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3 doors probability question / puzzler

  1. Nov 14, 2006 #1
    my brother in law posed the following to me :

    A contestant on a game show is presented 3 doors, one of which
    has a car behind it and two do not. The contestant is asked to pick one. After choosing a door, the host then opens one of the two doors not picked by the contestant. The door opened does NOT have a car behind it.

    The contestant is now given the option of keeping his original door choice, or trading for the other remaining unopened door.

    Now, here is the question :

    Is there a higher probability that the car will be behind the remaining unopened door, behind his original door choice, or neither ?

    ( we have differing opinions ) Can anyone answer this ? Please provide a clear scientific explanation, as we got into a rather heated debate over this. If either door does have a better probability, can it be quantified ?

    thanks in advance
    Last edited: Nov 14, 2006
  2. jcsd
  3. Nov 15, 2006 #2
    the car is behind the remaining unopened door if the original choice was wrong. and the probability of the original choics being wrong is 2/3. so the probability of the car being behind the unopened door is higher than being behind the original choice.

    you can try this: http://en.wikipedia.org/wiki/Monty_Hall_problem
    Last edited: Nov 15, 2006
  4. Nov 15, 2006 #3
    Thanks very much !! ( even though I was wrong !) I insisted
    the new odds were 50/50. But, it the link does explain the solution
    quite well.
  5. Dec 8, 2009 #4
    The answer is the door which he asked if he wants to switch with the second time.Cant tell you the exact probability,but I just know the door was offered twice.(Watch the movie "21")
  6. Dec 8, 2009 #5


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    This is the famous "Monty Hall" problem. That was a TV game show a long time ago which worked this way.
  7. Dec 9, 2009 #6
    manan.vora The answer is the door which he asked if he wants to switch with the second time.Cant tell you the exact probability,but I just know the door was offered twice.(Watch the movie "21")

    As mentioned by mathman, this is a famous problem. It was advanced by Joyce Brothers. The matter is basically very simple, though many mistakes have occurred.

    Of your original choice of three doors, what is the odds that you have actually picked the right door? Obviously the answer is 1/3 as murshid islam says. So changing doors in this case is a loss.

    But the chance that your choice was false is 2/3. So since the announcer has removed the other false choice, by switiching you will win.

    Thus the odds of winning are twice as high--2/3rds-- if you switch.

    If this proves to be a problem, as Joyce Brothers pointed out, you can construct pieces of paper on which you have printed the correct door--but do not show that side--and just run a probability experiment over and over again, hundreds of times if you wish. You pick a door and decide if you want to switch or not, and then turn over the paper. (Joyce said that no one ever did this and failed in time to understand the correct answer.)
    Last edited: Dec 9, 2009
  8. Dec 25, 2010 #7
    OK, I'm not getting this. My brother-in-law tried to explain it to me and I think he and everybody is all wet. Here's what I don't get: After the first door is opened, you, the contestant, has another choice. Basically, you have to disregard your initial choice and make a new choice. So now you have two doors and one prize, which will be behind one of the two doors. The chance is 50-50. It matters not what your original door was. What is the flaw in my logic?

    Another way of looking at this is to take it a step further. Let's say that you pick a door. The probability of that being the right door is one 1 in 3. The gamemaster then opens BOTH of the other doors, which are empty. What is the possibility of the door you chose being the correct door? 1-1, right? Then how is that differant than just one empty door being opened? It would HAVE to change the the probability of BOTH the other doors, wouldn't it?

    One more angle on this. If you choose door one, what is the probability of that door being correct? 1-3. Door 2? 1-3. Door 3? 1-3. Now, He opens door 2. Empty. If the probability on door one doesn't change from 1-3, then niether can the probability for door 3. What if you had chosen door three? Does it change if you don't tell the gamemaster your choice?

    Basically, you have three combinations at the beginning, but that is irrelivent after he opens an empty door. Now you only have two combinations: Door 1-prize, door 3-empty, and door 1-empty, door 3-prize. That is all you have, period!

    I am most interested in hearing you response. It could be that I am all wet.
  9. Dec 25, 2010 #8


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    Well before Dr. Brothers brought it up, I saw this problem as an exercise in an undergraduate probability text- in the first chapter!

    If you found that interesting, you might like to think about these:

    A new family moves into the neighborhood and, since you are interested in playmates for your children, you ask about their children.

    1) You find that they have two children and at least one of them is a girl. What is the probability that the other is also a girl?

    2) You find that they have two children and the older child is a girl. What is the probability that the other is also a girl?

    Explain why these two answers are NOT the same.
  10. Dec 25, 2010 #9


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    I don't know a really simple way to explain what the flaw is (maybe someone else does), but I know a really simple way to show you that what you're saying can't be right: If it really is a 50-50 situation, it means that you have the supernatural ability to get your first guess right 50% of the time. (Imagine that you get to play the game many times, and that you decide before you begin that you will always stick to your first choice).
  11. Dec 26, 2010 #10
    The difference between the two assumes we recognize birth order in the second question. Without conditions, there are 4 possibilities (assume equally probable in the absence other information): BB,BG.GB,GG. BB is ruled out so for the first question, there are three possibilities; two of which include boys so P(G=2|G=1)=1/3.

    In the second question we have information on birth order and we code BG for boy before girl. This is also ruled out so we have two possibilities: GB and GG so P(G=2|G*=1)=1/2 where * indicates first in birth order.
    Last edited: Dec 26, 2010
  12. Dec 26, 2010 #11
    I did some more thinking on this, and I think I have the explaination. You chose door 1, which has a 1/3 chance of being right. The gamemaster will open one of the other doors, but it has to be empty, so there is a 2/3 chance that the prize had something to do with his opening door 2, since there is a 2/3 chance that the prize is behind doors 2 or 3. So, if there is a 2/3 chance that where the prize is made him choose door 2, then there is a 2/3 chance that door 3 is the right door.

    I think that is about as concise an explaination as I've seen. Everyone else just says that there is still a 2/3 chance that you picked the wrong door, which doesn't say much. Either that or they just say go do the experiment.
  13. Dec 27, 2010 #12


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    Yes, 2 times out of 3, he couldn't have opened the other door, and in those cases you picked the wrong door to begin with. This explanation is correct, but it's not really different from the ones you're dismissing.

    Did you look at the Wikipedia page that murshid_islam linked to? It includes several simple solutions.
  14. Dec 27, 2010 #13


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    Note that, in order to be sure he opens a door that the prize is not behind, the host must know where the prize is. By switching after he does that, you are making use of his knowledge.
  15. Dec 28, 2010 #14
    Without conditions, the probability of choosing the right door is 1/3. Once you choose, the host will open one of two remaining doors. Since he knows where the car is, he must choose one of the wrong doors.

    If you chose the correct door the host has two choices. If you chose a wrong door the host has only one choice since he cannot open the door you chose or the right door.

    The probability that you chose a wrong door is 2/3. Therefore, when host opens the remaining wrong door, the probabilities of you winning are 2/3 if you switch, but only 1/3 if you don't switch.
    Last edited: Dec 28, 2010
  16. Dec 28, 2010 #15
    I'm double posting because, while I believe my reasoning in my previous post is correct IF this were a conditional probability problem, I seriously doubt it is. In post 10 I had all the information I needed to calculate the conditional probabilities of a second girl for for both of Halls of Ivy's questions. Here the chooser does not have the information needed to solve for a conditional probability. The chooser is presented with a new situation.

    The chooser only knows that the probability of having chosen correctly has changed from 1/3 to 1/2. While the host knows if he has only one choice of door to open, the chooser does not know that. Since probability is a measure of uncertainty, the chooser has no information as to whether the host had one or two choices of doors to open. Therefore, the chooser's uncertainty is best stated with a probability of 1/2. From the chooser's perspective, there is nothing to gain or lose from switching. I would recommend switching because of this. Calculated one way, there is a better chance of winning by switching. Calculated another way (I now believe the correct way), the chooser has nothing to lose.
    Last edited: Dec 28, 2010
  17. Dec 28, 2010 #16


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    The probability of having chosen correctly doesn't change. That's the point of the problem.

    The rules of the game imply that with probability 2/3 the host had only one choice. Those 2/3 of the time, the contestant will win if he switches. The other 1/3 of the time, the contestant will lose if he switches.
  18. Dec 29, 2010 #17
    One easy way to get more comfortable with the solution to the Monty Hall problem is to imagine that there are 100 doors, only one of which has a car behind it. You pick a door, then the host opens 98 doors, none of which lead to a car. Any argument I've seen for the theory that no advantage is gained in switching doors in the original 3-door problem would also work in this 100-door problem. It is obvious, however, that you should switch doors in the 100-door problem.
  19. Dec 29, 2010 #18
    Why is it obvious?
  20. Dec 29, 2010 #19
    You pick a door, and there's a 1/100 chance that it's the correct one.

    Do you really think you picked the right one, or do you think it's the one other door that the host didn't open?

    In this case there's a 1% chance you picked the right door, and a 99% chance that the other door is right.

    The idea is pretty simple: Everything must always add up to 100%. If I have a 10% chance of choosing a correctly out of 10 doors, the remaining doors each get 10%. Now that I've made that choice it's locked in at those odds. The remaining 90% must be divvied up between the remaining doors no matter how many get removed (as long as at least one unchosen door remains). Finally, if you're given a choice to choose again, as long as one door was removed, you have higher odds of winning by switching to any of the remaining doors (they all have equally greater odds of winning over the door originally chosen).
  21. Dec 29, 2010 #20
    Oh yes, I know that. But the same argument applies in the case where there are just three doors. So I was asking why it would be more obvious in case of 100 doors.
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