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Variable Change

  1. Aug 1, 2008 #1
    The movie "21" went over a statistics problem in an attempt to make the main character look smart. I'm not sure though that the Solution was correct, even though it does seem so.

    The problem is this: A game show host gives you a selection of 3 doors, two of which enclose goats, and the other encloses a car. Given that you selected the first door, the game show host reveals that door number 3 is a goat. He then offers you the option of changing your choice to door number 2.

    The proposed solution: You would want to switch because you're initial chance of being wrong was 66%, but when one of the doors was opened it placed a 66% probability on the 2nd door. It makes sense in theory, but I see a problem with it.

    In this situation, you can't place the 66% probability on door number 2, because a conflict arrises if you allow such an algorithm to determine probability. Please, tell me if I am wrong.

    Assume the movie is correct and when there is a 2/3 probability of one of two choices being accurate, the probability of either choice becomes 2/3 in the event that the other choice was eliminated as a possibility. That's essentially what the solutions suggests, but look what happens when you apply that concept to doors 1 and 3. Initially the probability of either door 1, or 3 being the door in front of the car is 66%. When Door three is eliminated, there is now a 66% chance that door 1 holds the car. but not both door 1 and 2 can have a 66%; that would be impossible, so the original assumption must not be true.

    Am I wrong?
     
  2. jcsd
  3. Aug 1, 2008 #2

    Mute

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    This is a well known problem called the "Monty Hall Problem". You can read more about it on the wikipedia article: http://en.wikipedia.org/wiki/Monty_hall_problem

    A typical way to get an intuitive feel for the problem is to consider the case of say, 1000 doors. You have a 1/1000 chance of picking the correct door the first time. If the host then opens 998 other incorrect doors, would you switch to the other remaining door that you didn't pick?

    In any event, I'm not entirely sure what your objection is. It looks to me like you're saying, "Well, suppose I picked door 1 initially - I conclude a 2/3 chance that it contains the car. But, suppose I'd picked door 1 first, then I'd conclude it had the 2/3 chance of containing the car." That's correct, but you need to realize that in switching your intial choice you've created two different scenarios which lead to the two different conclusions. Of course, the labels really aren't all that important, which appears to be confusing you a little. What the analysis of the problem should say is that no matter what the labels, you have a better chance of winning the car by switching your choice.
     
    Last edited: Aug 1, 2008
  4. Aug 1, 2008 #3

    nicksauce

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    This is known as the Monty Hall Problem.
    It is outlined well on wikipedia. http://en.wikipedia.org/wiki/Monty_Hall_problem


    Edit: Just barely beat me Mute
     
  5. Aug 2, 2008 #4

    HallsofIvy

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    That's not what they are saying. What they are saying is that, a priori, each door has probability 1/3 so together there is probability 2/3 that one of those two doors has the prize. Once one has been opened, the entire 2/3 shifts to the door that is left.

     
  6. Aug 2, 2008 #5

    I guess, I am just saying that you could use the same statistics to show that it would be best to stay with your first choice. It's the exact same reasoning as far as I can tell. Each door has a probability of 1/3 so together (door 1 and 3) there is a probability 2/3 that one of those doors has the prize. Once door number three is opened, there is a 2/3 chance that you have selected the correct door.

    Obviously, when the reasoning works that way, you end up with a 4/3 chance that the car is in door 1 or 2. It doesn't seem to work.
     
  7. Aug 2, 2008 #6

    nicksauce

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    Suppose you choose A, and C is shown to have a goat)

    Mathematically (| means given)

    Before: P(A) = 1/3, P(B or C) = 2/3
    After: P(B or C) = P(B or C | C shown to have goat) = P(B | C shown to have goat) = P(B and C shown to have goat) / P(C shown to have goat).

    what is P(C shown to have goat)? There are 3 equally likely possibilities for the goats, AB, BC, and AC. In the first case, there is 0 probability, in the second case there is 1/2 probability, and in the third case there is 1 probability. So P(C shown to have goat) is (1/3)*0 + (1/3)*(1/2) + (1/3)*1 = 1/2 (as expected). What is P(B and C shown to have goat)? There are 3 equally likely possibilites for the goats, AB, BC and AC. In the first case, there is 0 probability, in the second case there is 0 probability, and in the third case there is 1 probability, so P(B and C shown to have goat) is (1/3)*0 + (1/3)*0 + (1/3)*1 = 1/3. So P(B | C is shown to have goat) is (1/3)/(1/2)=2/3.

    Now what is P(A | C is shown to have goat)? As before, P(C is shown to have goat) is 1/2. What is P(A and C is shown to have goat)? There are 3 possibilities for the goats, AB, BC and AC. In the first case, the probability is 0, in the second case, the probability is the second case is 1/2 and the probability in the third case is 0. So P(A and C is shown to have goat) is 0*1/3 + 1/2*1/3 + 0*1/3 = 1/6. So P(A | C is shown to have goat) = (1/6)/(1/2) = 1/3.

    Thus we have mathematically shown that
    P(A | C is shown to have goat) = 1/3 and P(B | C is shown to have goat) = 2/3.

    Please let me know what, if any, steps in my reasoning you don't understand.
     
  8. Aug 2, 2008 #7
    Edit: Apparantly I screwed up the quote tag. It actually looks fine to me, but w/e.

     
  9. Aug 2, 2008 #8
    I understand the issue with my thinking now. The movie never really explains the algorithm. Ben simply says, "before my chances wer 33.3%, and now they are 66.7%. When I looked up the solution, I found explanations that act as though the 2/3 combined probability from doors 2 and 3 transfer completely to door 2 when door 3 is closed. That is not the reason it is best to switch. If it were, you could also prove that it would be best not to change your choice (I think I showed why already). The solution is much more simple.
    There are three possible Goat/car combinations:

    CGG
    GCG
    GGC

    Switching in the first case looses. Switching in the 2nd and 3rd case wins, so switching has a 2/3 chance of wining.
     
  10. Aug 3, 2008 #9

    HallsofIvy

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    No, because the MC NEVER opens the door with the prize behind it. You are using the extra knowledge the MC has.
     
  11. Aug 18, 2008 #10
    I agree with Before: P(A) = 1/3, P(B or C) = 2/3

    I think, for example, if car is behind A:

    P(win) = P(Ben chooses A, host shows C, Ben stays) + P(Ben chooses A, host shows B, Ben stays) + P(Ben chooses B, host shows C, Ben switches) + P(Ben chooses C, host shows B, Ben switches)
    P(win) = P(1/3*1/2*1/2) + P(1/3*1/2*1/2) + P(1/3*1*1/2) + P(1/3*1*1/2) = 1/2

    and:
    P(loose)= P(Ben chooses A, host shows C, Ben switches) + P(Ben chooses A, host shows B, Ben switches) + P(Ben chooses B, host shows C, Ben stays) + P(Ben chooses C, host shows B, Ben stays)
    P(loose) = P(1/3*1/2*1/2) + P(1/3*1/2*1/2) + P(1/3*1*1/2) + P(1/3*1*1/2) = 1/2

    Repet it for car in B and C and you'll get to P(win)=P(loose)= 1/3 * 1/2 + 1/3 * 1/2 + 1/3 * 1/2 = 1/2

     

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    Last edited: Aug 18, 2008
  12. Aug 19, 2008 #11
    HallsofIvy: Why the 2/3 shifts? Didn't you already know that the first door to be opened will be one with a goat? So who cares what door you're already choose?. There is a 50% chance between the door who you just pointed and the one the hosts just pointed. Am I wrong?
    I think the host showes you 3 doors and tells you "there's 2 goats", you say "open that one" and he opens another one, with a goat. So you must think: Thats good, now my chance is simply 2 agains 1, without taking chances. Host is basically telling you, before your final (and only one real) determination wich door not to choose

    Table of truth attached

    The rules, if you take the labels from the doors are:

    1) You are going to choose between 3 doors
    2) Host is going to open a door (not the one you've choose, useless information) and that door will have a goat inside. Then he'll make you choose between the others 2 doors, one of them will have a goat, the other a car.


     
    Last edited: Aug 19, 2008
  13. Aug 19, 2008 #12
    MC NEVER opens the door you choose either. So why there's a 1/3 chances of winning? You are afraid of your first choice, when you have 1/3 for car. But MC don't kick you if you choose the goat at fist stage either.

    Sorry for being this annoing, but I've read many threads about this, non of them has a clear logical explanation, and many doubt (specially those who simulate it).

     
  14. Aug 19, 2008 #13

    CRGreathouse

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    How about this. You and I simulate this game with dice.* We play 100 times. If the original choice has the car more than 44% of the time, I give you $100; otherwise you give me $100. Sounds like a good deal, right? That is, since you think the original has a 50% chance of having the car.

    * Dice: To ensure against cheating, we each roll a 6-sided die. We add the two values and reduce mod 3. A 0 means that the car is behind the original choice, while 1 and 2 are the other choices. If the car is not the original choice the revealed choice is the one out of 1 and 2 that isn't a car; otherwise we roll dice and reduce mod 2, then add one, to get the number of the choice that is revealed. The motivation behind this die-rolling procedure is that if either of us uses a fair die, the result is uniformly random (even if the other is cheating with loaded dice). That way we don't need to trust each other.
     
  15. Aug 19, 2008 #14
    Thanks for your explanation GreatHouse. I've finally get it (thanks to the explanations from physforums members). I was a hard night (i couldn't sleep until i've understood it), but thanks to this community now I'm just a little less ignorant ;).

    Thanks for your time, to all of you!
     
  16. Aug 19, 2008 #15

    CRGreathouse

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    Glad to have helped. It's a tricky problem, for sure -- don't feel bad for taking some time to think it through!
     
  17. Feb 15, 2009 #16
    I apologize if someone has said this already, I didn't read every post but think of it like this:

    -by choosing one door out of three, you have a 1/3 chance of being correct and a 2/3 chance of being wrong

    -by opening one incorrect door out of the two you did not choose, you still only have a 1/3 chance of being correct and a 2/3 chance of being wrong

    -by switching to the last remaining door the odds then flip flop in your favor

    -yes, in the end, only 2 doors remain but the only reason one of them is still there is b/c it was originally your choice and doesn't quite make it a 50/50 chance.

    -the example used with the 1000 doors is a great way to think of it as well, the fact that there are only 3 doors makes it what it the hot topic that this has become really.
     
  18. Feb 16, 2009 #17
    At that point, under those end circumstances, the odds are 50-50.
     
  19. Feb 16, 2009 #18
    Because of this, the value then becomes one of extended entertainment using statistical probability.
     
  20. Apr 4, 2009 #19
    One of three things can happen:
    1. player picks car, host shows goat A or goat B, switching loses
    2. player picks goat A, host shows goat B, switching wins
    3. player picks goat B, host shows goat A, switching wins

    2/3 times switching wins. Seems simple.

    it's harder to believe this, but it must be true:
    1/3 chance initial pick correct
    2/3 chance one of unchosen doors is correct
    0 chance revealed unchosen door is correct
    2/3 - 0 = 2/3 chance of remaining unchosen door being correct.
     
  21. May 11, 2009 #20
    You have all done an outstanding job of describing the solution to the problem but unless I missed reading a post (I apologize if this is the case) you have not properly addressed the most important part of this problem: the premises.

    Premise 1: The game show host knows what’s behind each door. (Otherwise he has a 1/3 chance to open the door with the car and the statistics change.)

    Premise 2: The game show host always opens a door with a goat behind it. (Otherwise he could use the offer to trade only when he felt he had an opportunity to trick you.)

    Premise 3: You are always given the opportunity to switch doors. (Otherwise he could only offer you a chance to switch to the remaining goat.)

    These 3 premises are what assure you that you get the benefit of the 2 door statistical switch instead of a 50/50 or worse.
     
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