Content deleted Content added
Line 59:
===Solutions using conditional probability===
<!---===Criticism of the simple solutions=== Any missing stuff to be re-incorporated
As already remarked, most sources in the field of [[probability]], including many introductory probability textbooks, solve the problem by showing the [[conditional probabilities]] the car is behind door 1 and door 2 are 1/3 and 2/3 (not 1/2 and 1/2) given 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 several that explicitly criticize the popularly presented "simple" solutions, saying these solutions are "correct but ... shaky" ([[#refRosenthal2005a|Rosenthal 2005a]]), or do not "address the problem posed" ([[#refGillman1992|Gillman 1992]]), or are "incomplete" ([[#refLucasetal2009|Lucas et al. 2009]]), or are "unconvincing and misleading" ([[#refEisenhauer2001|Eisenhauer 2001]]) or are (most bluntly) "false" ([[#refMorganetal1991|Morgan et al. 1991]]).
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" ([[#refGillman1992|Gillman 1992]], emphasis in the original).
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". However, the probability of winning by ''always'' switching is a logically distinct concept from the probability of winning by switching ''given the player has picked door 1 and the host has opened door 3''. As one source says, "the distinction between [these questions] seems to confound many" ([[#refMorganetal1991|Morgan et al. 1991]]). This 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 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. In this situation the following two questions have different answers:
# What is the probability of winning the car by ''always'' switching?
# What is the probability of winning the car ''given the player has picked door 1 and the host has opened door 3''?
The answer to the first question is 2/3, as is correctly shown by the "simple" solutions. 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. This is because Monty's preference for rightmost doors means 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). For this variation, the two questions yield different answers. 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. ([[#refMorganetal1991|Morgan et al. 1991]])
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 ([[#refRosenhouse2009|Rosenhouse 2009]]). Behrends ([[#refBehrends2008|2008]]) 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. One analysis for one question, another analysis for the other question. Several discussants of the paper by ([[#refMorganetal1991|Morgan et al. 1991]]), whose contributions were published alongside the original paper, strongly criticized the authors for altering vos Savant's wording and misinterpreting her intention ([[#refRosenhouse2009|Rosenhouse 2009]]). One discussant (William Bell) considered it a matter of taste whether or not one explicitly mentions that (under the standard conditions), ''which'' door is opened by the host is independent of whether or not 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 approaches using the concept of odds and Bayes theorem. It is based on the deeply rooted intuition that ''revealing information that is already known does not affect probabilities''. But knowing 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 initially chosen door. 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-randomly 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'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 ([[Monty Hall problem#refFalk1992|Falk 1992:207,213]]). The assertion therefore needs to be justified; without justification being given, the solution is at best incomplete. The answer can be correct but the reasoning used to justify it is defective.
Some of the confusion in the literature undoubtedly arises because the writers are using different concepts of probability, in particular, [[Bayesian probability|Bayesian]] versus [[frequentist probability]]. For the Bayesian, probability represents knowledge. For us and for the player, the car is initially equally likely to be behind each of the three doors because we know absolutely nothing about how the organizers of the show decided where to place it. For us and for the player, the host is equally likely to make either choice (when he has one) because we know absolutely nothing about how he makes his choice. These "equally likely" probability assignments are determined by symmetries in the problem. The same symmetry can be used to argue in advance that specific door numbers are irrelevant, as we saw [[Monty Hall problem#From simple to conditional by symmetry|above]].-->
The [[Monty Hall Problem#Simple Solutions|simple solutions]] above show that a player with a strategy of switching wins the car with overall probability 2/3 ([[#refGrinsteadandSnell2006|Grinstead and Snell 2006:137–138]] [[#refCarlton2005|Carlton 2005]]). In contrast most sources in the field of [[probability]] calculate the [[conditional probabilities]] that the car is behind door 1 and door 2 are 1/3 and 2/3 given the contestant initially picks door 1 and the host opens door 3 ({{Harvtxt|Selvin|1975b}}, [[#refMorganetal1991|Morgan et al. 1991]], [[#refChun1991|Chun 1991]], [[#refGillman1992|Gillman 1992]], [[#refCarlton2005|Carlton 2005]], [[#refGrinsteadandSnell2006|Grinstead and Snell 2006:137–138]], [[#refLucasetal2009|Lucas et al. 2009]]). The solutions in this section consider just those cases in which the player picked door 1 and the host opened door 3.
<!--====Refining the simple solution====-->
| |||