Diskussion:Monty Hall-problemet

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	Artiklen Monty Hall-problemet har været vist på Wikipedias forside som ugens artikel i uge 49, 2007.
Milepæle Tidsmarkering Milepæl Resultat 8. april 2008 Nomineret til lovende artikel Vedtaget 5. oktober 2007 Nomineret til ugens artikel Vedtaget

For mange eksterne henvisninger

Så er der igen tilføjet en simulering af problemet, og det bringer antallet af eksterne henvisninger op på 15, hvoraf omkring halvdelen består af eller indbefatter simuleringsprogrammmer. Det er efter min mening ganske enkelt for meget! Artiklen er i forvejen forsynet med rigtig mange fodnoter og (skriftlige) kilder, så derfor må man kunne skære ned i link-antallet uden at svække artiklens troværdighed og/eller dens informationsværdi. En til to simulatorer burde være tilstrækkeligt. --Arne (Amjaabc) 25. jan 2008, 16:14 (CET)

Explanation

I'm sorry I cannot write in your language, but I do enough understand it as to comprehend the article. As in most other languages 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):

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

$\frac{P(M=3|X=2)P(X=2)}{P(M=3,X=1 \cup M=3,X=2)} = \frac{\frac 13}{\frac 12 \times \frac 13 + 1\times \frac 13}=\frac 23$

Nijdam 7. feb 2009, 23:14 (CET)

As far as I can see the difference between the problem as it is stated here in Danish and your formulation is that the contestant makes his decision whether or not to change doors after the presenter has opened a door behind which a goat is found. How does this change the probability of the choice when the contestant always knows that the door opened by the presenter is a goat-door? --NiceGuy152 15. aug 2009, 12:30 (CEST)

Well, probabilities don't change. If an event has occurred, the probabilities of interest are the conditional ones given the occurred event. Nijdam 8. dec 2009, 22:16 (CET)

I'm unsure as to what needs to be corrected then. Should the problem be formulated such that the player knows both his chosen door and which door the host reveals to contain a goat, before making the choice as to which door he finally wants.--NiceGuy152 11. dec 2009, 15:19 (CET)