User:Martin Hogbin/The full Monty

This page includes original research. This means that anything discovered or proposed on this page that is not supported by a reliable sources does not have a right to be included in any WP article. On the other hand, it does not mean that things that are agreed by other editors to be correct and that are not contradicted by any reliable source cannot possible go in an article if they are considered routine calculations. WP policy says that 'all material challenged or likely to be challenged' is not suitable for inclusion in articles.

It is not my intention to try to force anything from this page into a WP article without full consensus or publication.

The aim of this page is not to directly create material for the Monty hall problem but to help editors to clarify their thoughts on the subject and, in particular, to separate the non-contentious mathematics, from the highly contested philosophy. I do, however, assert that any solution not based on the calculation shown below cannot be considered totally complete. =A complete solution to the Monty Hall problem based on probability theory=

The standard rules of the game are assumed throughout. These are that the host always offers the swap and always opens an unchosen door to reveal a goat.

This is a complete solution to the Monty Hall problem based on discreet distributions within probability theory. It comes from my analysis page, where its derivation can be found.

Pws is the probability of winning by switching. The stochastic variables C, X and H describe the problem: C = number of door with car, X = number of door of player's initial choice and H = number of door opened by host.

Pws =

( P(C=1) P(X=2) + P(C=1) P(X=3) + P(C=2) P(X=1) + P(C=2) P(X=3) + P(C=3) P(X=1) + P(C=3) P(X=2) )

/

( P(C=1) P(X=1) P(H=2|C=1,X=1) + P(C=1) P(X=1) P(H=3|C=1,X=1) + P(C=1) P(X=2) + P(C=1) P(X=3) +

+ P(C=2) P(X=1) + P(C=2) P(X=2) P(H=1|C=2,X=2) + P(C=2) P(X=2) P(H=3|C=2,X=2) + P(C=2) P(X=3) +

+ P(C=3) P(X=1) + P(C=3) P(X=2) + P(C=3) P(X=3) P(H=1|C=3,X=3) + P(C=3) P(X=3) P(H=2|C=3,X=3) )

This is a general solution to the MHP. Specific solutions may be obtained by applying distributions and conditions to the above. Note that in the general case the probability of winning by switching, with no assumed distributions and no conditions applied, is indeterminate and can be anything from 0 to 1. For example:

If the car is always behind door 1 and the player always picks door 1 then Pws = 0 If the car is always behind door 1 and the player always picks door 2 then Pws = 1

I believe that the above calculation (which may well be routine to some) is mathematically correct, others are welcome to check, and therefore essentially non-contentious.

Distributions
Before any useful solution can be found to the problem it is necessary to decide on three distributions. These are the initial car placement, the players initial door choice, and the host's choice of door to open to reveal a goat. This distributions respectively determine the values of terms like: P(C=c)P(X=x)P(X=3) P(H=h|C=c,X=3) representing the probability with thich the car is placed behind each door, the probability with which the player chooses each door and the probabilities with which the host opens the available doors given the car position and the player's initial choice of door.

Distributions may be determined either because they are expressly stated in the problem statement (as in the cars are initially placed uniformly at random) or by applying some philosophically-based rational such as the principle of indifference (as by saying that the player has no knowledge of the car's initial placement or the hosts policy regarding opening a goat-hiding door therefore we should take these distributions as uniform). These are the things that may be argued about and which ultimately determine the exact question that is being answered by any given solution.

Conditions
Conditions are the defined occurance or specification of events in addition to the game rules, for example that the player has chosen (or even will choose) door 1 and the host has opened (or will open) door 3. In some problem statements it is clear that a condition is specified but in others a degree of interpretation may be required, as in for example 'the host opens another door, say door 3'. Does this specify that the host has opened any other door or is it a condition that the host has opened door 3. Martin Hogbin (talk) 14:42, 24 July 2010 (UTC)

Examples
Here are some interesting and informative examples. For details of the calculations see the analysis page.

The player initially chooses a door uniformly at random - unconditional
This means that P(X=1) = P(X=2) = P(X=3) = 1/3 thus

Pws = ( P(C=1) + P(C=1) + P(C=2) + P(C=2) + P(C=3) + P(C=3) ) / ( P(C=1) + P(C=1) + P(C=1) + P(C=2) + P(C=2) + P(C=2) + P(C=3) + P(C=3) + P(C=3) )= 2/3

In other words, just taking it that the player initially chooses a door uniformly at random is sufficient for the answer (Pws) to be 2/3, regardless of how the car was initially placed or what the host's door opening policy is. This is a point that I have made before and that Gill has also made several times.