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

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

Date: Sun, 19 Sep 2004 14:48:08 +0200

On 18 Sep 2004 18:41:33 -0700, clintonz@prodigy.net (Clinton C
Zimmerman) wrote:

>gtsavdar@auth.gr (George) wrote in message news:<6caa1e8b.0409180150.64142efe@posting.google.com>...
>> First of all i want you to understand that i understand that by
>> changing the decision the probability of winning is higher.
>> (Here are my thoughts:
>> 1st case: He chooses to remain in his decision:
>> 1st subcase: He chooses a goat at the one door at the beginning.
>> 2nd subcase: He chooses a goat at the other door at the beginning.
>> (Although the 2 goats are not something different, we have to take 2
>> cases for the 2 goats, as the option the player has is to choose
>> between 3 boxes, the 2 of them contain 2 goats)
>> 3rd subcase: He chooses the car at the one door at the beginning.
>>
>> So 1st case+ 1st subcase = lost
>> 1st case+ 2nd subcase = lost
>> 1st case+ 3nd subcase = win
>>
>> So we have 1/3 probability of winning.
>> -------------------------------------------------
>> 2st case: He chooses to change his decision:
>> 1st subcase: He chooses a goat at the one door at the beginning.
>> 2nd subcase: He chooses a goat at the other door at the beginning.
>> (Although the 2 goats are not something different, we have to take 2
>> cases for the 2 goats, as the option the player has is to choose
>> between 3 boxes, the 2 of them contain 2 goats)
>> 3rd subcase: He chooses the car at the one door at the beginning.
>>
>
>
>Maybe I read this wrong but how did you get three subcases out
>of "changes decision" ( doesn't stay?). There are four.
>Chooses one of the remaining doors wins or loses (2 cases)
>chooses other remaining door and wins or loses (2 cases)

No, there are not four options. This is the Monty Hall problem. After
the contestant has made an initial choice, an unpicked door with a
goat behind it is opened. There are thus three options. Perhaps this
is more easily seen if there is a chicken and a goat (instead of two
goats). After the initial choice, a door with a chicken or a goat
behind it is opened.

1st subcase: He chooses the chicken at the beginning.
2nd subcase: He chooses the goat at the beginning.
3rd subcase: He chooses the car at the beginning.

If he chooses the chicken, Monty will show the goat. The contestant
switches and ends up by the car. (win)
If he chooses the goat, Monty will show the chicken. The contestant
switches and ends up by the car. (win)
If he chooses the car, Monty will show either the chicken or the goat.
Monty has no preference for either animal in this situation. The
contestant switches and ends up with the animal that wasn't shown.
(loss)

>> So 2st case+ 1st subcase = win
>> 2st case+ 2nd subcase = win
>>
> 2st case+ 3nd subcase = lost
>>
>> So we have 2/3 probability of winning.
>> -------------------------------------------------
>> So by having as a strategy that of changing our first decision we
>> would win 2 out of 3 times, instead of 1 out of 3 if we remain at our
>> first decision.)
>> /////////////////////////////////////////////////////////////////////
>
>???

This is the ordinary Monty Hall problem. Nothing strange here.

>> But............
>> Let suppose the player has choosen a door and Monty has taken out one
>> of the remaining doors, with a goat of course. And now Monty asks the
>> player: You remain or change your decision?
>
>If the player decides to switch before Monty eleminates a door or
>"no matter what he will switch" then the chances of success is 2/3.

The moment of making the decision to switch has no influence on the
problem. However, you might as well make the decision beforehand
because it is always advantageous to switch (provided you don't have
extra information about how the situation is setup/handled).

>> And let's name this sitation as XX. And the probability of the player
>> to win (the car) as P(A) before he answer(at the situation XX), P(B)
>> after he answer that he remains in his decision and P(C) after he
>> answer that he changes his decision.
>>

You know P(B) (=1/3) and P(C) (=2/3). Suppose the contestant has a
chance p of choosing to stick with the door he has chosen.
P(A) = (1/3)*p + (2/3)*(1-p)

Obviously P(A) is maximised by choosing p=0 (which means the
contestant should always switch).

>After Monty eleminates a door and the contestant says, "I still would
>like to switch as I had originally said" his chances of success cannot
>be determined without knowing information about the conditional
>probablity
>that Monty opens that door if there is a car behind the door initially
>chosen.

This is wrong.

>If after Monty opens a door and the contestant says "I would like to
>change
>my mind, I'm not going to switch after all", the chances of success
>cannot be determined without knowing the conditional probability
>of Monty opening the door given the car is behind the firt chosen
>door.

This is also wrong.

>> Now i have some questions (Q-J) that i'm not sure what is going on.
>> The important questions for me are Q-1, Q-8, Q-9 and Q-10.
>>
>> Q-1: How much is P(A)? <Remember: at the situation XX> (Undefined?)

depends on p.

>> Q-2: Is P(A) = P(B) = P(C)? (I believe no.)

P(A), P(B) and P(C) are not equal.

>> And suddenly when we are in situation XX, Conan the barbarian enters
>> the game and kills the player and takes his position without knowing
>> any of the previous facts. Then Monty asks Conan: what of the 2 doors
>> you choose? Let's say that the probability of Conan to win the car is
>> P(KA) in this situation.
>>
>> Q-3: Is P(KA)= 0.5? (In my opinion: It is definitely.)

This is correct.

>Do you mean if Conan makes the same decision as the original player
>would have made? Switch?

The OP states quite clearly that Conan does not have that information.

> In that case the probability does not change
>from P(B) if P(B) is the probability of the players success by
>switching after
>eleminatin of a door. P(KA)=P(B) may be .5 but not necessarily. It
>probably
>isn't.

The probability of Conan winning the car (which he can then trade for
a horse) is 1/2 regardless (since Conan has no information to begin
with).

>> Q-4: What is the difference between the situation Conan is, after
>> Monty asked him to take a decision, and that of the player at
>> situation XX?
>
>Conan may himself have a greater probability of deciding not to
>switch but if he makes the same decision as the first player,
>there is no differnce in his success rate.

Conan's success rate is 1/2 (Remember, Conan doesn't know what door
the original contestant picked). The original contestant had a success
rate of 2/3. The difference is the amount of information the two have.
The original contestant knows his chances of picking a wrong door were
larger than picking the right door. Conan doesn't have this
information.

>>
>> After Monty asks Conan, the second answers: "I choose this door"(It
>> doesn't matter what door he said-That's why i use the word "this").
>> Let's call the probability of Conan to win the car P(KB).
>>
>> Q-5: Is P(KB) = P(KA) (It obviously is.)
>>
>>
>If you mean P(KA) is the probability of succes if Conan chooses one of
>the remaining doors and P(KB) is the chance of success if he chose the
>other
>remaining door, they could be , but probably aren't equal.

Suppose p is the probability of the car being behind the first
remaining door. The probability of the car being behind the other
remaining door is (1-p). The probability of Conan picking the first
remaining door is q. The probability of Conan picking the other door
is (1-q).

P(KB) = p*q + (1-p)*(1-q) = 1 + 2*p*q - p - q

If Conan has no preference for any door: q = 1/2

P(KB) = 1/2 (regardless of p)

>> But Monty now comes with another question: "Do you want to change
>> your decision?" Let's now call in this situation, the probability of
>> Conan to win the car, P(KC).
>
>I must be confused, do you mean Conan changes between doors a third
>time to the originl door or that P(KA) was the probability if the
>player was replaced by Conan before a door was eleminated?
>>
>> Q-6: How much is P(KC). (0.5 of course in my opinion.)
>> Q-7: What is the difference between the situation Conan is, after
>> Monty asked him to change his decision, and that of the player at
>> situation XX?
>>
>Whenever the player is replaced by Conan, if Conan makes the same
>selection as the player was about to make the chance of success are
>the same.

Yes, but the probabilty that Conan makes the same choice is only 1/2
(provided Conan doesn't have extra information).

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