User talk:Martin Hogbin/Monty Hall analysis

I like to help you with this page of yours. Is it okay if I delete irrelevant parts? Nijdam (talk) 14:31, 31 December 2009 (UTC)[reply]

Thanks for your help, it is most welcome. What do you mean by irrelevant parts? Martin Hogbin (talk) 15:15, 1 January 2010 (UTC)[reply]

The parts you already deleted. Nijdam (talk) 19:13, 1 January 2010 (UTC)[reply]

When to add conditions

The purpose of this page is to produce as general a solution to the standard game rules version of this problem in order to separate questions about what exactly the real question is from the mathematics. For that reason I do not want to add any assumptions that are not part of the standard game rules until a solution has been worked out. Martin Hogbin (talk) 09:50, 2 August 2010 (UTC)[reply]

Combining doors

I hoped you would quote the appropriate sources, saying:

The car is with probability 2/3 behind the not chosen doors. As it is not behind the opened door, the probability it is behind the remaining door must be 2/3.

Do you agree this is what sources mention to be the combined doors solution? Nijdam (talk) 10:45, 15 August 2010 (UTC)[reply]

Yes. Martin Hogbin (talk) 10:53, 15 August 2010 (UTC)[reply]

Then, please, put this in formulas. Nijdam (talk) 08:17, 16 August 2010 (UTC)[reply]

P(C=1) = 1/3

Okay, if the car is placed randomly (uniform).

Nijdam, we are talking about the standard formulation of the problem where the producer, player, and host, all choose uniformly from their legal options.

Nowhere is mentioned and noway it is needed the player choses uniformly.

P(C=1|H=2 or H=3) = P(C=1) = 1/3 By symmetry

Where does the combined doors solution mention symmetry?

It is obvious, natural, and intuitive. Martin Hogbin (talk) 17:04, 16 August 2010 (UTC)[reply]

So the answer is: nowhere does this solution mention the symmetry! It simply does not use it!

P(C<>1| H=2 or H=3) = 2/3.

Your insistence that every problem must be explained using a method of your choice is not justified. The disagreements here result from ambiguities and disagreements in how to formulate the problem statement precisely.

I do not understand this remark. You yourself admit the existence of the combining door solution. You agreed on the formulation. Why do you object to formalize it?Nijdam (talk) 10:52, 16 August 2010 (UTC)[reply]

See below.

The combining doors solution is best explained using simple verbal logic.

1. The probability that the car is behind door 1 is 1/3

Meaning: P(C=1)=1/3

2. When the host opens either door 2 or door 3 to reveal a goat this does not affect that probability.

Of course not, it is and will be forever: P(C=1)=1/3

This why I do not like to formalise the problem the way that you wish. I mean the probability that the car is behind door 1 after the host has opened on of the two other doors to reveal a goat. You can show this formally if you wish.

You may not like it, but if it is impossible to formalize, something is wrong. What should be meant by the phrase: ... does not affect that probability ... is: the conditional probability has the same value as the unconditional. And although simply and easy: this has to be proven.

Which door did the host open? And which one is chosen by the player: Let us call them h and x.

Do you mean: P(C=1|X=x and H=h)?

>>>Please, continue here.Nijdam (talk) 21:27, 20 August 2010 (UTC)[reply]

How could it? The problem is obviously completely symmetrical with respect to door numbers. The result of opening one legal door must be the same as the result of opening the other legal door. In fact the result must, by simple, obvious, and intuitive symmetry be the same if the player originally chose door 3 and the host opened door 1.

Conclusion?

As shown below.

The probability that the car is behind one of the unchosen doors must be 2/3

Meaning: P(C<>X|H=x) = 2/3?? Or what?

Again, my English statement is quite clear. I mean that the probability that the car is behind one of the unchosen doors is 2/3 after the host has opened one to reveal a goat. I you want to formalise this you can.

How about P(C=2 or C=3| H=2 or H=3)?

Has the player chosen door 1?

The car is not behind the door opened by the host.

Meaning: P(C<>H)=1? Or what else?

Yes.

It is therefore behind the unchosen, unopened, door with probability 2/3. Martin Hogbin (talk) 08:32, 16 August 2010 (UTC)[reply]

==>>P(????) = 2/3.Nijdam (talk) 10:52, 16 August 2010 (UTC)[reply]

Therefore P(C=2|H=3) =2/3 Equally P(C=3|H=2) =2/3

What exactly is the point of all this? Martin Hogbin (talk) 17:04, 16 August 2010 (UTC)[reply]