Re: Oh no! Monty Hall problem again.....

From: No Way (Not_at_real.com)
Date: 09/19/04 

Date: Mon, 20 Sep 2004 00:59:44 +0200

On 19 Sep 2004 13:06:07 -0700, clintonz@prodigy.net (Clinton C
Zimmerman) wrote:

>No Way <Not@real.com> wrote in message news:<pbpqk011q3juqeungqkh5ftd9v0270num6@4ax.com>...
>> On 18 Sep 2004 18:57:38 -0700, clintonz@prodigy.net (Clinton C
>> Zimmerman) wrote:
>>
>> >First the chances of the first player when one door is removed are not
>> >necessarily 2/3.
>>
>> 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.

>> Since the contestant doesn't know what p and q are, the best he can do
>> is to pick one of the doors randomly. The probability for picking a
>> door is obviously 1/3.
>>
>> The probability that he has picked the door with the car behind it is:
>>
>> (P door 1)*(P car behind door 1) + (P door 2)*(P car behind door 2) +
>> (P door 3)*(P car behind door 3) =
>>
>> (1/3)*p + (1/3)*q + (1/3)*(1-p-q) = 1/3
>>
>
>Wrong the probability (P door1 ),(P door 2), and (P door3) are not
>necessarily 1/3 if you meant (P door1) is the probability that door 1
>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?

> 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. 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.)

>> > Only the weighted average between doors is 2/3.
>> >
>> >.
>>
>> Conan is not aware of anything that has happened before he arrived. He
>> doesn't know which door the contestant chose. All he sees are two
>> doors. One of these doors has probability m of having the car behind
>> it, the other has probability (1-m). m is not known to Conan. The
>> probability that Conan picks the door with the car behind it is:
>>
>> (1/2)*m + (1/2)*(1-m) = 1/2.
>
>That is the average succes rate for both doors, or the success rate
>of conan if he chooses randomly.
>
>Read what I said above again, carefully:
>
>"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.

>The contestant IS NOT choosing randomly,
>he is always switching. The rate of Succes for Conan if he chooses the
>SAME door as the contestant is m, or (1-m) which was computed above because
>m is a funtion of p. If after one door was eliminated the
>contestant decided to choose randomly and abandon the always switch
>strategy, he too would have an average rate of success of 50%.

Exactly, which is why Conan has a 50% chance of winning. Conan does
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.

Relevant Pages

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