NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/alt/folklore/science/4481

Internet Message Format | 1992-12-23 | 3.9 KB

Path: sparky!uunet!cis.ohio-state.edu!rutgers!concert!sas!mozart.unx.sas.com!sasjec
From: [e-mail] (Jerry M. Cox)
Newsgroups: alt.folklore.science
Subject: Re: Monty Hall/Shell Game..
Message-ID: <[e-mail]>
Date: 23 Dec 92 23:29:32 GMT
References: <[e-mail]> <[e-mail]>
Sender: [e-mail] (Noter of Newsworthy Events)
Organization: SAS Institute Inc.
Lines: 60
Originator: [e-mail]
Nntp-Posting-Host: asimov.unx.sas.com


In article <[e-mail]>, [e-mail] (Cliff Heller) writes:
|>
|> I can't believe we are going through this again, but I suppose it is
|> somewhat educational.
|>
|> In <[e-mail]> [e-mail] (Donald Byrne) writes:
|> >     He was asked to change only after the elimation of one of the other
|> >doors. Now we know that there are two non-win doors in the game, and the
|> >object of Monty Hall is to elimanite one by the rules. If our player has
|> >any smarts he knows that he will either have a winner or a looser, and
|> >that Monty Hall will asure him of that. The reason is that of the three
|> >doors he will only really deal with two. Monty Hall will elimaniate the
|> >other. So unless I'm not seeing the game properly it seems to me the
|> >final results should be 50%.
|> >     Now explain why all of you say 66% of a chanch if he changes to the
|> >other one of the remaining two??
|>
|> Think about it this way:  If the person guessed correctly initially,
|> then switching would not benefit him.  Since he selected one door from
|> three, the odds of having a correct initial guess are 1/3.  There should be
|> no dispute on this point or a review course in probability is necessary.

Odds are that I could stand a review course in probability. But hear me
out anyway :-) If the contestant selects one door of three and that was
all there was to it then the odds of initially selecting the winning door
are one out of three. I have no problem with that, but I don't think
this describes the situation. If Monty is *always* going to reveal a
goat behind one of the doors, that door is removed from consideration.
The final choice then becomes a choice between one of two doors. If
Monty *always* removes one of the doors from consideration, then the
final choice (as well as the initial choice) is between a door with a
car behind it, and a door with the "other" (the one Monty doesn't show)
goat behind it. The odds of picking the car are 50/50 whether you switch
or not because Monty is *always* going to take away one of the losing
choices.

|> Now, this isn't quite as easy to see, but if the person guessed
|> INCORRECTLY initially, then he automatically wins by switching.  Why?
|> Because Monty always reveals a non-winning door.  He equivalently says to
|> you: you can have the door that you picked, or the TWO doors that you
|> didn't pick.  Since He always reveals a goat, if the initial guess was
|> incorrect, then the remaining door contains the prize.  The odds of the
|> initial guess being incorrect are 2/3 by the same reasoning as above.

If he said "You can have the door you picked or the TWO doors that you
didn't pick", and didn't reveal anything about the two doors, then I
would agree that the best thing to do would be to switch. But, Monty
changes the odds when he shows you one of the goats behind one of the
two doors. He removes that door from consideration completely. Unless
you would like to take the goat Monty just showed you, you are left with
a one out of two choice. He equivalently says to you: "You can have the
car or the goat that I didn't show you", or "You can have the door you
initially chose or the door I didn't show you". That sounds like 50/50
odds to me.

To sum up the initial choice is not a one of three choice between
a car or goat #1 or goat #2. The initial (and final) choice is between a
car and one of the 2 goats. The other goat will be removed as a choice
by Monty.