Talk:Monty Hall problem

Attempt

I tried to make this. Some help would be nice. Images are really, really needed. I would do so, but I'm not very good with Wikipedia code. GSGold 23:32, 24 December 2006 (UTC)

Full Revamp

I'm attempting a revamp of the topic. I'm not very familiar with the 'Simple English' so a little help would be great. --Code Adder 13:16, 21 November 2007 (UTC)

Completed revamp. Added picture from en.wikipedia.org with a few changes. I've only left the first paragraph of the first author. I understand that the text is a little hard to understand, but the Monty Hall Problem is not an easy concept to grasp. As far as I can see, most of my words fall under BE 850 and the rest under BE 1500. If you feel you can add anything or change something to make it sound simpler, I'd urge you to do the same. Thankyou.

--Code Adder 13:43, 21 November 2007 (UTC)

Explanation

The given explanation is not correct. It gives the solution to a slightly, but essential other problem. The real problem as stated has the condition that the door that is chosen and the door that is opened and revealing a goat are both known to the player. This excludes possibilities in which the other door is opened. Many people does not see the difference with the problem, in which the chosen door is known, but the presentator explains his plans to the player, and before opening one of the othere doors, asks the player what he intends to do if a door is opened. The presented solution is the right one for the last case, but not for the real problem.

In more formal mathematics: Let X be the door behind which the car is, Y the door chosen by the player and M the door opened by the presentator, then when Y=1 (conditional that door 1 is initially chosen):

$P(X=2|M=3,X\neq 3)={\frac {P(M=3|X=2)P(X=2)}{P(M=3,X\neq 3)}}=$

${\frac {P(M=3|X=2)P(X=2)}{P(M=3,X=1\cup M=3,X=2)}}={\frac {\frac {1}{3}}{{\frac {1}{2}}\times {\frac {1}{3}}+1\times {\frac {1}{3}}}}={\frac {2}{3}}$ Nijdam (talk) 22:30, 7 February 2009 (UTC)

Adding clarity

In my opinion, the mathematical explanation in the prior section is not appropriate for this project. The spirit of this project ought to to apply to the talk as well as the article pages so that we all can follow along and so I ask Nijdam to keep that in mind when posting within this project. I have made several edits to the article in order add clarity and understanding within the guidelines of this project. -Hydnjo (talk) 03:18, 18 February 2009 (UTC)

Proposed new solution

I suggest the following solution be considered in place of the existing solution. This relates to an ongoing controversy at en:talk:Monty Hall problem regarding the problem. My idea is that if the problem can be explained simply here, it can be explained simply there as well. The proposed solution is:

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This problem seems simple but is very hard for most people because it is a conditional probability problem. At first, the chance of the car being behind any of the doors is 1/3. The sum of these chances equals 1. If the player's door is door 1, the chances that the host opens door 2 are equal to the chances that the host opens door 3 - so the sum of the chances after the host opens one of these doors is only 1/2. For example (looking at the picture), the chances of the car being behind each door after the host opens door 3 are 1/6 for door 1, 1/3 for door 2, and 0 for door 3.

                   1/3           +          1/3         +         1/3       = 1
                   /\                       /\                    /\
                  /  \                     /  \                  /  \
                 /    \                   /    \                /    \
host opens:  door 2   door 3          door 2   door 3       door 2   door 3
               /        \               /        \            /        \ 
              /          \             /          \          /          \
             /            \           /            \        /            \
            1/6    +      1/6    +   0       +     1/3  +  1/3     +      0 = 1

The conditional chances, given the host opens door 3, are (1/6) / (1/2) (which equals 1/3) for door 1, (1/3) / (1/2) (which equals 2/3) for door 2, and 0 for door 3. So changing choices increases the chances of getting the car from 1/3 to 2/3.

A different way some people look at the problem is based on the idea that when the host opens a door, the chance that the car is behind the player's choice of door does not change. This is true because the host has to open a door and has to open a door with equal chances if the car is behind the player's door. If the chance that the car is behind the player's door is 1/3, the chance it is behind the only other closed door is 2/3.

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Please comment. -- Rick Block (talk) 04:30, 26 March 2009 (UTC)