The Blind Door Experiment and Statistics

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SUMMARY

The discussion revolves around the Monty Hall problem, a probability puzzle involving three doors, where one door conceals a grand prize and the others contain junk. Participants initially have a 1/3 chance of selecting the correct door. After one incorrect door is revealed, switching doors increases the probability of winning to 2/3. This counterintuitive result is explained by the fact that the initial choice is more likely to be wrong, thus making the other unopened door the better option.

PREREQUISITES
  • Understanding of basic probability concepts
  • Familiarity with the Monty Hall problem
  • Knowledge of conditional probability
  • Basic mathematical reasoning skills
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  • Study the Monty Hall problem in detail
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  • Explore variations of the Monty Hall problem with more doors
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Students of statistics, educators teaching probability concepts, and anyone interested in understanding counterintuitive results in probability theory.

Lyuokdea
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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, let's 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.
 
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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.
 
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).
 
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 let's make a deal many years ago) as an exercise in chapter 1 of a prob and stats book.
 

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