Talk:Monty Hall problem/draft2: Difference between revisions

Stibel et al. ([[#refStibeletal2008|2008]]) propose working memory demand is taxed during the Monty Hall problem and that this forces people to "collapse" their choices into two equally probable options. They report that when increasing the number of options to over 7 choices (7 doors) people tend to switch more often; however most still incorrectly judge the probability of success at 50/50.

 

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 ([[#refMueserandGranberg1999|Mueser and Granberg, 1999]]). Out of 228 subjects in one study, only 13% chose to switch ([[#refGranbergandBrown1995|Granberg and Brown, 1995:713]]). In her book ''The Power of Logical Thinking'', vos Savant ([[#refvosSavant1996|1996:15]]) quotes cognitive psychologist [[Massimo Piattelli-Palmarini]] as saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that 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." Interestingly, pigeons make mistakes and learn from mistakes, and experiments, [[#refHerbransonandSchroeder2010|Herbranson and Schroeder, 2010]], show that they rapidly learn to always switch, unlike humans.

 

In addition to the "equal probability" intuition, a competing and deeply rooted intuition is that ''revealing information that is already known does not affect probabilities''. Although this is a true statement, it is not true that just knowing the host can open one of the two unchosen doors to show a goat necessarily means that opening a specific door cannot affect the probability that the car is behind the initially-chosen door. If the car is initially placed behind the doors with equal probability and 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 door are very persuasive, but lead to the correct answer only if the problem is completely symmetrical with respect to both the initial car placement and how the host chooses between two goats ([[Monty Hall problem#refFalk1992|Falk 1992:207,213]]).

 

Another approach showing switching wins with probability 2/3 is to determine the [[conditional probability]] the car is behind Door 2 given that the player has initially picked Door 1 and the host has opened Door 3 ({{Harvnb|Selvin|1975b}}; [[#refMorganetal1991|Morgan et al. 1991]]; [[#refGrinsteadandSnell2006|Grinstead and Snell 2006:137]]). Referring to the [[decision tree]] as shown to the right ([[#refChun1991|Chun 1991]]) or the equivalent figure below, and considering only cases where the host opens Door 3 after the player picks Door 1, switching loses in a case with probability 1/6 but wins in a case with probability 1/3. The conditional probability the car is behind Door 2 is therefore 2/3 = 1/3 / (1/6 + 1/3), while the conditional probability the car is behind Door 1 is only 1/3 = 1/6 / (1/6 + 1/3).