monty hall problem history

1 Morgan et al[38] and Gillman[35] 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 paradox (apparent, but not actual, contradiction) holds because it is equally true that initially door 1 had chance 1/3 to hide the car, while after the player has chosen door 1 and the host has opened door 3, door 1 still has chance 1/3 to hide the car. In the latter case you keep the prize if it's behind either door. N First, why aren’t the odds 50-50 after the host opens the door? If you picked the door with the car, the two other doors are guaranteed to have a goat. From this point of view, one has to remember that the player has two opportunities to make choices: first of all, which door to choose initially; and secondly, whether or not to switch. A common variant of the problem, assumed by several academic authors as the canonical problem, does not make the simplifying assumption that the host must uniformly choose the door to open, but instead that he uses some other strategy. Dominance is a strong reason to seek for a solution among always-switching strategies, under fairly general assumptions on the environment in which the contestant is making decisions. A stayer only wins one third of the time. P(H∣E)=P(E∣H)×P(H)P(E∣H)×P(H)+P(E∣notH)×P(notH)=1×131×13+12×23=1323=12. That’s the Bayesian method: whenever the evidence is “more consistent” with world A than with world B (meaning, the evidence would be more likely if A is true than if B is true), you revise your probabilities to believe A is more likely than you thought it was before. You pick a door, and before the host opens it, he opens a door that you did not pick, which has nothing behind it. 1 For this variation, the two questions yield different answers. By the rules of conditional probability: = Pr(C3∩M2) / [Pr(M2∩C1) + Pr(M2∩C2) + Pr(M2∩C3)], = Pr(C3∩M2) / [Pr(M2|C1) Pr(C1) + Pr(M2|C2) Pr(C2) + Pr(M2|C3) Pr (C3)], = Pr(C3) / [p Pr(C1) + 0 Pr(C2) + 1 Pr(C3)], So the chance that you’d win by switching is 1/(p+1). Click on the door that you think the car is behind. In order to explain why the numbers are suggesting that it is better Following Gill,[56] a strategy of contestant involves two actions: the initial choice of a door and the decision to switch (or to stick) which may depend on both the door initially chosen and the door to which the host offers switching. A contestant is given the choice of three doors: behind one door is a car and behind the other two doors are goats. Monty Hall decides to perform his final episode of the Let's Make a Deal series with a little twist, and Calvin is elated when he is the first contestant to be called to the stage. "Anything else is a different question. [38], Sasha Volokh (2015) wrote that "any explanation that says something like 'the probability of door 1 was 1/3, 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. After selecting a door, Monty would then proceed to open one of the doors you didn't select. [25], 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.

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