Moreover, the host is certainly going to open a (different) door, so opening a door (which door unspecified) does not change this. This means even without constraining the host to pick randomly if the player initially selects the car, the player is never worse off switching. "Angelic Monty": The host offers the option to switch only when the player has chosen incorrectly. The best I can do with my original choice is 1 in 3. Go beyond details and grasp the concept (, “If you can't explain it simply, you don't understand it well enough.” —Einstein However, if the show host has not randomized the position of the prize in a fully quantum mechanical way, the player can do even better, and can sometimes even win the prize with certainty.[64][65]. Steve Selvin posed the Monty Hall problem in a pair of letters to the American Statistician in 1975. I'm trying to simulate the Monty Hall Problem where someone chooses a door, and a random one is removed--in the end it must be one with a car and one without, one of which someone must have chosen. You can read more about this problem, and the controversy, on Marilyn Vos Savant's website www.marilynvossavant.com 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. 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. Switching wins the car two-thirds of the time. 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. The rules can be stated in this language, and once again the choice for the player is to stick with the initial choice, or change to another "orthogonal" option. N Morgan et al[37] and Gillman[34] both show a more general solution where the car is (uniformly) randomly placed but the host is not constrained to pick uniformly randomly if the player has initially selected the car, which is how they both interpret the statement of the problem in Parade despite the author's disclaimers. The name refers to its famous presenter, and his well-known 3-door game . ", 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".[38]. 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 considerable number of other generalizations have also been studied. The host must always open a door that was not picked by the contestant. 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. No. The earliest of several probability puzzles related to the Monty Hall problem is Bertrand's box paradox, posed by Joseph Bertrand in 1889 in his Calcul des probabilités. − 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. You’re hoping for the car of course. In Morgan et al.,[37] four university professors published an article in The American Statistician claiming that vos Savant gave the correct advice but the wrong argument. The warden obliges, (secretly) flipping a coin to decide which name to provide if the prisoner who is asking is the one being pardoned. 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. The simple solutions show in various ways that a contestant who is determined to switch will win the car with probability 2/3, and hence that switching is the winning strategy, if the player has to choose in advance between "always switching", and "always staying". When you select the user's door, you need to determine what the other_door is then. ), the player is better off switching in every case. He closes the door and mixes all the prizes, including your door. Later in their response to Hogbin and Nijdam,[43] they did agree that it was natural to suppose that the host chooses 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 his choice) has the same value, 2/3, as the unconditional probability of winning by switching (i.e., averaged over all possible situations). The more you test the old standard, the less likely the new choice beats it. [57][58] Three cards from an ordinary deck are used to represent the three doors; one 'special' card represents the door with the car and two other cards represent the goat doors. In particular, vos Savant defended herself vigorously. The problem is actually an extrapolation from the game show. The Infamous Monty Hall Problem If you have taken a statistics course at an upper level or seen the movie “21”, then you have probably heard of the Monty Hall problem. Look at your percent win rate. 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. The latter strategy turns out to double the chances, just as in the classical case. 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. For example: A Bayesian Filter improves as it gets more information about whether messages are spam or not. There are 3 doors, behind which are two goats and a car. The Monty Hall Problem has perplexed and angered people for years. 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. The key to this solution is the behavior of the host. There are 3 doors, behind which are two goats and a car. However there are a number of different sites explaining the solution to the Monty Hall problem. Here’s the key: Monty does not try to improve your door! Happy math. Let’s think about other scenarios to cement our understanding: Suppose your friend walks into the game after you’ve picked a door and Monty has revealed a goat — but he doesn’t know the reasoning that Monty used. There's plenty more to help you build a lasting, intuitive understanding of math. Monty could add 50 doors, blow the other ones up, do a voodoo rain dance — it doesn’t matter. However neither source suggests the player knows what the value of q is so the player cannot attribute a probability other than the 2/3 that vos Savant assumed was implicit. The version of the Monty Hall problem published in Parade in 1990 did not specifically state that the host would always open another door, or always offer a choice to switch, or even never open the door revealing the car. Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. Is it above 50% Is it closer to 60%? Your decision: Do you want a random door out of 100 (initial guess) or the best door out of 99? Monty gives you 6 doors: you pick 1, and he divides the 5 others into a group of 2 and 3. [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. After the player picks his card, it is already determined whether switching will win the round for the player. The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975. So what should you do? Is it to your advantage to switch your choice? Take a look Wikipedia's page for the Monty Hall Problem Online simulation of the Monty Hall Problem 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. You’re hoping for the car of course. The problem is a paradox of the veridical type, because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true. [3] She received thousands of letters from her readers – the vast majority of which, including many from readers with PhDs, disagreed with her answer. Said another way, do you want 1 random chance or the best of 99 random chances? You'd rather have a two-in-three shot at the prize than one-in-three, wouldn't you? 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. 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. 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. Caveat emptor. If the card remaining in the host's hand is the car card, this is recorded as a switching win; if the host is holding a goat card, the round is recorded as a staying win. This is what happens with the 100 door game. The switch in this case clearly gives the player a 2/3 probability of choosing the car. No, he is only “pulling the weeds” out of the neighbor’s lawn, not yours. The main confusion is that we think we’re like our buddy — we forget (or don’t realize) the impact of Monty’s filtering. [44] One discussant (William Bell) considered it a matter of taste whether one explicitly mentions that (under the standard conditions), which door is opened by the host is independent of whether one should want to switch. After a reader wrote in to correct the mathematics of Adams's analysis, Adams agreed that mathematically he had been wrong. 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? You pick the name that sounds cooler, and 50-50 is the best you can do. [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. Shame! This video was created for University of Waterloo's STAT 333 Winter 2012 Youtube Assignment. But the goal isn’t to understand this puzzle — it’s to realize how subsequent actions & information challenge previous decisions. [20][4][23] Krauss and Wang conjecture that people make the standard assumptions even if they are not explicitly stated. The solution to the Monty Hall problem is not intuitive. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'". 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. The given probabilities depend on specific assumptions about how the host and contestant choose their doors. In November 1990, an equally contentious discussion of vos Savant's article took place in Cecil Adams's column "The Straight Dope". As Monty starts removing the bad candidates (in the 99 you didn’t pick), he “pushes” the cloud away from the bad doors to the good ones on that side. Rephrased, the most glaring problem is that there is actually only one chance, 100%, that the Car door is the Car Door. The analysis also shows that the overall success rate of 2/3, achieved by always switching, cannot be improved, and underlines what already may well have 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. Your uninformed friend would still call it a 50-50 situation. 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. But, these two probabilities are the same. Vos Savant suggests that the solution will be more intuitive with 1,000,000 doors rather than 3. Filtered is better. (If both doors have goats, he picks randomly. Enjoy the article? These are general cases, but the message is clear: more information means you re-evaluate your choices. One problem, however, caused by far the most dissent: The Monty Hall Problem, a problem inspired by the famous game show Let's Make A Deal. You're a contestant on a game show--and you're given 3 doors to choose from.Behind one door is a shiny new sports car--behind the The Monty Hall Problem Explained in Excel Monty Hall, the game show host, examines the other doors (B & C) and opens one with a goat. Have a look at that Wikipedia page if you're not familiar with it. The controversy began in 1990 when Marilyn vos Savant posed the question in her column. The simulation can be repeated several times to simulate multiple rounds of the game. 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. As one source says, "the distinction between [these questions] seems to confound many". Vos Savant's response was that the contestant should switch to the other door. [69] As a result of the publicity the problem earned the alternative name Marilyn and the Goats. A restated version of Selvin's problem appeared in Marilyn vos Savant's Ask Marilyn question-and-answer column of Parade in September 1990. If you switch doors you’ll win 2/3 of the time! In the show, the contestants are shown three doors and told behind one of the doors is a brand new car. Here's the new challenge, late as usual! For this variation, the two questions yield different answers. The Monty Hall Problem is a famous (or rather infamous) probability puzzle. "[59] The answer follows if the car is placed randomly behind any door, the host must open a door revealing a goat regardless of the player's initial choice and, if two doors are available, chooses which one to open randomly. 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. Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans. (Try this in the simulator game; use 10 doors instead of 100). 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. "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. The Parade column and its response received considerable attention in the press, including a front-page story in the New York Times in which Monty Hall himself was interviewed. I remember this problem from watching an episode of numbers. Ron Clarke takes you through the puzzle and explains the counter-intuitive answer. If the player picks door 1 and the host's preference for door 3 is q, then the probability the host opens door 3 and the car is behind door 2 is 1/3 while the probability the host opens door 3 and the car is behind door 1 is q/3. Assuming that “two choices means 50-50 chances” is our biggest hurdle. 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. Monty reveals the goat, and then has a seizure. [13] Adams initially answered, incorrectly, that the chances for the two remaining doors must each be one in two. 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. 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. You’re probably muttering that two doors mean it’s a 50-50 chance. The host opens a door and makes the offer to switch 100% of the time if the contestant initially picked the car, and 50% the time otherwise. "Monty from Hell": The host offers the option to switch only when the player's initial choice is the winning door. In general, more information means you re-evaluate your choices. to begin with, and the host will always give away the location of the other one, there is a 66% chance that switching was within the player's best interest all along (as opposed to the less likely 33% chance the player had chosen the car). 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). Vos Savant wrote in her first column on the Monty Hall problem that the player should switch. It is based on the deeply rooted intuition that revealing information that is already known does not affect probabilities. Monty Hall Problem --a free graphical game and simulation to understand this probability problem. Numerous readers, however, wrote in to claim that Adams had been "right the first time" and that the correct chances were one in two. On and on it goes — and the remaining doors get a brighter green cloud.
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