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Can anyone explain the solution of the Monty Hall problem? (probability)

  1. Jan 13, 2009 #1
    The problem comes down to this:

    At a gameshow there are three doors. Behind two of them there is a goat, and behind one is a car. You pick a door. The gameshow host, who knows where the car is, then opens one door that you didn't pick, but contains a goat. You are then allowed to change your pick to the other door that is left unopened. Does this increase your chance of winning?

    For a more detailed description plus solution, read here:


    I've read it and the solution makes sense when you look at the possibilities and write them out, but i don't really get it on an intuitive level.

    There is this bit to understand it better:

    "It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats—imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one."

    This does explain on how you have a 1/3 chance of selecting the right one from the start, and how the probability of the car being behind the other one is 1 - 1/3 = 2/3, just like it is 1 - 1/1.000.000 in the other case. Okay. I kind of understand the solution of the problem.

    But i still can't get my head around the fact that one door is shown to contain a goat, let's say, door 3, the car is either behind door 1 or door 2, i.e. a 50% chance. Why is this not true? Is it because our mind tends to ignore additional information that makes it less than 50%, as the 1.000.000 goats example shows?
  2. jcsd
  3. Jan 13, 2009 #2
    That's the trick to it, and why it's such an interesting problem. Human intuition seems to universally fail for the problem. If you ever think you understand it on an intuitive level, you've probably misunderstood it.
  4. Jan 13, 2009 #3


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    And NEVER reveals it.

    That's the important thing and the point which people who have only learnt this for interview questions forget. If the host just opens one of the remaining doors at random then the intuitive answer, that there is no advantage in swapping, is correct.
  5. Jan 13, 2009 #4


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    I used a text for an introductory probability course over twenty years ago which had the "Monty Hall" problem as an exercize in chapter one!
    Last edited by a moderator: Jan 20, 2010
  6. Jan 13, 2009 #5
    They sure knew how to motivate students. :surprised
  7. Jan 14, 2009 #6


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    The most intuitive solution, in my opinion, is to imagine that you play this game over and over, using the strategy to never switch when the host opens the second door. If the probability really is 1/2, you'll win the prize 50% of the times you play the game, and that means that you have the supernatural ability to get your first guess right 50% of the time.

    Without supernatural abilities, the probability must be 1/3.
  8. Jan 14, 2009 #7
    Another way of thinking about it is this:

    Suppose you choose a door out of the three. Then the host asks you if you want to stick with your door, or if you want to pick both of the remaining doors. Clearly, you should switch to the two doors. Then, since there is one car, you know that at least one of those two doors has a goat, and the host just tells you which one. Does it really matter that the host asked you to switch before opening one of the other two doors, rather than afterwards?
  9. Jan 14, 2009 #8
    It does. This is why you have to pick lottery numbers BEFORE the drawing.

    I think a key point in the scenario is that your first choice of door directly affects what options are available to the host. You initially have three choices. You pick one. In two of those three cases, you'll initially pick the wrong door, leaving the host with no choice (he MUST open the other wrong door). In the third case, (that is, when you initially pick the correct door), you allow the host a choice he did not have. He can now choose between either of the two bad doors.

    If instead we are playing a game where the host simply reveals a bad door at the start, he is ALWAYS given a choice between two alternatives.
  10. Jan 14, 2009 #9
    I think it is easier to see in the following:

    In your first pick there are two possibilies:
    1) you got the car (odds 1/3)
    2) you got a goat (odds 2/3)

    after the doors are opened, and there is one door left, the odds reverse themselves. this is because left behind the two doors is one goat and one car. in other words, if you picked a goat the first time around, swapping gets a car, and if you got a car the first time, you get a goat.
    since the odds switch places, there is a 2/3 chance that you will get the car after swapping.
  11. Jan 15, 2009 #10
    That's an entirely different situation. In this game, we already know that we can choose to switch or stay. We could make up our minds to switch before we even pick the first door! If the host asks, "After you pick your door, would you like to switch to another door, after I open one with a goat behind it (which you didn't pick at first)?" before you've chosen your door, you could tell him, "Yes," and the game would continue the same way.

    The problem really boils down to: Do you think you initially picked the right door? Or do you think one of the other doors is the right one? Two out of three times it will be one of the other doors.
  12. Jan 15, 2009 #11
    Best explanation?

    At late 80-ies, when the "Car and goats" issue was vividly discussed, there was a competition announced by a popular technology paper in "my country" for best explanation of the "paradox". After thinking hard, I arrived at following:

    "Likelihood the car is behind first selected door is 1/3 and 2/3 it is behind one of the two remaining doors. After the gamekeeper (knowing where the car is) has opened a goat-door among those remaining doors, the likelihood 2/3 is transferred to the only
    remaining door - and therefore should be selected".

    But I never posted that explanation, for some reasons. But some years later a university mathematician thanked me for this explanation that he said made him really grasp this "paradox" for the first time. This explanation is maybe more immediate and "intuition-friendly" than other ways of explaining. :cool:
    Last edited: Jan 15, 2009
  13. Jan 16, 2009 #12
    It makes sense if you assume the host always picks the door without the car and with the goat. Then you should always change.

    It's a matter of assuming the host always picks a door with a goat, because he knows where the car is. Thus you have now two choices stick or change.

    The host can't pick the door you chose so the only option is to pick the one without the car if indeed you picked it. You cash in on your knowledge and kaching!!! If you see a goat then you know that the car is either behind your door or behind the other door thus 1/2.
  14. Jan 19, 2010 #13
    The Monty Hall Problem (Parade Magazine Version)
    This is a problem in logic, with very little probability knowledge required.

    There are three doors, one with a car, two with goats.

    1) You guess a door.
    2) The host opens one of the other doors.
    3) You may switch to the other unopened door.

    1/3 of first guesses are correct, and a switch to any other door awards a goat.
    So, 1/3 of the time a switch is a losing move.

    2/3 of first guesses are wrong, and the only other door to switch to is the car door.
    (You guessed a goat door, and Monty opened the other goat door.)
    So, 2/3 of the time, a switch is a winning move.

    It's that simple.

    (By the way, probabilities can change but they don't transfer or move.)
    Last edited: Jan 19, 2010
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