User:Antaeus Feldspar/MHP

Monty Hall problem

The conclusions drawn in the article appear to be based on a misunderstanding or misstatement of the factors involved. There's no edit war (yet) but the last two items in the talk section are verging on the acrimonious.

Mathematicians particularly encouraged.

The disputation of the conclusions drawn in the article are quite clearly based on ignoring information stated in the first paragraph of the article, namely, that a party in the problem who is removing incorrect choices the player might otherwise make knows they are incorrect choices -- he is not randomly picking choices to remove which only happen to be incorrect choices. Acrimony is because the only disputant is basing his disputation on non-facts which reading the problem statement would make very clear are non-facts and yet insists he is not trolling. -- Antaeus Feldspar 04:17, 18 Feb 2005 (UTC)

The problem statement is ambiguous. Someone needs to adjust it so that it describes a game in which the host always opens a door that the contestant did not pick. Leaving this constraint out admits the possibility that the two are playing a game in which the host sometimes opens the same door that the contestant picked. For example, he may have chosen the door to open at random among all three doors, or at random between the two doors with goats. These seem to be the games that some reviewers have in mind, when they conclude that the two remaining doors are equally likely to conceal the car. -- Wmarkham

"... after Jane has selected a door but before she actually opens it, the host (who knows what is behind each door) opens one of the other doors ..." How is "one of the other doors" ambiguous? How does it admit the possibility of the host choosing to open the same door that the contestant picked? -- Antaeus Feldspar 23:30, 1 May 2005 (UTC)

It is ambiguous as to whether or not the host will always do this, when the game is played multiple times. Probabilities can only be interpreted in the context of a repeatable process. The question does not clearly describe any particular repeatable process. One way to construct one from the question, as stated, is to suppose that every time the game is played, the host will choose another door, and that that in doing so, he will reveal a goat. However, this is not the only possible game.

The fact that some people interpret the question differently than you do is the very definition of ambiguity. -- Wmarkham 23:37, 3 May 2005 (UTC)

It's completely irrelevant. Even if you choose to read in an implication that was never made that the host is choosing which door to open by chance and that we are concentrating only on those occasions when chance made it a door other than the player's -- it doesn't make any difference!

Let me illustrate with a problem from Lewis Carroll: You take a normal coin, and a double-headed coin, and shake them up in a bag. You reach in and pick one out, and look at one side of that coin, which happens to be heads. What is the chance that the other side is tails?

The key to this problem is that there are two coins you could be drawing out of the bag, and there are two ways you could be holding the coin -- making a total of four different coin faces you could be staring at. However, one of those four coin faces is a tails, and the problem statement states that the coin face you look at after drawing the coin from the bag is heads. Therefore, out of those four outcomes, one of them has been proven by the evidence not to have happened. Out of the three remaining outcomes, only one of them results in tails being on the other side.

A legitimate problem with some attempts to state the Monty Hall problem (not the current description given in our article, however) is that it does not make clear that Monty knows what is behind each door, and opens the door he does because it contains a goat. Someone could legitimately think "Oh, Monty's opening doors at random, but we're discounting those outcomes where he opens the door and the car is behind the door he opens." Since random guessing will pick the car 1/3 of the time, that means that we are discarding one of the three possible outcomes and that the probability of either of the remaining outcomes is 1/2.

But if someone misreads the problem statement and gets the idea that a) Monty is choosing between the two doors which have goats behind them at random, and b) we are eliminating those outcomes where he picks the player's door, just as we eliminated those outcomes where the player was staring at a tails, it doesn't matter. What it brings us back to is "