From: seyyed@wuee1.wustl.edu (Seyyed Mahdavian)
Subject: Re: Physics-1A Question
Organization: Washington University, St. Louis Mo.
Date: Mon, 12 Oct 1992 21:59:18 GMT

In article <31443@sophia.inria.fr> navab@sophia.inria.fr writes:

>I am sorry Seyyed, but it seems that you didn't understand the "conditional"
>probability. The choice of the door to be opened after your first choice is 
              ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>also conditioned by your first choice!!
 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

This statement doesn't make sense. "Conditioning" changes the "probability",
i.e. you ask: "what is the probability of ... given that ...". You don't ask
"what is your choice of ... give that ... ".

>Behzad answer is correct. 

Then you have to share the prize that I gave to him.   :-)

>Keeping always your first choice means betting on the correctness of your 
>first choice of probability 1/3.

You have to present a proof for this. As I mentioned above "conditioning"
changes the "probability", unless you present a proof that the apriori and
aposteriori probabilities are equal.

>Changing always your choice means betting on your wrong first choice of 
>probability 2/3. 

See above.

>And  therefore Behzad is correct. 

His and your explanation prove that you didn't understand the problem, though
your answer is correct if you state the problem otherwise.

>If you need more explanation here are all the different possibilities:
>    First choice             Change/Keep             Prize/empty
>     Prize	        	Change			Empty
>     Empty1			Change			Prize
>     Empty2			Change			Prize
>     Prize			Keep			Prize
>     Empty1		        Keep			Empty
>     Empty2		        Keep			Empty
>
>
>Result: over six situation 3 are winning and 3 are not; but
>2/3 of situations are winners if you always change your choice and
>1/3 of situations are winners if you always keep your first choice.

I have two objections here:
1) These are not really all the possibilities. After the first choice,
   when Monty opens one of the remaining two doors, it might happen that
   there is a prize behind it. You implicitely are assuming that Monty
   KNOWS where the prize is, and he ALWAYS picks the empty one.
2) You are trying to divide the probability space into disjoint events,
   and are trying to make deductions about the probability of those
   events by just counting them. This is a very common strategy in
   probability theory, however you must make sure that those events
   are EQUIPROBABLE. In other words, if I forget my first objection, you
   have to prove that all of the above situations are equiprobable.
   Otherwise I could simply say: I have only two possibilities: either
   there is a prize or there isn't. So the probability of having a prize
   is 1/(1+1) = 1/2 :-)  a reasoning which is obviously wrong.

>Did you understand Seyed ?

I understood that your knowledge is about 50-100 Sh (Sh is the smallest unit of
knowledge, after the great contempory mathematician and Farsi expert, ostAd
Shahrokh Mortazavi). That's a good figure I'd say, compared to what others
become after they suffer being in exile.

Seyyed.

Seyyed.