(Unless the game has a rule where Monty Hall would open both doors if car was not behind any of them). Your first pick is a random door (1/100) and your other choice is the champion that beat out 99 other doors (aka the MVP of the league). With the Japanese baseball players, you know more than your friend and have better chances. When he’s done, he has the top door out of 99 for you to pick. So what should you do? 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. This probability does not change after the host reveals the location of the other goat. The more you test the old standard, the less likely the new choice beats it. The player initially chooses door i = 1, C = X1 and the host opens door i = 3, C = H3. [4] Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy. For example, strategy A "pick door 1 then always stick with it" is dominated by the strategy B "pick door 1 then always switch after the host reveals a door": A wins when door 1 conceals the car, while B wins when one of the doors 2 and 3 conceals the car. I remember this problem from watching an episode of numbers. The host must always open a door that was not picked by the contestant. That’s the hard (but convincing) way of realizing switching works. A show master playing deceitfully half of the times modifies the winning chances in case one is offered to switch to "equal probability". "[5] "Virtually all of my critics understood the intended scenario. Ambiguities in the Parade version do not explicitly define the protocol of the host. ", The host opens a door, the odds for the two sets don't change but the odds move to 0 for the open door and, "You blew it, and you blew it big! In general, more information means you re-evaluate your choices. If the car is behind door 2 (and the player has picked door 1) the host must open door 3, so the probability the car is behind door 2 AND the host opens door 3 is 1/3 × 1 = 1/3. Today let’s get an intuition for why a simple game could be so baffling. He sees two doors and is told to pick one: he has a 50-50 chance! Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. You now have a once-in-a-lifetime chance of winning a fantastic sports car which is hidden behind one of these three doors. Stibel et al[17] proposed that working memory demand is taxed during the Monty Hall problem and that this forces people to "collapse" their choices into two equally probable options. Among the simple solutions, the "combined doors solution" comes closest to a conditional solution, as we saw in the discussion of approaches using the concept of odds and Bayes theorem. 3, which has a goat. Monty Hall, the game show host, examines the other doors (B & C) and opens one with a goat. Since it is always more likely the player has chosen a goat Would this change your guess? The best I can do with my original choice is 1 in 3. You pick a door (call it door A). Thus, the posterior odds become equal to the Bayes factor 1 : 2 : 0. However, as long as the initial probability the car is behind each door is 1/3, it is never to the contestant's disadvantage to switch, as the conditional probability of winning by switching is always at least 1/2.[37]. Information affects your decision that at first glance seems as though it shouldn't. In the problem, you are on a game show, being asked to choose between three doors. Moreover, the host is certainly going to open a (different) door, so opening a door (which door unspecified) does not change this. This equality 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. The Monty Hall problem is one of the most famous problems of probability and comes from the television contest of the 70’s Let’s Make a Deal. The three doors are replaced by a quantum system allowing three alternatives; opening a door and looking behind it is translated as making a particular measurement. However, vos Savant made it clear in her second follow-up column that the intended host's behavior could only be what led to the 2/3 probability she gave as her original answer. As already remarked, most sources in the field of probability, including many introductory probability textbooks, solve the problem by showing the conditional probabilities that the car is behind door 1 and door 2 are 1/3 and 2/3 (not 1/2 and 1/2) 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. As Keith Devlin says,[14] "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 2/3. [2] The problem is mathematically equivalent to the Three Prisoners problem described in Martin Gardner's "Mathematical Games" column in Scientific American in 1959[7] and the Three Shells Problem described in Gardner's book Aha Gotcha.[8]. This code was a bit hard to follow, so I'll go over some issues I found while attempting to find your issue. Vos Savant asks for a decision, not a chance. They report that when the number of options is increased to more than 7 choices (7 doors), people tend to switch more often; however, most contestants still incorrectly judge the probability of success at 50:50. Even if the host opens only a single door ( The general principle is to re-evaluate probabilities as new information is added. But I find it intriguing not for how to solve it, but for how widespread having trouble with it is. One of the prisoners begs the warden to tell him the name of one of the others to be executed, arguing that this reveals no information about his own fate but increases his chances of being pardoned from 1/3 to 1/2. One group has a single door; the other has two doors. Welcome to the most spectacular game show on the planet! If I rigidly stick with my first choice no matter what, I can’t improve my chances. Does switching help? Your decision: Do you want a random door out of 100 (initial guess) or the best door out of 99? The given probabilities depend on specific assumptions about how the host and contestant choose their doors. Other possible behaviors of the host than the one described can reveal different additional information, or none at all, and yield different probabilities. Pick a door, Monty reveals a goat (grey door), and you switch to the other. Many probability text books and articles in the field of probability theory derive the conditional probability solution through a formal application of Bayes' theorem; among them books by Gill[50] and Henze. Only when the decision is completely randomized is the chance 2/3. [44] Behrends concludes 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. {\displaystyle p=1} {\displaystyle {\frac {1}{N}}\cdot {\frac {N-1}{N-p-1}}} On average, in 999,999 times out of 1,000,000, the remaining door will contain the prize. He’s letting us choose between a generic, random choice and a curated, filtered choice. Given that the host opened door 3, the probability that the car is behind door 3 is zero, and it is twice as likely to be behind door 2 than door 1. Your original guess has 1/6 (16%), and the group that had 2 has a 2/6 = 33% of being right. "Mind-reading Monty": The host offers the option to switch in case the guest is determined to stay anyway or in case the guest will switch to a goat. He said he was not surprised at the experts' insistence that the probability was 1 out of 2. Without any evidence, two theories are equally likely. The host must always offer the chance to switch between the originally chosen door and the remaining closed door. He is purposefully not examining your door and trying to get rid of the goats there. There’s a chance the stay-and-hold strategy does decent on a small number of trials (under 20 or so). = Ok bub, let’s play the game: Try playing the game 50 times, using a “pick and hold” strategy. This video was created for University of Waterloo's STAT 333 Winter 2012 Youtube Assignment. In particular, if the car is hidden by means of some randomization device – like tossing symmetric or asymmetric three-sided die – the dominance implies that a strategy maximizing the probability of winning the car will be among three always-switching strategies, namely it will be the strategy that initially picks the least likely door then switches no matter which door to switch is offered by the host. [56] No matter how the car is hidden and no matter which rule the host uses when he has a choice between two goats, if A wins the car then B also does. (If both doors have goats, he picks randomly. Does it matter? Under the standard assumptions, the probability of winning the car after switching is 2/3. He then removes goats until each group has 1 door remaining. If he has a choice, he chooses the leftmost goat with probability, If the host opens the rightmost door, switching wins with probability 1/(1+. [22], Most statements of the problem, notably the one in Parade Magazine, do not match the rules of the actual game show [10] and do not fully specify the host's behavior or that the car's location is randomly selected. Go beyond details and grasp the concept (, “If you can't explain it simply, you don't understand it well enough.” —Einstein The typical behavior of the majority, i.e., not switching, may be explained by phenomena known in the psychological literature as: Experimental evidence confirms that these are plausible explanations that do not depend on probability intuition. Since you seem to have difficulty grasping the basic principle at work here, I'll explain. Then, if the player initially selects door 1, and the host opens door 3, we prove that the conditional probability of winning by switching is: From the Bayes' rule, we know that P(A,B) = P(A|B)P(B) = P(B|A)P(A). 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".[42]. For instance, one contestant's strategy is "choose door 1, then switch to door 2 when offered, and do not switch to door 3 when offered". "They'd think the odds on their door had now gone up to 1 in 2, so they hated to give up the door no matter how much money I offered. Determining the player's best strategy within a given set of other rules the host must follow is the type of problem studied in game theory. By definition, the conditional probability of winning by switching given the contestant initially picks door 1 and the host opens door 3 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". Steve Selvin posed the Monty Hall problem in a pair of letters to the American Statistician in 1975. Whether you change your selection or not, the odds are the same. Therefore, whether or not the car is behind door 1, the chance that the host opens door 3 is 50%. Solutions based on the assertion that the host's actions cannot affect the probability that the car is behind the initially chosen appear persuasive, but the assertion is simply untrue unless each of the host's two choices are equally likely, if he has a choice. Before the host opens a door there is a 1/3 probability the car is behind each door. The confusion as to which formalization is authoritative has led to considerable acrimony, particularly because this variant makes proofs more involved without altering the optimality of the always-switch strategy for the player. 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 1. After the player picks a door, the host opens 999,998 of the remaining doors. "That's the same assumption contestants would make on the show after I showed them there was nothing behind one door," he said. Have a look at that Wikipedia page if you're not familiar with it. In general, the answer to this sort of question depends on the specific assumptions made about the host's behavior, and might range from "ignore the host completely" to "toss a coin and switch if it comes up heads"; see the last row of the table below. I personally read nearly three thousand letters (out of the many additional thousands that arrived) and found nearly every one insisting simply that because two options remained (or an equivalent error), the chances were even. Instead of the regular game, imagine this variant: Do you stick with your original door (1/100), or the other door, which was filtered from 99? After all the filtering, there’s your original door (still with a pale green cloud) and the “Champ Door” glowing nuclear green, containing the probabilities of the 98 doors. The explanation may make sense, but doesn’t explain why the odds “get better” on the other side. The Monty Hall problem is a puzzle about probability and even though is simple to understand, the answer is counterintuitive. Caveat emptor. 1 [9] This "equal probability" assumption is a deeply rooted intuition. A restated version of Selvin's problem appeared in Marilyn vos Savant's Ask Marilyn question-and-answer column of Parade in September 1990. [51] Use of the odds form of Bayes' theorem, often called Bayes' rule, makes such a derivation more transparent.[33][52]. Reading time: ~10 min Reveal all steps. According to Bayes' 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 Bayes 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). The Monty Hall Problem is a famous (or rather infamous) probability puzzle. See more ideas about monty hall, hall, problem. (Try this in the simulator game; use 10 doors instead of 100). The main confusion is that we think we’re like our buddy — we forget (or don’t realize) the impact of Monty’s filtering. "Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car. [2][37][49][34][48][47][35] The solutions in this section consider just those cases in which the player picked door 1 and the host opened door 3. On and on it goes — and the remaining doors get a brighter green cloud. . The key to this solution is the behavior of the host. 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. On those occasions when the host opens Door 3. Look at your win rate. The problem continues to attract the attention of cognitive psychologists. Now, let’s say Pitcher A is a rookie, never been tested, and Pitcher B won the “Most Valuable Player” award the last 10 years in a row. Monty Hall Problem: Solution Explained Simply - Statistics How To How to solve the puzzle in easy steps! N You don’t want to stay static with your initial training set of data. 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." The player picks one of the three cards, then, looking at the remaining two cards the 'host' discards a goat card. They consider a scenario where the host chooses between revealing two goats with a preference expressed as a probability q, having a value between 0 and 1. If the car is behind door 1 the host can open either door 2 or door 3, so the probability the car is behind door 1 AND the host opens door 3 is 1/3 × 1/2 = 1/6. As previous, but now host has option not to open a door at all. 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. Extending this logic to multiple events, for example A, B and C, we get that we can play with the different subsets of {A, B, C} to calculate the probability of the intersection, as a tool to simplify the calculation of our conditional probability: In our case, since we know that P(H3|C2,X1) = 1, we are in luck: Going back to Nalebuff,[54] the Monty Hall problem is also much studied in the literature on game theory and decision theory, and also some popular solutions correspond to this point of view. The Monty Hall problem is a famous, seemingly paradoxical problem in conditional probability and reasoning using Bayes' theorem. The Monty Hall problem is a famous probability puzzle with a counter-intuitive solution. Is it to your advantage to switch your choice? [30][31] Another possibility is that people's intuition simply does not deal with the textbook version of the problem, but with a real game show setting. Jan 23, 2016 - Explore Mohammad Saleh's board "Monty Hall Problem" on Pinterest. The formulation is loosely based on quantum game theory. Look at your percent win rate. At the start, every door has an equal chance — I imagine a pale green cloud, evenly distributed among all the doors. 1 You choose a door. the-monty-hall-problem www.sasimasuk.com ภายใต้การบริหารของบริษัท เอชซีดี อินโนเวชั่น จำกัด (สำนักงานใหญ่) เลขประจำตัวผู้เสียภาษี 0215561004032 [3] Under the standard assumptions, contestants who switch have a .mw-parser-output .sfrac{white-space:nowrap}.mw-parser-output .sfrac.tion,.mw-parser-output .sfrac .tion{display:inline-block;vertical-align:-0.5em;font-size:85%;text-align:center}.mw-parser-output .sfrac .num,.mw-parser-output .sfrac .den{display:block;line-height:1em;margin:0 0.1em}.mw-parser-output .sfrac .den{border-top:1px solid}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance. Said another way, do you want 1 random chance or the best of 99 random chances? 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. Intuitively, the player should ask how likely it is that, given a million doors, he or she managed to pick the right one initially. Filtered is better. The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based 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. Your uninformed friend would still call it a 50-50 situation. [19], The discussion was replayed in other venues (e.g., in Cecil Adams' "The Straight Dope" newspaper column[13]) and reported in major newspapers such as The New York Times.[4]. [63] 1/3 must be the average probability that the car is behind door 1 given the host picked door 2 and given the host picked door 3 because these are the only two possibilities. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 than was available at the beginning of the game when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. They believed the question asked for the chance of the car behind door 2 given the player's initial pick for door 1 and the opened door 3, and they showed this chance was anything between 1/2 and 1 depending on the host's decision process given the choice. Vos 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 correspondents had correctly understood the problem assumptions, and were still initially convinced that vos Savant's answer ("switch") was wrong. Ron Clarke takes you through the puzzle and explains the counter-intuitive answer. Evaluating theories. After the player picks his card, it is already determined whether switching will win the round for the player. 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 1/3) or if the car is behind door 2 (also originally with probability 1/3). Now reset and play it 20 times, using a “pick and switch” approach. Richard Gill[53] analyzes the likelihood for the host to open door 3 as follows. [9] Out of 228 subjects in one study, only 13% chose to switch. Monty Hall, the game show host, examines the other doors (B & C) and opens one with a goat. N clear, insightful math lessons. [18] Numerous examples of letters from readers of Vos Savant's columns are presented and discussed in The Monty Hall Dilemma: A Cognitive Illusion Par Excellence. [54], "The Monty Hall Trap", Phillip Martin's 1989 article in Bridge Today, presented Selvin's problem as an example of what Martin calls the probability trap of treating non-random information as if it were random, and relates this to concepts in the game of bridge.[68]. It’s a bit clearer: Monty is taking a set of 99 choices and improving them by removing 98 goats. Vos Savant's response was that the contestant should switch to the other door. the newsletter for bonus content and the latest updates. [49][48][47] The conditional probability of winning by switching is 1/3/1/3 + 1/6, which is 2/3.[2]. Elementary comparison of contestant's strategies shows that, for every strategy A, there is another strategy B "pick a door then switch no matter what happens" that dominates it. The errors of omission vs. errors of commission effect, What is the probability of winning the car by, What is the probability of winning the car, This page was last edited on 3 May 2021, at 17:43. [10] Some authors, independently or inclusively, assume that the player's initial choice is random as well. Then I ask you to put your finger on a shell. After the host reveals a goat, you now have a one-in-two chance of being correct. "If the host is required to open a door all the time and offer you a switch, then you should take the switch," he said. For example, assume the contestant knows that Monty does not pick the second door randomly among all legal alternatives but instead, when given an opportunity to pick between two losing doors, Monty will open the one on the right. 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. The problem is actually an extrapolation from the game show. The simulation can be repeated several times to simulate multiple rounds of the game. Twelve such deterministic strategies of the contestant exist. To 66%? Another way to understand the solution is to consider the two original unchosen doors together. [20] In her book The Power of Logical Thinking,[21] cognitive psychologist Massimo Piattelli Palmarini [it] 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 in print those who propose the right answer." [37] The fact that these are different can be shown by varying the problem so that these two probabilities have different numeric values. In an attempt to clarify her answer, she proposed a shell game[8] to illustrate: "You look away, and I put a pea under one of three shells. Switching wins the car two-thirds of the time. The Monty Hall Problem has perplexed and angered people for years. The Monty Hall problem is a strange result arising from a very simple situation. I may be autistic or something, but I still don't get it. 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 has opened door 3 (the door on the right) is 1/2. The host knows what lies behind the doors, and (before the player's choice) chooses at random which goat to reveal. Monty could add 50 doors, blow the other ones up, do a voodoo rain dance — it doesn’t matter. Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. Enjoy the article? This problem involves three condemned prisoners, a random one of whom has been secretly chosen to be pardoned. Nalebuff, as later writers in mathematical economics, sees the problem as a simple and amusing exercise in game theory. [46] The assertion therefore needs to be justified; without justification being given, the solution is at best incomplete. Then I simply lift up an empty shell from the remaining other two. Here’s the key points to understanding the Monty Hall puzzle: The fatal flaw in the Monty Hall paradox is not taking Monty’s filtering into account, thinking the chances are the same before and after. My first guess is 1 in 3 — there are 3 random options, right? [25] People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not.[26]. Similarly, strategy A "pick door 1 then switch to door 2 (if offered), but do not switch to door 3 (if offered)" is dominated by strategy B "pick door 3 then always switch". The odds that your choice contains a pea are 1/3, agreed? 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 1 and the host has opened door 3. − Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans. There is disagreement in the literature regarding whether vos Savant's formulation of the problem, as presented in Parade magazine, is asking the first or second question, and whether this difference is significant. You’ll see it settle around 1/3. 1 Vos Savant wrote in her first column on the Monty Hall problem that the player should switch. The information "host opens door 3" contributes a Bayes factor or likelihood ratio of 1 : 1, on whether or not the car is behind door 1. Monty helps us by “filtering” the bad choices on the other side. [32] There, the possibility exists that the show master plays deceitfully by opening other doors only if a door with the car was initially chosen. In words, the information which door is opened by the host (door 2 or door 3?) [66] In this puzzle, there are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each. A version of the problem very similar to the one that appeared three years later in Parade was published in 1987 in the Puzzles section of The Journal of Economic Perspectives. If the host picks randomly q would be 1/2 and switching wins with probability 2/3 regardless of which door the host opens. ), the player is better off switching in every case. Some individuals become very emotional about this problem. In this situation, the following two questions have different answers: The answer to the first question is 2/3, as is correctly shown by the "simple" solutions. [13][14][15][16][17] As Cecil Adams puts it,[13] "Monty is saying in effect: you can keep your one door or you can have the other two doors." (Several readers have left their own explanations in the comments — try them out if the 1/3 stay vs 2/3 switch doesn’t click).