchosen and unrevealed door. as opposed to the one-third chance of the car being behind the door the contestant chose initially. The given probabilities depend on specific assumptions about how the host and contestant choose their doors. An important insight is that with these standard conditions, there is more information about Doors 2 and 3 than there was available at the beginning of the game

when Door 1 was chosen by the player. The host's action adds value to the door not eliminated, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action from choosing between the two remaining doors at random, as the former action uses the previous information and the latter does not. Other possible behaviors of the host than the one described can reveal different additional information or none at all, leading to different possibilities.

In her response, Savant states, Suppose there are a million doors, and you pick door number one. And the host who knows what's behind the doors, and will always avoid the one with the prize, opens them all except door number 77,77,77. You'd switch to that door pretty fast, wouldn't you? Many readers of Savant's column refused to believe switching is beneficial and rejected her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine,

most of them calling Savon wrong. Even when given explanations, simulations, and formal mathematical proofs, many people still did not accept that switching is the best strategy. Paul Erdis, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation, demonstrating Savant's predicted result. The problem is a paradox of the veritical type, because the solution is so counterintuitive, it can seem absurd, but is nevertheless demonstrably true. The Monty Hall problem is mathematically related closely to the earlier three prisoners problem and the much older Bertrand's Box paradox. Here's the paradox.

Stephen Selvin wrote a letter to the American statistician in 1975, describing a problem based on the game show Let's Make a Deal, dubbing it the Monty Hall problem in a subsequent letter. The problem is equivalent mathematically to the, the three prisoners problem described in Martin Gardner's mathematical games column in Scientific American in 1959, and the three shells problem described in Gardner's book, Aha, Gotcha. Standard Assumptions By the standard assumptions, the probability of winning the car after switching is two-thirds.

This solution is due to the behavior of the host. ambiguities in the parade version do not explicitly define the protocol of the host. However, Marilyn Vos Savant's solution printed alongside Whitaker's question implies, and both Selvin and Savant explicitly define, the role of the host as follows. One, the host must always open a door that was not selected by the contestant. Two, the host must always open a door to reveal a goat and never think. car. Three, the host must always offer the chance to switch between the door chosen originally and the closed door remaining. When any of these assumptions is varied, it can change the

probability of winning by switching doors as detailed in the section below. It is also typically presume that the car is initially hidden randomly behind the doors, and that if the player initially chooses the car, then the host's choice of which goat-hiding door to open is random. Some authors independently or inclusively assume that the player's initial choice is random as well. Simple solutions. The solution presented by Savand and Parade shows the three possible arrangements of one car and two goats behind three doors, and the result of staying or switching after initially picking door 1 in each case. Behind door 1, goat.

Behind door 2, goat. Behind door 3, car. Result of staying at door number 1, Wynn's goat. Result of switching to the door offered, Wins' car. Behind door 1, goat, behind door 2, car,

behind door 3, goat. Result of staying at door number 1, Wins' goat. result of switching to the door offered, wins car. Behind door one, car, behind door two, goat, behind door three, goat. Result of staying at door number one, wins car. Result of switching to the door offered, wins goat. A player who stays with the initial choice wins in only one out of three of these equally likely possibilities,

while a player who switches wins in two out of three. An intuitive explanation is that if the contestant initially picks a goat, two of three doors, the contestant will win the car by switching because the other goat can no longer be picked. The host had to reveal its location, whereas if the contestant initially picks the car, one of three doors, the contestant will not win the car by switching. Using the switching strategy, winning or losing, thus only depends on whether the contestant has initially chosen a goat, two-thirds probability, or the car, one-third's probability. The fact that the host subsequently reveals a goat in one of the unchosen doors

changes nothing about the initial probability. Most people conclude that switching does not matter because there would be a 50% chance of finding the car behind either of the two unopened doors. This would be true if the host selected a door to open at random, but this is not the case. The host open door depends on the player's initial choice. So the assumption of the independence does not hold. Before the host opens a door, there is a third probability that the car is behind each door. If the car is behind door one, the host can open either door 2 or 3, so the probability that the car is behind door 1

and the host opens door 3 is 1 3 3 3rd times 1 1 1⁄2. If the car is behind door 2 was the player having picked door 1, the host must open door 3, such the probability that the car is behind door 2 and the host opens door 3 is 1 3 3 3 3 3 3 3 3rd times 1 equals 1 3rd.

These are the only cases where the host opens door 3. So if the player has picked door 1 and the host opens door 3, the car is twice as likely to be behind door 2 as door 1. The key is that if the car is behind door 2, the host must open door 3. But if the car is behind door 1, the host can open either door. Another way to understand the solution is to consider together the two doors initially unchosen by the player. As Cecil Adams put it,

Monty is saying in effect, you can keep your one door, or you can have the other two doors. The two-thirds chance of finding the car has not been changed by the opening of one of these doors, because Monty, knowing the location of the car, is certain to reveal a goat. The player's choice after the host opens a door is no different than if the host offered the player the option to switch from the original chosen door to the set of both remaining doors.

The switch in this case clearly gives the player a two-thirds probability of choosing the car. As Keith Devlin says, By opening his door, Monty is saying to the contestant, There are two doors you did not choose, and the probability that the prize is behind one of them is two-thirds. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of one in three of being the winner. I have not changed that, but by eliminating Door C, I have shown you that the probability that Door B hides the prize is 2 and 3.

Savant suggests that the solution will be more intuitive with 1 million doors rather than 3. In this case, there are 999,99 doors with goats behind them and one door with a prize. After the player picks a door, the host opens 99,998 of the remaining doors. On average, in 99,99 times out of 1 million, the remaining door will contain the prize. Intuitively, the players should ask how likely it is that, given a million doors, they managed to pick the right one initially. Stibel et al proposed that working memory demand is taxed during the Monte Hall problem and that this forces people to collapse their choices into two equally probable options. They report that when the number of options is increased to more than seven, people tend to switch more often.

However, most contestants still incorrectly judged the probability of success to be 50%. Savant and the Media Furer You blew it, and you blew it big. Since you seem to have difficulty grasping the basic principle at work here, I'll explain. After the host reveals a goat, you now have a one-and-two chance of being correct. Whether you change your selection or not, the odds are the same. There is enough mathematical literacy in this country, and we don't need the world's highest IQ propagating more.

shame. Scott Smith, University of Florida. Savant wrote in her first column on the Monty Hall problem that the player should switch. She received thousands of letters from her readers, the vast majority of which, including many from readers with PhDs, disagreed with her answer. During 1990 to 1991, three more of her columns in parade were devoted to the paradox. numerous examples of letters from readers of savants columns are presented and discussed in the Monty Hall Dilemma, a cognitive illusion par excellence. The discussion was replayed in other venues, e.g. in Cecil Adams, the Straight Dope newspaper column, and reported in major newspapers such as the New York Times.

In an attempt to clarify her answer, she proposed a shell game to illustrate You look away, and I put a P under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a P are a third, agreed? Then I simply lift up an empty shell from the remaining other two. As I can and will do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger. She also proposed a similar simulation with three playing cards.

Savant commented that though some confusion was caused by some readers not realizing they were supposed to assume that the host must always reveal a goat. Almost all her numerous correspondence had correctly understood the problem assumptions, and were still initially convinced that Savant's answer, switch, was wrong. Confusion and criticism sources of confusion. When first presented with the Monty Hall problem, an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter.

Out of 228 subjects in one study, only 13% chose to switch. In his book, The Power of Logical Thinking, Cognitive psychologist Massimo Piatali-Palmerini writes, No other statistical puzzle comes so close to fooling all the people all the time, and even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate and print those who propose the right answer. Pigeons repeatedly exposed to the problem shows that they rapidly learn to always switch, unlike humans.

Most statements of the problem, notably the one in parade, do not match the rules of the actual game show and do not fully specify the host's behavior, or that the car's location is randomly selected. However, Krauss and Wang argue that people make this standard assumptions, even if they are not explicitly stated. Although these issues are mathematically significant, even when controlling for these factors, nearly all people still think each of the two unopened doors has an equal probability,

and conclude that switching does not matter. This equal probability assumption is a deeply rooted intuition. People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not. The problem continues to attract attention of cognitive psychologists. The typical behavior of the majority, i.e., not switching, may be explained by far. phenomenon known in the psychological literature as one, the endowment effect, in which people

tend to overvalue the winning probability of the door already chosen, already owned. 2. The status quo bias, in which people prefer to keep the choice of door they have made already. 3. The errors of omission versus errors of commission effect. In which all other things being equal, people prefer to make errors by, inaction, stay, as opposed to action, switch. Experimental evidence confirms that these are plausible explanations that do not depend on probability intuition. Another possibility is that people's intuition simply does not deal with a textbook version of the problem, but with a real game show setting. There, the possibility

exists that the showmaster plays deceitfully by opening other doors, only if a door with a car was initially chosen. A showmaster playing deceitfully half of the time modifies the winning chances in case one is offered to switch to equal probability. Criticism of the simple solutions. As already remarked, most sources in the topic of probability, including many introductory probability textbooks, solves the problem by showing the conditional probabilities that the car is behind door 1 and door 2 are 1 3rd and 2 3, not 1 half and 1 half, given that the contestant initially picks door 1 and the host opens door 3. Various ways to derive and understand this result were given in the previous subsections. Among these sources are separate,

several that explicitly criticize the popularly presented simple solutions, saying these solutions are correct but shaky, or do not address the problem posed, or are incomplete, or are unconvincing and misleading, or are, most bluntly, false. Sasha Volok, 2015, wrote that any explanation that says something like the probability of Door One was one-third,

and nothing can change that, is automatically fishy. Probabilities are expressions of our ignorance about the world, and new information can change the extent of our ignorance. Some say that these solutions answer a slightly different question. One phrasing is, you have to announce before a door has been opened whether you plan to switch. The simple solutions show in various ways that a contestant who is determined to switch will win the car with probability two-thirds, and hence that switching is the winning strategy if the player has to choose in advance between always switching and always staying. However, the probability of winning by always switching is a logically distinct concept

from the probability of winning by switching, given that the player has picked door one, and the host has opened door three. As one source says, the distinction between the these questions seems to confound many. The fact that these are different can be shown by varying the problem so that these two probabilities have different numeric values. For example, assume the contestant knows that Monty does not open the second door randomly among all legal alternatives, but instead, when given an opportunity to choose between two losing doors, Monty will open the one on the ride. In this situation, the following two questions have different answers.

1. What is the probability of winning the car by always switching? 2. What is the probability of winning the car by switching, given the player has picked door 1, and the host has opened door 3? The answer to the first question is 2 thirds, as is shown correctly by the simple solution. But the answer to the second question is now different. The conditional probability the car is behind door 1 or door 2, given the host is open door 3, the door on the right, is 1 half. This is because Monty's preference for rightmost doors means that he opens door 3 if

the car is behind door 1, which it is originally with probability a third, or if the car is behind door two, also originally was probability one-third. For this variation, the two questions yield different answers. This is partially because the assumed condition of the second question that the host opens door three would only occur in this variant with probability two-thirds. However, as long as the initial probability the car is behind each door is one-third, it is never to the contestants' disadvantage to switch, as the conditional probability of winning by switching is always at least one-half. In Morgan at all, four university professors published an article in The American Statistician,

claiming that savant gave the correct advice but the wrong argument. They believe the question asked for the chance of the car behind Door 2, given the player's initial choice of door one and the game host opening door three, and they showed this chance was anything between one-half and one, depending on the host's decision process given the choice. Only when the decision is completely randomized is the chance two-thirds. In an invited comment and in subsequent letters to the editor,

Morgan et al were supported by some writers, criticized by others. In each case, a response by Morgan et al is published alongside the letter or comment in the American statistician. In particular, Savant defended herself vigorously. Morgan at all complained in their response to Savant that Savant still had not actually responded to their own main point. Later, in their response to Hogbin and Nijam, they did agree that it was natural to suppose that the host choose a door to open completely at random when he does have a choice, and hence that the conditional probability of winning by switching, i.e. conditional given the situation the player is in when he has to make the choice, has the same value, two-thirds, as the unconditional probability of winning by switching, i.e., averaged over all

possible situations. This equally was already emphasized by Bell, 1992, who suggested that Morgan et al's mathematically involved solution would appeal only to statisticians, whereas the equivalence of the conditional and unconditional solutions in the case of symmetry was intuitively obvious. There is disagreement in the literature regarding whether Savant's formulation of the problem, as presented in Parade, is asking the first or second question, and whether this difference is significant. Barron's conclusion that one must consider the matter with care to see that both analyses are correct, which is not to say that they are the same. Several critics of the paper by Morgan et al, whose contributions were published along with the original paper,

criticized the authors for altering savon's wording and misinterpreting her intention. One discussant William Bell considered it a matter of taste, whether one explicitly mentioned, that, by the standard conditions, which door is open by the host, is independent of whether one should want to switch. Among the simple solutions, the combined doors solution comes closest to a conditional solution, as we saw in the discussion of the methods using the concept of odds and base theorem. It is based on the deeply rooted intuition that revealing information that is already known does not affect probabilities.

But knowing that the host can open one of the two unchosen doors to show a goat does not mean that opening a specific door would not affect the probability that the car is behind the door chosen initially. The point is, though we know in advance that the host will open a door and reveal a goat, we do not know which door he will open.

If the host chooses uniformly at random between doors hiding a goat, as is the case in the standard interpretation, this probability indeed remains unchanged. But if the host can choose non-randomely between such doors, then the specific door that the host opens reveals additional information. The host can always open a door revealing a goat, and, in the standard interpretation of the problem, the probability that the car is behind the initially chosen door does not change.

but it is not because of the former that the latter is true. Solutions based on the assertion that the host actions cannot affect the probability that the car is behind the initially chosen appear persuasive, but the assertion is simply untrue unless both of the host's two choices are equally likely if he has a choice. The assertion therefore needs to be justified. Without justification being given, the solution is at best. incomplete. It can be the case that the answer is correct, but the reasoning used to justify it is

defective. Solutions using conditional probability and other solutions. The simple solutions above show that a player with a strategy of switching wins the car with overall probability two-thirds, i.e. without taking account of which door was opened by the host. In accordance with this, Most, most sources for the topic of probability calculate the conditional probabilities that the car is behind door 1 and door 2 to be 1 3rd and 2 3, given the contestant initially picks door 1 and the host opens door 3. The solutions in this section consider just those cases in which the player picked door 1 and the host open door 3, refining the simple solution. If we assume that the host opens a door at random, when given a choice, then which door the host opens gives us no information at all as to whether or not the car is behind door one.

In the simple solutions, we have already observed that the probability that the car is behind door one, the door initially chosen by the player, is initially one-third. Moreover, the host is certainly going to open a different door, So opening a door, which door is unspecified, does not change this. One-third must be the average of, the probability that the car is behind door one, given that the host picked door two, and the probability of car behind door one, given the host picked door three.

This is because these are the only two possibilities. However, these two probabilities are the same. Therefore, they are both equal to one-third. This shows that the chance that the car is behind Door 1, given that the player initially chose this door and given that the host opened Door 3 is 1 3. And it follows that the chance that the car is behind Door 2, given that the player initially chose Door 1

and the host Open Door 3 is 2 thirds. The analysis also shows that the overall success rate of 2 thirds, achieved by always switching, cannot be improved. and underlines what already may have well been intuitively obvious. The choice facing the player is that between the door initially chosen and the other door left closed by the host. The specific numbers on these doors are irrelevant. Conditional probability by direct calculation.

By definition, the conditional probability of winning by switching, given the contestant initially picks door one and the host opens door three, is the probability for the event car is behind door 2 and host opens door 3 divided by the probability for host opens door 3. These probabilities can be determined referring to the conditional probability table below in this article, or to an equivalent decision tree.

BASE theorem Many probability textbooks and articles in the field of probability theory derive the conditional probability solution through a formal application of Bay's theorem, among them books by Gill and Hensey. Use of the odds form of Bay's theorem, often called Bay's rule, makes such a derivation more transparent. Initially, the car is equally likely to be behind any of the three doors. The odds on door 1, 2, and 3 are 1 to 1 to 1.

This remains a case after the player has chosen Door 1 by Independence. According to Bay's rule, the posterior odds on the location of the car, given that the host opens Door 3, are equal to the prior odds multiplied by the Bay's factor or likelihood, which is, by definition, the probability of the new piece of information, host opens Door 3, under each of the hypotheses considered location of the car. Now, since the player initially chose door 1, the chance that the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus, the base factor consists of the ratios 1⁄2 to 1 to 0, or equivalently 1 to 2, while the prior odds were 1 to 1.

Thus, the posterior odds become equal to the base factor 1 to 2 to 0. Given that the host open door 3, the probability that the car is behind door 3 is 0, and it is twice as likely to be behind door 2 than door 1. Richard Gill analyzes the likelihood for the host to open door 3 as follows. Given that the car is not behind door 1, it is equally likely that it is behind door 2 or 3. Therefore, the chance that the host opens door 3 is 50%. Given that the car is behind door 1, the chance that the host opens door 3 is also 50%, because when the host has a choice, either choice is equally likely.

Therefore, whether or not the car is behind door 1, the chance that the host opens door 3 is 50%. The information, host opens door three, contributes a base factor or likelihood ratio of one-to-one on whether or not the car is behind door one. Initially, the odds against door one hiding the car were two-to-one. Therefore, the posterior odds against door one hiding the car remain the same as the prior odds, two-to-one. In words, the information which door is opened by the host's. door 2 or door 3, reveals no information at all about whether or not the car is behind

door 1. And this is precisely what is alleged to be intuitively obvious by supporters of simple solutions, or using the idioms of mathematical proofs, obviously true, by symmetry. There is so much more to go over in this article, so if you want to hear more about the Monty Hall problem, let me know and I'll continue on. Otherwise, thanks for listening. to this episode of the podcast, and feel free to share and follow the podcast to get the latest updates on each episode. Happy sleeping.