From: glitto@iastate.edu (Scott M Klabacha)
Subject: Re: Game show contestant puzzle. SPOILER!
Organization: Iowa State University, Ames, IA
Distribution: usa
Date: Thu, 30 Jul 1992 19:50:00 GMT

echarne@ics.uci.edu (Eli B. Charne) writes:
>
>  I recall seeing about a year ago discussion on this puzzle where
>  someone is a contestant on a game show.
>
>  There are three curtained rooms.  One of which contains a prize.  The
>  contestant picks a curtain.  The game show host then reveals one of
>  the other rooms to be empty.  The question is, should the contestant
>  now choose the final curtained room or stick with his/her original
>  choice.
>
>  I hope I got the situation correct.  I would appreciate the solution
>  to this problem.  It seems to me that it doesn't matter if the
>  contestant switches or not, but I recall the discussion being
>  different.
>

(haven't figured out ^L yet, so my response is about forty lines down...)










































Okay. This is one you hit (in some or another variation) when you start
taking classes in probability.

To make matters simple, let's call:
            A = room which contestant chooses
            B = room which host reveals is empty
            C = remaining room (which contestant can switch over to)

When the contestant first chooses a room, he knows *nothing* about which
room contains a prize. Therefore, all rooms are equally likely (from the
contestant's viewpoint) to contain the prize. This means room A has a 1/3
chance of containing the prize.

Now room B is revealed. Room A is still fixed at 1/3...the probability does
not change with added information *because the choice was made without that
information*. However, the probability of room C containing the prize now
jumps up to 2/3...because the contestant never reached a decision about it.

To help understand the 2/3 probability:  when room A is fixed at 1/3, you
know (from the Law of Total Probability) that...

    1 = (probability of A having the prize) + (probability of B or C having it)

         ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
         this is fixed at 1/3                  by subtraction, this is 2/3

So we know 2/3 = (probability of B having it) + (probability of C having it)
                  ^^^^^^^^^^^^^^^^^^^^^^^^^^     ^^^^^^^^^^^^^^^^^^^^^^^^^^
                  opening the room makes this
                  probability zero...
                                                 ...so by subtraction, this is 2/3

Does that make it clearer?

Essentially, the question could be rephrased in another way:
   You choose a room. Now the announcer says, "There are two situations. First,
   the room you chose could contain the prize. Second, the two left over rooms
   could contain the prize...and one of those rooms is certainly empty, of course.
   Now, you can keep the room you chose, or switch to the two remaining rooms.
   If you switch to the two remaining rooms and *either one* has the prize, you
   win it."
This makes the contestant's choice way easier (oh boy...my chances are twice as
good if I've got two rooms), even though it's essentially the same problem as
initially presented.

Scott M Klabacha
***gLITTOfERN*** glitto@iastate.edu
just trying to waste some more time...
Newsgroups: rec.puzzles