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The Blind Door Experiment and Statistics

  1. Sep 4, 2003 #1
    Ok, so my statistics teacher talked to our class about this as a thought experiment, there are three doors, behind two of the doors are junk, the other door has some grand prize behind it, for the sake of simplification, lets say the prize is behind door two.

    So you pick a door, you have a 1/3 chance of picking the correct door. Now they open one of the other wrong doors (if you picked a wrong door, they show you the other one, if you picked the right door they could show you either wrong one) and then give you the chance to repick, Now, there are two doors and one is right, so your chances would appear to be 50%, but if you picked that door out of three, they really are still 33%. Now you are given the chance to pick the other door if you want to, upon making a choice of weather to switch or not, it would appear that your odds are now 50%, yet how did the odds suddenly change if you decide to stay with the door you were originally on?

    Also, my teacher says that if do choose the other door, you will be correct 2/3 of the time, how is this true? Since you are picking again from two, it would seem that you only have a 50% chance of choosing the right door. Can anybody prove that it would be 66% of the time mathematically?

    Thanks for your help.
  2. jcsd
  3. Sep 4, 2003 #2
    ok, well, 5 seconds after I posted this it clicked, actually it clicked right after I walked down to play madden 2004, maybe I should try playing that game more often, is this the correct reason?

    When you are choosing either door, you are actually choosing the other two doors, the chances at the beginning are 2/3 that you will chose the wrong door, and thus whenever you do that, the other wrong door will be shown and the other door closed must be the right one, so according to those statistics, the other door will be right 2/3 of the time. For instance if it is door 2, and you choose 3, then three will be revealed, you switch to 2 and are correct, and if you choose 3, then 1 will be revealed and you will be right if you switch again, and if you pick to and then switch, you will be wrong, so you are right 2/3 of the time and only wrong 1/3.
  4. Sep 4, 2003 #3


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    Yes your right, the imporant thing to remeber is that the right door will not be closed if you don't pick it.

    One easier way to understand it is to imagine the same situation, but with 100 doors (i.e. you pick one out of 100 doors, and 98 out of the 99 doors you didn't pick are closed).
  5. Sep 5, 2003 #4


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    I believe the "consider the case of 100 doors and the m.c. opens 99 of them" was Marilyn VosSavant's suggestion. I first saw the "Monty Hall" problem (he was m.c. of lets make a deal many years ago) as an exercise in chapter 1 of a prob and stats book.
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