Re: Oh no! Monty Hall problem again.....
From: Clinton C Zimmerman (clintonz_at_prodigy.net)
Date: 09/20/04
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Date: 19 Sep 2004 21:51:10 -0700
No Way <Not@real.com> wrote in message news:<dr0sk0lvccea4nl1q8reo7ansk45d35jb7@4ax.com>...
> On 19 Sep 2004 13:06:07 -0700, clintonz@prodigy.net (Clinton C
> Zimmerman) wrote:
>
> >>
> >> In the Monty Hall problem the car is placed behind one of the three
> >> doors. Suppose that the probability of the car being behind door
> >> number 1 is p, the probability of the car being behind door number 2
> >> is q and the probability of the car being behind door number 3 is
> >> (1-p-q).
> >>
> >these are each 1/3, BEFORE a door is opened. that is tivial
>
> No, these are not each 1/3. p is the probability that Monty puts the
> car behind door number 1. Maybe Monty never puts the car behind door
> number 1... maybe he puts it there everytime... maybe he throws a die
> to determine where the car is put. What is important is that p and q
> are not known (by us or the contestant). They might be 1/3, but they
> don't have to be.
This is the Monty hall problem. The initial distribution is
1/3,1/3,1/3
changing this distribution will create an even more complicated
solution,
far more complex than your solution.
> >is opened by the contestant.
>
> The contestant will do this if he wants to maximise his chances of
> winning the car.
>
> > The number of times each door is opened in the
> >game by contestant in an always switch strategy are not necessarily
> >equal!!
>
> How is that relevant?
Because for each door the number of WINS (the number of times
switching
with the door open wins is the same). # of wins for each door
the same. # of times each door is open may be different. Get it?
>
> > Suppose the contestant always chooses door 1 initially and always
> >switches.
>
> Bad strategy. Suppose the car is always behind door number 1. This
> strategy now always loses. You are acting as if p cannot be 1.
p as in the probability that Monty puts the car behind door 1 as
defined by you above? It can't be. p=1/3 by definition. Your badly
confused.
This is
> false. p can be 1 (or 0 or 0.324151, the point is that you don't
> know). p (and q) are unknown. Given that they are unknown, you must
> find the best strategy. This one (always picking one and switching) is
> not the best strategy (given the information provided).
>
> <snipped a whole story about a situation that the OP didn't talk about
> at all>
>
> >> The probability that he has picked a door which doesn't have the car
> >> behind it is thus 2/3. This is true regardless of the values of p and
> >> q.
> >>
> >> Monty now opens a door that has a goat behind it. This doesn't change
> >> the probabilities calculated above. The probability that the
> >> contestant is standing in front of a door that has the car behind it
> >> is still 1/3. Since the car has to be behind either this door or the
> >> other unopened door, the probability of the car being behind the other
> >> door is 1-(1/3) = (2/3). Since the probability that the car is behind
> >> the other door is greater, the contestant should always switch to the
> >> other door.
> >
> >Completely wrong. Each door has a very real and not necessarily
> >equal chance of success.
>
> You have never done this in real life now have you? Get three empty
> matchboxes. Put a coin in one of them. Give these matchboxes to a
> friend. He will place them in some order (out of your sight). You pick
> one. He will tell you which of the other ones is empty. Switch to the
> one that is left. Repeat the process 50 times (or more). Notice that
> you have won about 2/3 of the time. (Or write a computer simulation
> that simulates this process.)
That's the average probability for all trials. Okay, instruct you
friend
to place each matchbox under a label with the number 1,2 or 3 on it.
Now instruct your friend to randomly place the matchbox with coin,
which
is understood in the gameshow. Choose a label. Now instruct you
friend,
if the box under label 1 is empty and I did not choose that door, do
not elminate that box if the coin is in the box I choose. You winning
chances for "label 1" are not 2/3!
>
> >
> >"Whenever Conan replaces the player, the chances of success of Conan
> >are the same if he makes the same choice the player would have made"
>
> It won't become correct by repeating it... You keep ignoring that
> Conan knows nothing about the situation. Conan does not know what door
> of the three doors the contestant chose. All Conan sees are two doors.
Soooo what. Run one of the many computer simulations on the net where
the doors are labeled. Chose door1 and eleminate a door. Then have
a friend switch with you and choose one of the remaining to doors.
Keep a note of the winning percentage for each door. It is not 1/2!
> not know what the contestant did.
>
> >The point is that given three doors 1,2,3 if say door 3 is removed
> >the chances of success for choosing door 2 DO NOT change if the contestant
> >is replaced by Conan.
>
> Get the matchboxes out again. Get another friend. Your new friend has
> to wait out in the hall. Your other friend does his coin placing
> again. You pick one of the matchboxes like before. Your friend opens
> an empty matchbox. At this point you leave the room (hide under the
> table, whatever, just as long as you don't communicate anything to the
> friend in the hall). Your hall friend comes in. He has to pick a
> matchbox. Then you come back and pick a matchbox (the one you would
> have switched to). Do this 50 times (or more). Notice that you have
> one about 2/3 of the time and that your hall friend has only won about
> 1/2 of the time.
That's not the question. I said given three labeled doors 1,2,3. Your
scenario
is like removing the label from each door between each rounds and
randomly rearranging them and then computing the probablity for each
"number".
Get out a notepad.
Place three pieces of tape on the floor, labeled 1,2,3. Have your
1st friend places the boxes under the tape and always first choose
the box under label 1. Have you friend eleminate an empty box
randomly.
Now have you second friend come in with no knowledge of what
transpired
and have him open one of the remaining boxes.
Record which label # the box was under and whether it was a win or
loss. The win/loss percentage for each label is not 50%
Furthermore if you instruct your first friend to never eleminate
the box under label 2 when the coin is in the box under label #1
(which you always choose first), you will find that your win
percentage
for label 2 and 3 is not 2/3 though the average is.
If you want, you can even get out a six sided die and tape over x
number of faces. Always choose label 1 and instruct you first friend
to only eleminate the box under label 3 if the coin is in the box
under label number 1 and if after a roll of the die he gets a face
with
tape over it. Otherwise eleminate the other box.
You will find that the chances of success for always switch
when the box under lable 3 is eleminted is 1/1+(x/6)
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