the best door out of a set of random doors will always have better odds.This solution, given in Parade Magazine, shows all of the possible results of staying or switching.Although the problem was made famous in the Ask Marilyn column in 1990, the earliest mention of the problem was in a letter Steve Selvin wrote to the American Statistician.


Viewed 2 times 0 $\begingroup$ Let us take 3 doors A, B, C. Now, let us say car is in A. But with all of those Ph.D.s being wrong, don’t feel bad if you’re still stumped.On the other hand, you can find comfort in the fact that pigeons may be smarter than mathematicians: they perform better when it comes to the Monty Hall Dilemma. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 than was available at the beginning of the game when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally.

Google “Monty Hall Problem” and you’ll get several hundred thousand pages. A simple way to demonstrate that a switching strategy really does win two out of three times with the standard assumptions is to simulate the game with The simulation can be repeated several times to simulate multiple rounds of the game.

They include letters from the Deputy Director of the Center for Defense Information and a Research Mathematical Statistician from the National Institutes of Health.

(Vos Savant wrote in her first column on the Monty Hall problem that the player should switch (The discussion was replayed in other venues (e.g., in In an attempt to clarify her answer, she proposed a shell game (Vos Savant commented that, though some confusion was caused by 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 (Most statements of the problem, notably the one in 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 (Some say that these solutions answer a slightly different question – one phrasing is "you have to announce The simple solutions show in various ways that a contestant who is determined to switch will win the car with probability Four university professors published an article (Morgan et al., 1991) in There is disagreement in the literature regarding whether vos Savant's formulation of the problem, as presented in 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. Monty goes wild. (Under the standard assumptions, the probability of winning the car after switching is When any of these assumptions is varied, it can change the probability of winning by switching doors as detailed in the A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. When you pick one of the three doors, you truly have a 0.33 probability of picking the correct door.

Many probability text books and articles in the field of probability theory derive the conditional probability solution through a formal application of Initially, the car is equally likely to be behind any of the three doors: the odds on door 1, door 2, and door 3 are Then, if the player initially selects door 1, and the host opens door 3, we prove that the conditional probability of winning by switching is: 2009). This time however conditioning on red being preferred to green reduced the original probability of 1/2 to 1/3, whereas in the Monty Hall problem the probability was initially 1/3 and did not change.

This means that the probability of the car being behind door 3 is 1 – (1/3) = 2/3. Since you seem to have difficulty grasping the basic principle at work here, I'll explain. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not.
You pick a door, say No.

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