The Monty Hall Dilemma

Discussion in 'Physics & Math' started by jrc, Jul 14, 2004.

  1. jrc Registered Member

    Messages:
    17
    I've been having a very heated and time consuming debate with some individuals, who are having problems grasping the Probability aspect of the Monty Hall Dilemma.

    Only two or three of them can grasp the math and concept, to be honest with you, i have run out of ways to try getting its concept across.

    Perhaps its the way i am explaining it?

    Is there some kind person hear, who can think of a way to demonstrate this principal and perhaps save me from banging my head on a wall in frustration.

    Thanks for any help.
     
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  3. Nasor Valued Senior Member

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    6,231
    It might help to demonstrate the same sort of thing with larger numbers. Imagine you had a million doors and that the prize was behind one of them. You pick a door, then 999,998 of the doors that don't have a prize behind them are eliminated. Should you keep your door, or switch to the only remaining door? It should be obvious to anyone that even though there are only 2 doors left, the odds are far greater that the prize is behind the only remaining door that the player didn't pick - since if it wasn't, it would mean that the player had randomly picked the one door out of a million that had the prize behind it.

    Then explain that it's the same sort of thing in the classic monty hall problem, just with fewer doors.
     
    Last edited: Jul 15, 2004
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  5. Pete It's not rocket surgery Registered Senior Member

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    10,167
    Have you tried doing it by experiment?

    Play a few rounds of modified "Find the Lady":

    You are the host. You have three cards, including one queen.
    You lay the three cards face down, with the Queen in a random position known only to you.
    Your contestant guesses where the Queen is.
    You turn over a non-Queen card that they didn't pick, and ask if they want to switch.
    A few rounds should demonstrate that switching wins more often.


    You could also use the physical demonstration to more clearly explain particular scenarios at each stage of the game.
    For example, after the contestant guesses a card, ask them what they expect you to do if Card 1 is the Queen, if Card 2 is the Queen, and if Card 3 is the Queen.
     
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  7. jrc Registered Member

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    17
    Thank you Nasor and Pete. I will apply both solution's and Waite for that Ha HA! factor, when the penny drops.
     
  8. DOS Registered Senior Member

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    37
  9. jrc Registered Member

    Messages:
    17
    Thanks every one, they finally get it..

    I think?
     
  10. DOS Registered Senior Member

    Messages:
    37
    The catch is that Monty KNOWS which door is a bad prize. If he didn't then he might open the door to the good prize.

    I would like to expand Nasor's example.

    Suppose there are a Billion doors. the chance of picking the correct one is very, very slim;almost impossible.

    You pick a door.

    Now Monty (with knowledge of which is the good door) opens 999,999,998 bad doors.

    You can now use his knowledge to your advantage.

    Since at the outset your chances were so slim, Monty has almost told you which door has the prize by his not opening that door.
     
  11. jrc Registered Member

    Messages:
    17
    Its quite interesting. No matter how many times i demonstrate this to them, they insist that the probability of choosing correctly, 'with only two doors available out of the original three doors', is 50 - 50.

    It truly is counter intuitive to the Human mind. I suppose there is one consolation, I'm not having to discuss the model for the Interference experiment or the the quantum neater of the electron..
     
  12. metacristi Registered Senior Member

    Messages:
    92
    The math behind this alleged 'dillema' is quite simple,here is my solution:

    Let

    A,B,C =doors
    'X'=1=the prize is behind door 'X'
    'X'=0=there is nothing [or the goat] behind door 'X' ; where 'X'=A,B or C


    There are possible 4 distinct events after choosing initially a door,let it be A.The underlined doors are those shown later as having the goat [or nothing in some variants of the problem] behind:

    event1: A=1 AND B=0 AND C=0

    event2.: A=1 AND B=0 AND C=0

    event3: A=0 AND B=1 AND C=0

    event4: A=0 AND B=0 AND C=1


    We have:

    P[event1]=P1=(1/3)*(1/2)=1/6

    P[event2]=P2=(1/3)*(1/2)=1/6

    P[event3]=P3=(1/3)*1=1/3

    P[event4]=P4=(1/3)*1=1/3


    Therefore the probabilities to have chosen a door having or not the prize behind are respectively:

    P[the prize behind]=P1+P2=2*(1/6)=1/3

    P[nothing behind]=P3+P4=2*(1/3)=2/3

    From here it's clear why switching increase the chances to find something behind the door from 1/3 to 2/3.
     
  13. metacristi Registered Senior Member

    Messages:
    92
    Regarding the above demonstration I have to mention that we must accept first the 'frequentist' interpretation of probability.This interpretation is the default,in fact,in science,being widely used,as of now at least.In this acception the new piece of data (a door is 'eliminated' by Monty) does not change the initial (apriori) probabilities (another good example for that is in the 'the prisoners' dilemma',see http://mathworld.wolfram.com/PrisonersDilemma.html).

    Still some people might argue,having some rational grounds in my opinion,that we do not have the right to use the frequentist approach since we deal with a single,unique,event.They argue that a bayesian approach,notwithstanding its subjectivity,is the only accepted in this case (the bayesian interpretation of probabilities is a promising but still contested approach).Thus the new piece of information acquired does change the initial probabilties though the aposteriori probabilties are subjective.Before Monty eliminates one of the doors the probabilties,'objective',are those computed in the previous post indeed but after Monty's action they change,thus we (better said,at least the choser) are entitled to assign a 50/50 aposteriori 'subjective' chance (use Bayes' formula to compute the aposteriori subjective probabilties).

    But as I've already underlined the frequentist approach is the default currently and moreover in this case it is better suited since there is no sensibility at the choser's identity.
     
    Last edited: Jul 18, 2004
  14. Tracker00 Registered Senior Member

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    51
    well i still like the bayesian approach... but if you were given this situtation in real life what would you do?
     
  15. Facial Valued Senior Member

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    2,225
    I like Marilyn vos Savant's quantitative exagerration, much like DOS' explanation. If you had a million doors, and Monty opens every single one of them except for door number 777,777 then you would switch pretty fast wouldn't you? She lives up to her name.
     
  16. metacristi Registered Senior Member

    Messages:
    92
    With the assumption that Monty makes random choices-when eliminating one of the doors-it's clear that this problem is amenable to an 'objectivist' (frequentist) approach.Indeed,presumably,if we repeat in practice the same situation many times (in order to obtain a statistically relevant sample) we would regain the previously calculated probabilities.Thus,in the light of this,switching is the most rational choice.

    However since in the problem we deal with a single event it is also rational to prefer a bayesian interpretation,entirely subjective,on purely personal grounds,counting as a personal degree of confidence (a personal,fallible,'belief').I'd say however that overall the subjective approach has a lower degree of priority (notwithstanding its rationality) compared with the 'objectivist' frequentist solution.We are rationally entitled to choose as the standard of knowledge the results given by the frequentist approach.Preferring the subjective alterantive,entirely on a subjective basis,is still rational,though it cannot be considered as being the standard of knowledge.

    Personally I would take in account both approaches,if both are applicable,the choice depending on the results obtained.For example if both indicates a certain possibility as the most probable clearly I would choose it.I would give however a higher degree of preference to the frequentist approach and if the subjective approach is nonconclusive (as it is the case here) I would choose the frequentist solution.So I would switch.However if the frequentist approach is not possible I would certainly accept the bayesian approach,I consider it a very valuable 'tool',though in many cases it can give us only a subjective rational support.Bayesianism is a very serious field having many applications in science,ranging from establishing (with a very great deal of confidence by the way,close to certainty) if a book has been plagiarized till accepting/rejecting the hypotheses explaining the results of tightly controlled experiments studying paranormal phenomena.Though we might not know the apriori probabilties (the most frequent case in practice) it is perfectly reasonable to assign some provisional apriori probabilities,depending on the situation,and in the light of new data to compute the revised aposteriori probabilites representing our subjective degree of confidence.In some cases the bayesian interpretation of probabilities enables us to obtain quasi-'objective' results which to be accepted by all,would be rational,people.
     
    Last edited: Jul 25, 2004
  17. MacM Registered Senior Member

    Messages:
    10,104
    A point seems to be missing I think. All this only shows the falacy of some mathematical concepts.

    Why. because I can reinterprete the puzzle and cause the results to reverse.

    That is if I suddenly say "Hey, I don't want a car, I want the goat"; what supposedly is behind the doors reverses.

    The mathematics are therefore arbitrary and meaningless in reality. The actual odds must therefore remain 50/50.
     
  18. Nasor Valued Senior Member

    Messages:
    6,231
    You're quite wrong about this. If you try it in real life, you'll see that it works out exactly the same as statistics predict. If you don't have a buddy who's willing to help you do the experiment and tabulate the results, you can write a program to do the experiment automatically a large number of times and compile the results for you.

    If you wanted the goat, then your best course of action would be to not switch doors, because by doing so you would decrease your odds of winning the goat from 2/3 to 1/3.
     
  19. MacM Registered Senior Member

    Messages:
    10,104
    Well, you can argue all you want but until you can explane how mathematics knows the difference between a goat and a car your arguement fails. Your claim comes down to an arguement that your first chice is wrong regardless of which item you choose. That doesn't fly.
     
  20. Nasor Valued Senior Member

    Messages:
    6,231
    No, not at all. Your first choice is probably right if you want the goat, but wrong if you want the car. That's because there are two doors with goats, but only one door with a car.
     
  21. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    Hi Mac,
    I don't understand what you mean.

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    When you first pick a door, the chance you have a goat is 2/3, and the chance you have a car is 1/3.

    When the host opens door to reveal a goat, the chance there is a car behind the third door is 2/3, and the chance there is a goat behind the third door is 1/3.

    To summarize:
    <table border=1 cellpadding=5><tr align=center><b><td></td><th>Your door</th><th>Open door</th><th>Other door</th></tr></b><tr align=center><th>Goat probability</th><td>2/3</td><td>1</td><td>1/3</td></tr><tr align=center><th>Car probability</th><td>1/3</td><td>0</td><td>2/3</td></tr></table>

    How can these results be changed by the contestant's interpretation?
     
    Last edited: Aug 7, 2004
  22. capslock Registered Member

    Messages:
    5
    The Monty Hall dilemma (MHD)is given a pretty good treatment in Paul Hoffman's biography of Paul Erdos. The book is called "The Man Who Loved Only Numbers" published by Hyperion 1998 (USA) and by Fourth Estate 1998 (Great Britain).

    The biography is quite an exraordinary story of the eccentric Erdos, who was perhaps one of the more able and prolific mathematician of the last century, yet who, like many, had real problems in coming to terms with the MHD.

    The book recounts great debates and controversy generated in 1990 by the MHD among mathematicians of considerable standing many of whom eventually retreated somewhat red-faced. So if you find the concepts dificult, despair not, you are in good company.

    One quote from this remarkable book re the dilemma...

    "When reality clashes so violently with intuition....people are shaken."
     
  23. Silas asimovbot Registered Senior Member

    Messages:
    1,116
    It's perfectly simple: You have a 1 in 3 chance of picking the million-dollar door the first go out. You therefore have a 2 in 3 chance of picking the wrong door.

    One of the other doors is eliminated.

    Does that change your original odds? No. Therefore you still have just a 1 in 3 chance of having picked the right door, and (here's the kick) the whole of the 2 in 3 chance of being the million dollar door is given to the remaining door.

    Therefore switch your choice. You will win twice as often.
     

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