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Question on the Logic of the Monty Hall Problem

  1. Apr 11, 2015 #1
    Consider this example:

    A player in a game show was asked to choose between three doors, two doors contain a faulty prize, while the other contains a million dollars. As the player wanted the money, he chose one (in this case, the probability of choosing the door with the money is 33.33%). Since the host knows where the prize money is, he opened one of the doors with a faulty prize. The player is then asked if he wants to switch doors. According to the solution of the problem, it is more likely to have the prize money if the player were to switch doors (66.66%).

    Here is my problem:

    If, before the player decides to switch doors, the player received urgent news and was permitted to leave the game show. His brother, who wasn't paying attention to the game, came as his replacement. The host asked him to choose between the two available doors, where one of them contains the prize money. (I couldn't think of a better example :P) Now, doesn't the probability of acquiring the money is 50%, which contradicts the 66.66%?

    Which of the two probabilities is correct? I have been thinking that there might be physical systems with similar situations (like a system where we thought of just having two possible states but actually contains three).
     
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  3. Apr 11, 2015 #2

    Simon Bridge

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    If the brother does not know the original choice then he us basically choosing from two options for 50% but this does not contrafict the original calculation.
    The brother cannot complete the assignment as asked (do you want to change your mind?) since he does not know the original choice. If he us told the original choice, then the odds follow as before.
     
  4. Apr 12, 2015 #3
    So, does this mean that probability is relative, and not inherent on the current situation of the subject?
     
  5. Apr 12, 2015 #4

    Simon Bridge

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    The probability depends on how you choose as well as the physical arrangements.
    Whats happened is that the idds change as more information about the setup comes in.

    In Monty Hall, you know the first choice is 2/3rds likely to be wrong... so, when one wrong alternate is removed, you are best advised to change your mind. Notice that if the host revealed the correct choice, the probability you chose right just dropped to 0?
     
  6. Apr 12, 2015 #5

    PeroK

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    The Monty Hall problem depends critically on Monty knowing where the prize is and always opening an empty door. There is an alternative problem where Monty opens a door at random (and sometimes spoils the game). It also depends on the contestant knowing that Monty knows.

    Anyway, as long the brother has the same information, the odds are the same. But, if the brother has no idea what has happened, then he will calculate different odds. If he's simply told that there are two doors and a car is behind one. Then, to him, it's 50-50.

    Each person can only calculate odds based on what they know. So, if two people know different amounts of information about a situation, they may calculate different odds.

    A simpler example is where you draw a card from a pack and tell one friend it is red but don't tell the other. Then you ask each to calculate the probability it is the Ace of Hearts. One says 1/26 and the other 1/52. They are both correct, given the information they have, which perhaps says something important about probabilities.
     
  7. Apr 12, 2015 #6

    Simon Bridge

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    It goes beyond what you calculate... those are the actual odds.
    It is not a matter of calculating the wrong odds due to not enough information.
     
  8. Apr 12, 2015 #7
    I heard in a class before that there are systems which have a total probability of less than one, and they said that this is because there are still missing information about the system. Doesn't this mean that there are ways to know that there are missing information? Which, I think, indicates that probability depends on the system, not on the given information.
     
  9. Apr 12, 2015 #8

    PeroK

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    What if you ask Monty himself? What are the odds, according to him, that the car is behind each door? It's 0 and 100%, because he has more information than the contestant.
     
  10. Apr 12, 2015 #9

    PeroK

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    I've never heard of a total probability of less than 1.

    In the real world, the probabilities you calculate depend on how much information you have about a system. You will calculate different probabilities depending on how much you know.

    In fact, in many cases, if you know everything, there are no longer any options, as you know the answer: to the person who draws the card at random (and looks at it), there is no longer any question of probabilities: and, to Monty (who knows where the prize is), there are no probabilities about where it is.
     
  11. Apr 12, 2015 #10

    Doug Huffman

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  12. Apr 13, 2015 #11

    Simon Bridge

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    ... that's right - and those are the correct odds for him ... it is not a matter of calculating erroneous odds due to insufficient knowledge.
    I want to make a distinction between the probability and the calculation to get a probability - both will change with knowledge, but the calculation can also be in error due to insufficient data.

    ... this is the situation where you have erroneous information about the "system" - i.e. you don't know about the whole system.
    For instance, you may have a double-headed coin and not know it ... so you will miscalculate the odds of each possible result. You could have an 8-sided die but assume there are only 6 sides ... then the sum of the probabilites from 1 through 6 will be less than one because you didn't count the "other" result.

    In the Monty Hall problem, the odds being calculated are correct odds and sum to 1.

    You can figure out that there is something wrong with your calculations as you collect the statistics ... see the "Beyesian Inference" link in post #10.
    It is also an informative formalism for how to modify a model as new information comes to light... i.e. as the result of an experiment.
     
  13. Apr 16, 2015 #12

    FactChecker

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    The first brother knows that the door he didn't pick has a higher probability than his door (This assumes that Monte would never open a door with the prize and that he is not more likely to make this offer if you picked the door with the prize). So he should switch. Suppose the second brother doesn't know which door was originally picked. He knows less than the first brother and can not take advantage of the additional information. His odds are only 50/50 because he knows less.
     
    Last edited: Apr 16, 2015
  14. Apr 17, 2015 #13

    Simon Bridge

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    Note: the brother does not get to "change his mind" since he does not know the original choice.
    The option to "change his mind" is not the same as the option to "pick a door any door"... so, basically, the brother is playing a completely different game.
     
  15. Apr 17, 2015 #14
    The system I was talking about is anomalous diffusion (as my former teacher reminded me), but she said that the system has a total probability of less than one because the system is not in equilibrium.

    Is this analogous to the "erroneous information"? I suppose the Monty Hall problem as an analogy to this problem is not that convenient.
     
  16. Apr 17, 2015 #15
    Im new but the 66.66% is for the original three doors and it consists of choosing between two options (to stick with the first door or to switch doors). It makes sense that the probability to have the right door when he changes his decision is 66.66%. But when his brother who has no knowledge of what happened has to chose, it has to be stated wheather the question is what the probability is of choosing the right door from the three doors or what the probability is of choosing the right door from two doors, given it might or might not be one of them.
     
  17. Apr 17, 2015 #16

    FactChecker

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    Yes. So many probability questions are really about the theory of what you should guess, given information, than it is about the odds of what happened. After all, once an experiment has been done, there is no random probability. The outcome is already determined.
     
  18. Apr 17, 2015 #17

    lavinia

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    Here is a modified Monty Hall that may help to see the logic.

    Suppose after the contestant chooses a door, Monty randomly chooses another door and says, "That door is mine. But if you like, you can switch to the third door. "

    What are the probabilities then?
     
  19. Apr 17, 2015 #18
    Back in the day, we actually tried this physically to see what would happen. We used a random number generator to pick the door that had the prize. Why don't you try it and see what happens if you always switch doors? The odds are so favorable for switching doors that it won't take more than 10 or 20 realizations to convince you.

    Chet
     
  20. Apr 27, 2015 #19
    The easiest way to be convinced is to assume that the game becomes so successful they turn it into a spectacular and have 1,000,000 boxes instead of 3, one of which contains a million dollars and the other 999,999 bananas. After the contestant makes his choice the host opens 999,998 doors to reveal 999,998 bananas. He then gives the contestant the option of switching doors. I don't think many people would then think there was no advantage in switching..

    Of course if the brother is asked to choose one of the closed doors (as opposed to being asked if wants to switch) his chances are still 50 50. The original contestant on the other hand knows which door represents a switch, so the probability that the brother will get the money is less than the probability that the original contestant would have had he stayed and been asked to choose a door (assuming he would have chosen the sensible one).
     
    Last edited: Apr 27, 2015
  21. May 9, 2015 #20
    What about if i think : if I had made the opposite choice after the doors are opened to reveal losing gains ? Then the probabilities were exchanged and the first door i chose had a bigger probability ?
     
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