Results 1 to 6 of 6

Math Help - Monty Hall Problem

  1. #1
    Member
    Joined
    Dec 2005
    Posts
    117

    Monty Hall Problem

    Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

    Can someone explain why the new probablilities of winning are not 1/2? I've read every article on this problem and still don't understand it! It's a paradox!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,843
    Thanks
    320
    Awards
    1
    Quote Originally Posted by c_323_h
    Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

    Can someone explain why the new probablilities of winning are not 1/2? I've read every article on this problem and still don't understand it! It's a paradox!
    I can explain it (at least the way I was told) but I'm not sure I can convince anyone because I'm not convinced myself.

    Let's change this from 3 doors to, say, 10.
    You pick, say, door 5.
    Monty says, "I'll tell you it's not behind door 7. Do you want to make a new choice?" You say no.

    On and on it goes until there are two doors left, door 3 and door 5.

    It is better to pick door 3 as your final choice.

    The reason is simple enough: Your original choice of a door had a 1 in 10 chance of being correct. Your final choice has a 1 in 2 chance of being correct. The odds are with the other door being right.

    I don't like it, but that's how it was explained to me.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by c_323_h
    Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

    Can someone explain why the new probablilities of winning are not 1/2? I've read every article on this problem and still don't understand it! It's a paradox!
    I love this classic problem.

    Think of this, if you use the strategy of chosing the other door then let us analyze what happens. If you pick the wrong door out of the three then you win. HOW?? Because if you pick the wrong door the other bad is revealed and hence the good door remains. But the probability of selecting a bad door is 2/3 thus, the probability of success is also 2/3. You can also think of it like this. If you pick the good door then you lose. WHY? Because the bad door is revealed and by changing you pick the bad door. Now the probability of the good door is 1/3. Thus, the probability of failure is 1/3, which means the probability of winning is 1-1/3=2/3.
    -----------------
    Another explanation. Since there are equal chances each door is of probability 1/3. Hence, when one door is open and revealed its probability is now zero, because it is definelty bad. But the door you chose is of 1/3 hence the remaining door is 2/3 (sum of probabilities is one i.e. 0+1/3+x=1).
    -----------------
    Think of this. There are 1,000,000,002 doors. You chose one (highly unlikey to win) and the other billion are shown to be false and only one remains. Do you really think by staying the probability is 1/2? No way!
    -----------------
    We can generalize this as, if there are n>2 doors then using this strategy the probability is \frac{n}{n+1}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2005
    Posts
    117
    Quote Originally Posted by ThePerfectHacker
    Think of this, if you use the strategy of chosing the other door then let us analyze what happens. If you pick the wrong door out of the three then you win. HOW?? Because if you pick the wrong door the other bad is revealed and hence the good door remains. But the probability of selecting a bad door is 2/3 thus, the probability of success is also 2/3. You can also think of it like this. If you pick the good door then you lose. WHY? Because the bad door is revealed and by changing you pick the bad door. Now the probability of the good door is 1/3. Thus, the probability of failure is 1/3, which means the probability of winning is 1-1/3=2/3.
    But the contestant doesn't know if he picked the wrong door or not.

    Quote Originally Posted by ThePerfectHacker
    Think of this. There are 1,000,000,002 doors. You chose one (highly unlikey to win) and the other billion are shown to be false and only one remains. Do you really think by staying the probability is 1/2? No way!
    What!? How are your chances of winning not not 1/2? There are only 2 doors remaining. One door wins and the other door loses. This is really counterintuitive. Is there some kind of math proof that you can share to show that your chances are not 1/2?
    Last edited by c_323_h; April 15th 2006 at 05:30 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by c_323_h
    But the contestant doesn't know if he picked the wrong door or not.
    But if he did chose the bad door then he win. Picking the bad door is 2/3
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2006
    Posts
    32

    think about it like this

    Here is the way I learnt to view the problem:

    Put yourself in the position of the host:

    The contestant has just picked a door, and you must now eliminate another door.
    Now, since 2 out of 3 doors are incorrect you would agree that MOST of the time- you (the host) will be left with ONE incorrect door which you HAVE to eliminate hence leaving only the remaining correct door.
    It follows therefore that if the contestant switches, then MOST of the time he will be left with the remaining correct door.

    (Of course, the contestant may by chance have picked the correct door to start with, but there is only 1/3 chance of this happening!)

    Try: http://www.stat.sc.edu/~west/javahtm...MakeaDeal.html which has a Monty hall java applet.

    Hope this helps.
    Last edited by jedoob; April 23rd 2006 at 05:03 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Monty Hall problem with 4 doors
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 9th 2011, 04:24 PM
  2. Monty Hall Problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 5th 2010, 03:50 AM
  3. Monty Hall goes Car-Goat-Key
    Posted in the Math Puzzles Forum
    Replies: 4
    Last Post: October 4th 2009, 05:09 PM
  4. The Monty Hall Paradox
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: December 27th 2007, 05:26 PM
  5. Monty Hall Variation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 6th 2007, 08:14 PM

Search Tags


/mathhelpforum @mathhelpforum