ny additional thousands that arrived) and found nearly every one insisting simply that because two options remained (or an equivalent error), the chances were even. Very few raised questions about ambiguity, and the letters actually published in the column were not among those few." (vos Savant, 1996) 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 (Mueser and Granberg, 1999). The table below shows a variety of OTHER possible host behaviors and the impact on the success of switching.

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. For example, if the host is not required to make the offer to switch the player may suspect the host is malicious and makes the offers more often if the player has initially selected the car. In general, the answer to this sort of question depends on the specific assumptions made about the host's behaviour, 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.

Morgan et al. (1991) and Gillman (1992) both show a more general solution where the car is randomly placed but the host is not constrained to pick randomly if the player has initially selected the car, which is how they both interpret the well known statement of the problem in "Parade" despite the author's disclaimers. Both changed the wording of the "Parade" version to emphasize that point when they restated the 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. If the host picks randomly "q" would be 1/2 and switching wins with probability 2/3 regardless of which door the host opens. If the player picks Door 1 and the host's preference for Door 3 is "q", then in the case where the host opens Door 3 switching wins with probability 1/3 if the car is behind Door 2 and loses with probability (1/3)"q" if the car is behind Door 1. The conditional probability of winning by switching "given the host opens Door 3" is therefore (1/3)/(1/3 + (1/3)"q") which simplifies to 1/(1+"q"). Since "q" can vary between 0 and 1 this conditional probability can vary between 1/2 and 1. This means even without constraining the host to pick randomly if the player initially selects the car, the player is never worse off switching. However, it is important to note that 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.

D. L. Ferguson (Selvin 1975b'deki alıntıya göre 1975'te Selvin'e yazdığı bir mektubunda) orijinal oyunun bir "N" sayıda kapı şeklinde genelleştirilmişini önermektedir. Bu durumda sunucu "p" sayıda kaybeden kapıyı açar ve oyuncuya değiştirme fırsatını verir; bu şekilde değiştirmeyle kazanma olasılığı 
("N"−1)/["N"("N"−"p"−1)] olur. Sunucu sadece bir kapıyı açmışsa bile oyuncunun değiştirmesi yine de yararına olur ancak "N" sayısı büyüdükçe avantaj da sıfıra yaklaşır (Granberg 1996:188). Diğer uç durumda sunucu bir kaybeden kapı dışında diğerlerinin tamamını açarsa, değiştirmeyle kazanma şansı 1'e yaklaşır.

Bapeswara Rao ve Rao (1992) farklı bir "N" sayıda kapı versiyonu önermektedir. Bu versiyonda sunucu oyuncunun seçiminden faklı bir kaybeden kapı açmaktadır. Sunucu, geriye yalnızca iki kapı kalana kadar her kapı açışında oyuncuya tercihini değiştirmesi için bir fırsat verir. Dört kapı ile yapılabilecek en uygun strateji önce kapıyı seçmek ardından sadece iki kapı kalınca tercihi değiştirmek olacaktır. "N" sayıda kapı ile bu stratejinin kazanma olasılığı ("N"−1)/"N"dir ve bunun en uygunu olduğu iddia edilmektedir.

A quantum version of the paradox illustrates some points about the relation between classical or non-quantum information and quantum information, as encoded in the states of quantum mechanical systems. The formulation is loosely based on Quantum game theory. 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. 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. The latter strategy turns out to double the chances, just as in the classical case. 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 (Flitney and Abbott 2002, D'Ariano et al. 2002).

Monty Hall problemine benzer birkaç olasılık bulmacasından en eskisi Bertrand'ın kutu paradoksudur. Bu 1889'da Joseph Bertrand tarafından "Calcul des probabilités" adlı kitabında ortaya atıldı (Barbeau 1993). Bu bulmacada üç kutu bulunur: bir kutu iki altın madeni para, bir kutu iki gümüş madeni para ve bir kutu da her birinden birer tane madeni para içerir. Bir kutunun rastgele seçilmesinin ve bir madeni paranın rastgele çekilmesinin ardından, çekilen paranın altın olması durumunda geriye kalan paranın altın olma olasılığı nedir. Monty Hall probleminde olduğu gibi burada da sezgisel yanıt 1/2'dir, halbuki olasılık 2/3'tür.

Martin Gardner'ın "Scientific American"daki "Mathematical Games" köşesinde 1959'da yayınlanan üç mahkûm problemi (1959a, 1959b), Monty Hall probleminin izomorfudur. Problem, rastgele içlerinden biri gizlice seçilip affedilecek olan üç mahkûmu içerir. Mahkûmlardan biri, diğer mahkûmlardan idam edilecek olanın adını söylemesi için gardiyana yalvarır. Bu mahkûmun kendi kaderi hakkında hiçbir bilgiyi açığa çıkarmasa da kendi affedilme şansını 1/3'ten 1/2'ye çıkarır. The warden obliges, (secretly) flipping a coin to decide which name to provide if the prisoner who is asking is the one being pardoned. The question is whether knowing the warden's answer changes the prisoner's chances of being pardoned. This problem is equivalent to the Monty Hall problem; the prisoner asking the question still has a 1/3 chance of being pardoned but his unnamed cohort has a 2/3 chance.

Steve Selvin 1975'te "American Statistician" dergisine gönderdiği bir çift mektup ile Monty Hall problemini ortaya attı (1975a, 1975b). İlk mektup, 15 yıl sonra "Parade" dergisinde yayınlanan gösterime yakın bir problem içeriyordu. İkici mektup "Monty Hall problemi" teriminin ilk kullanımı gibi görünmektedir. Problem aslında yarışma programının bir ekstrapolasyonudur. Monty Hall heyecan yaratmak için yanlış bir kapıyı açar ve yarışmacıya, kapı değiştirme fırsatı yerine 100 $ gibi daha küçük bir ödül teklif eder. Monty Hall'ün Selvin'e yazdığı bir mektuba göre:
And if you ever get on my show, the rules hold fast for you—no trading boxes after the selection. (Hall 1975)

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" (Nalebuff 1987).

Phillip Martin's article in a 1989 issue of "Bridge Today" magazine titled "The Monty Hall Trap" (Martin 1989) 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.

A restated version of Selvin's problem appeared in Marilyn vos Savant's "Ask Marilyn" question-and-answer column of "Parade" in September 1990 (vos Savant 1990). Though vos Savant gave the correct answer that switching would win two-thirds of the time, she estimates the magazine received 10,000 letters including close to 1,000 signed by PhDs, many on letterheads of mathematics and science departments, declaring that her solution was wrong (Tierney 1991). Due to the overwhelming response, "Parade" published an unprecedented four columns on the problem (vos Savant 1996:xv). As a result of the publicity the problem earned the alternative name Marilyn and the Goats.

In November 1990, an equally contentious discussion of vos Savant's article took place in Cecil Adams's column "The Straight Dope" (Adams 1990). Adams initially answered, incorrectly, that the chances for the two remaining doors must each be one in two. After a reader wrote in to correct the mathematics of Adams' analysis, Adams agreed that mathematically, he had been wrong, but said that the "Parade" version left critical constraints unstated, and without those constraints, the chances of winning by switching were not necessarily 2/3. Numerous readers, however, wrote in to claim that Adams had been "right the first time" and that the correct chances were one in two.

The "Parade" column and its response received considerable attention in the press, including a front page story in the "New York Times" (Tierney 1991) in which Monty Hall himself was interviewed. He appeared to understand the problem, giving the reporter a demonstration with car keys and explaining how actual game play on "Let's Make a Deal" differed from the rules of the puzzle.

Over 40 papers have been published about this problem in academic journals and the popular press (Mueser and Granberg 1999). Barbeau 2000 contains a survey of the academic literature pertaining to the Monty Hall problem and other closely related problems.

The problem continues to resurface outside of academia. The syndicated NPR program "Car Talk" featured it as one of their weekly "Puzzlers," and the answer they featured was quite clearly explained as the correct one (Magliozzi and Magliozzi, 1998). An account of the Hungarian mathematician Paul Erdős's first encounter of the problem can be found in "The Man Who Loved Only Numbers"—like many others, he initially got it wrong. The problem is discussed, from the perspective of a