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Why doesn't the solution to the Monty Hall problem make sense?

  1. Feb 11, 2019 at 2:39 AM #1
    1. The problem statement, all variables and given/known data
    You have 3 doors, with 2/3 chance of being wrong.

    A host opens a door, and there is no prize.

    You are now left with two doors. I would like an explanation why the car is still equally likely to be behind any three doors still after the host opens a door.

    I have a hard time that Paul Erdos could not logically come to this conclusion as well. Something is off about this problem

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Feb 11, 2019 at 2:45 AM #2

    PeroK

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    You could write a computer program to simulate it.

    Or, you could simulate the experiment using playng cards. Probablity is relative frequency, so if you simulate the experiment a large number of times, you can estimate the probability.
     
  4. Feb 11, 2019 at 2:51 AM #3
    Is there any logical way of explaining of why the probability of picking the car doesn't change
     
  5. Feb 11, 2019 at 3:00 AM #4

    PeroK

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    That's been discussed on here about a hundred times. The key point is that the host knows where the car is and never accidentally reveals it. The problem is different if one show in three on average Monty ruins things by revealing the car by mistake.

    Note that, as a student of mathematical probability, there is nothing funny about this problem. It's only controversial because it's in the public domain and debated by people with no knowledge of how to analyse probabilities. The sort of people who think that if a coin comes up heads 3-4 times in a row it is more and more likely to be heads next time.

    It's important, therefore, to analyse both problems:

    1) The actual scenario in the show.

    2) The scenario where Monty opens a door (without knowing where the car is) and by luck gets an empty door.

    In terms of scenario 1, I would imagine there are 100 doors. You pick one. Monty then opens 98 doors, all empty. That leaves two doors. Then I think it's more obvious that there is still only a 1% chance that you picked the right door first time and a 99% chance that the car is behind the one door that Monty didn't choose to open! It's fairly obvious to me where the car is likely to be in that scenario.
     
  6. Feb 11, 2019 at 3:11 AM #5
    Yes, I am questioning the situation in the show, where Monty knows where everything is.

    Your post didn't really answer my question though: why wouldn't the probability of picking the car not change?

    If he opens 98 doors, leaving two doors, you say there is a 1% chance that he chose the right door from the start. How could it possibly be 99% to 1% if he could have chosen that other door that is left, instead of the one he chose, then it would be 99% to 1% the other way.
     
  7. Feb 11, 2019 at 3:15 AM #6

    PeroK

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    That makes no sense.

    Simulate the experiment if you don't believe me. Or, bet on it! I'd be happy to take your money!
     
  8. Feb 11, 2019 at 3:15 AM #7

    stockzahn

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    In the scenario you described you can start the game with just two doors left, since the host opens one empty door before you even chose one. He could have done that before you enter the room and it wouldn't change anything, so in your described scenario the chance of getting the car should be 50-50. The original problem is based on the assumption that you have to choose one door before the host opens an empty one:

    1. Choose a door
    2. The host opens an empty door
    3. You have to decide if you switch to third door or stick with your initial choice

    In that case you can choose one of three doors: the car door ##C## or one of the empty doors ##E1##, ##E2## - each of the choices is made with the same probability:

    1 x 1/3) You choose ##E1 \rightarrow## the host opens ##E2\rightarrow## change = win, stick = lose
    1 x 1/3) You choose ##E2 \rightarrow## the host opens ##E1\rightarrow## change = win, stick = lose
    1 x 1/3) You choose ##C \rightarrow## the host opens an empty door ##\rightarrow## change = lose, stick = win

    In two thirds of your initial choices changing leeds two winning the car, in one third you lose. Therefore you should change.
     
  9. Feb 11, 2019 at 3:16 AM #8

    PeroK

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    ... Monty Hall rides again!
     
  10. Feb 11, 2019 at 3:22 AM #9
    Sorry I meant the original monty hall problem, I didn't feel like it needed to be written out since I thought everyone knows about it.

    My question was: Is there any logical way of explaining of why the probability of picking the car doesn't change

    and your post doesn't really answer it.

    You choose ##E2 \rightarrow## the host opens ##E1\rightarrow## (why doesn't the probability chance to 1/2, 1/2 here, since we only have two doors left?) change = win, stick = lose
     
  11. Feb 11, 2019 at 3:27 AM #10

    PeroK

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    The simplest answer is that the host can always open an empty door. What he does, therefore, does not affect the existing probability that you picked the correct door.

    The second event, therefore, carries no information about the door you picked. It carries information only about the doors you did not pick.
     
  12. Feb 11, 2019 at 3:29 AM #11

    stockzahn

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    It's that: The original problem is based on the assumption that you have to choose one door before the host opens an empty one. But I just can agree with @PeroK. Take three cards and play the game hundred times - you are done within one hour.

    EDIT: Following these possible paths seems to me as logical as it can be:

    1 x 1/3) You choose ##E1 \rightarrow## the host opens ##E2\rightarrow## change = win, stick = lose
    1 x 1/3) You choose ##E2 \rightarrow## the host opens ##E1\rightarrow## change = win, stick = lose
    1 x 1/3) You choose ##C \rightarrow## the host opens an empty door ##\rightarrow## change = lose, stick = win
     
    Last edited: Feb 11, 2019 at 3:39 AM
  13. Feb 11, 2019 at 3:37 AM #12
    The host will always open an empty door. If it doesn't change the existing probability that I picked the correct door, why doesn't the probability of the remaining door change?

    Mathematically it would be because door 1 is 1/3 then the remaining door would have to be 2/3 to satisfy probability laws.

    Which is why I feel as though the probability of the first door changing would have to happen, or the entire model would have to change.
     
  14. Feb 11, 2019 at 3:39 AM #13
    I mean I guess I will have to do it as I've searched all over math stack exchange and there was a question word for word like mine and I haven't found a single answer that satisfied me. I will try out the experiment but it is hardly mathematical and I wouldn't consider it a proof
     
  15. Feb 11, 2019 at 3:41 AM #14

    stockzahn

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    What's mathematically wrong with decision trees?
     
  16. Feb 11, 2019 at 3:42 AM #15

    PeroK

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    This is what I warned you about: people who say "it's more likely to come up heads next time". Then, when you do an experiment with a real coin and it's shown that it's 50-50 every time, they say "it's mathematically more than 50%, but in an experiment it's 50%".

    What you're saying is that "mathematically the probabilty is 50-50". It's just that reality differs from your mathematics. In which case your mathematics is wrong!
     
  17. Feb 11, 2019 at 3:48 AM #16
    I was talking about the act of carrying an experiment out. I could carry out experiments until I die and it would not constitute a proof under "mathematics" am i right?

    I completely agree, but this is a mathematical problem. It's in a textbook, not the real world. So it only feels like it should be able to be worked out through logic and proven.

    I will have to go talk to my prof in a couple of hours here since I don't feel like I convert my doubts and my understanding from reading text is limited I guess but hopefully I can move along soon. I have wasted a lot of time on this already -_-
     
  18. Feb 11, 2019 at 3:57 AM #17

    stockzahn

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    You are right. Then differently: Why don't you accept the decision tree as mathematical method?
     
  19. Feb 11, 2019 at 4:03 AM #18
    Are you talking about this:

    1 x 1/3) You choose ##E1 \rightarrow## the host opens ##E2\rightarrow## change = win, stick = lose
    1 x 1/3) You choose ##E2 \rightarrow## the host opens ##E1\rightarrow## change = win, stick = lose
    1 x 1/3) You choose ##C \rightarrow## the host opens an empty door ##\rightarrow## change = lose, stick = win

    If so, the problem to me doesn't seem like "if you choose E1," "if you choose E2" .... and so on (if we look at this problem with the 100 door example)

    It's: "You've chosen a door. A door with no prize is eliminated. There are now two doors, you can have the same choice or change it"

    I'm looking for something that satisfies the question "why does the probability not change"

    We can look at this problem after a trail of 100, even 1 million, but this problem is mathematically a trail of just 1. (Actually, on a game show it is too. Youre probably not going to get invited again.)
     
    Last edited: Feb 11, 2019 at 4:13 AM
  20. Feb 11, 2019 at 4:14 AM #19

    PeroK

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    That's not true of applied mathematics and especially probability theory and computer science.

    It's not necessarily about proving something rigorously but about informing your logic.
     
  21. Feb 11, 2019 at 4:39 AM #20

    stockzahn

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    It's: You have three choices of equal probability (paths). Two of the three paths lead to the car, if you change. The third path leads to the car, if you don't change. Therefore in two out of three choices, changing is the winning strategy.

    Because you made your first door choice before the host opens one of the empty doors. The choice was made at 1/3 probability - that doesn't change even another door is opened afterwards (consider the 100 doors example of @PeroK).
     
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