Q: How does the Monty Hall Problem work?

For those of you who aren’t familiar with The Monty Hall Problem: You’re on a game show where there is a prize hidden behind one of three doors (A, B, or C), and the objective is to guess the correct door.  After you make a guess the host of the show always opens another door (that is not the one you picked) that has no prize behind it.  You are now given the option to stay with your original guess or switch to the remaining unopened door.  The “paradox” is that if you stay you’ll have a 1 in 3 chance of getting the prize, but if you switch to the remaining door you’ll have a 2 in 3 chance.


Mathematician: In my opinion, the easiest way to understand the Monty Hall problem is this: Suppose there are three doors, A, B and C and you originally chose door A. If you stay with your original door, then the only way that you win is if originally the prize was behind A, which has a chance of 1 in 3. If the prize was originally behind door B on the other hand (which also has a chance of 1 in 3), then when you pick door A, door C will be removed. Hence, if you switch you will be switching to door B, and therefore you will win. Finally, if the prize was originally behind door C (which again has a chance of 1 in 3) then door B will be removed, and if you switch you will be switching to door C and therefore will win. Hence, if you stay with your original door, you win if and only if the prize was originally behind door A. If you switch though, you win if it was originally behind either door B or door C. Since the chance of the prize being behind door A from the get go is 1 in 3, whereas the chance of it being behind either B or C from the get go is 2 in 3, you are better off switching!

This entry was posted in -- By the Mathematician, Brain Teaser, Math. Bookmark the permalink.

13 Responses to Q: How does the Monty Hall Problem work?

  1. john says:

    You omit a critical condition in the statement of the problem. The contestant must know that the host’s decision to give him a second chance (by opening another door without the prize and allowing a swich) was in no way influenced by the contestants original choice. This condition is rarely stated and the problem is even sometimes presented with the oposite implication (as in the movie “21”) which makes your standard answer wrong.

    When the problem is stated such that the standard answer (switching doubles the chance of winning) is correct, the extra condition given is usually that the host always opens another door and that door never has the prize, and the host always gives the contestant the option to switch to the remaining unopened door.

    Without the extra condition, the host could end the game after the first choice whenever he wants (say except when the contestant’s first choice is already a winner). By sticking, the contestant is assured of 1 chance in 3 of winning (whether it be with or without the chance to switch). By switching every time the oportunity is presented, the contestant gives the host complete control of the outcome. If the host wants the contestant to lose, the contestant will never win (but if the host is benevolent and wants the contestant to win, the contestant will win every time).

  2. The Mathematician The Mathematician says:

    Thanks for your comment. You’re absolutely right that this is important, and the way I stated the problem didn’t make it completely clear. I added the word “always” which should help, I think.

  3. Again, the set-up must be precise to clarify the problem.
    After the contestant selects a door the hosts says;”I will open one of the other doors that does not have the new car.” The host then opens a door with a goat behind it.
    Now the odds favor the contestant 2:1 if he switches his choice to the other door.

    I first saw this problem forty years ago in a different form. It has been one of the most challenging (and frustrationg) simple problems I have ever seen . Science and math professors usually got it wrong and argued forever. I finally stopped doing it because we would both leave thining the other person was an idiot.

    The version I used was two black cards and one red card shuffled and dealt face down.
    The contestant selects a card but does not look at it. They agree they have a one in three chance of having the red card. I say, “I am going to show you a black card.” I
    look at the other two cards, turn a black card up, put the other card face down. They then claim they have a one in two or 50% chance of having a red card. Of course they still have a one in three chance–33%.

    Then try it with a whole deck. Deal the contestant one card. What are the ods it is the ace of hearts? “One in fifty-two” they say. “Now I will show you fifty cards that are not the ace of hearts.” I fumble through the deck and expose fifty cards, keeping (usually) the ace of hearts face down in front of me. “Now I have a 50% chance for the ace of hearts”, they say. Me? EXASPERATION!!

  4. Ron says:

    In simple terms….

    The ONLY way to loose when you switch, is if you were right to begin with, which clearly happens 1 out of 3 times. This leaves us with winning 2 out of three times.

  5. Anthony Rose says:

    Wow! Educational! Thank you Mathematician and especially David Thorndill!
    The odds of the prize being behind the 2 doors is 2 out of 3, and when the presenter removes one of his doors, he is helping the contestant with information he did not have before.

  6. fabrice bresil says:

    I dont beleave that , the probability 3/2 is a fake hh! look wy (but in french sorry) .

    -Bon , en gros voila l’esbrouffe selon moi :
    le systeme est a trois case donc au 2ieme coup la probabilité est à 2/3 mais comme le premier coup est un indicateur qui fausse les probabilité réel sa implique un re-équilibré dans se systeme à 3 case c’est à dire que la probabilité 2/3 au 2ieme coup decend à 1/2 se qui corespond finalement au systeme indépendant à 2 cases (conventionel pour conventionel la réalité reprend le dessus). voila un truc oral populaire dans le pdf avec l’iddée de la démonstration :
    http://www.fichier-pdf.fr/2012/12/12/monty-hall/

  7. fabrice bresil says:

    bon maintenant faut pas non plus dire que le principe du jeux monty hall ne fonctione pas ! les 2/3 contre 1/3 si le candidat ne change pas de case est facile à comprendre et c’est vrai puisque le choix de sa case se fait dans l’ensemble des n cases du systeme donc sa probabilité est de 1/n alors que la case du gain est ramener progréssivement par le présentateur dans l’ensemble des 2 cases restantes et forcément sa probabilité est de (n-1)/n se qui conseille fortement au candidat de permuter…. se sont des problemes différent par rapport au conventions initial…

  8. Anthony Rose says:

    fabrice bresil, I think we can agree that the odds of one door being right is 1/3.
    So after choosing a door, the odds of the right door being in the set of the other two doors is 2/3.
    Now, when the presenter takes away the wrong door from the set of two, (not touching the right door, just preventing you from choosing the wrong door), the odds of the right door being there in the set is still 2/3.
    So your door is still 1/3 – because when you chose, you chose out of 3,
    and so the other door is 2/3.
    If you did not have the opportunity to choose 1/3, but only had the chance to choose after one wrong door was removed, then it would be 50/50, because you are choosing out of 2 now.
    It is a help to you when you choose out of 3 and then the wrong door is removed from the set of two. It is not the same as choosing out of two doors.

  9. robert bristow-johnson says:

    i think that Marilyn vos Savant put it in simple and stark terms.

    suppose there were 10 doors, 1 door with a prize and 9 duds. so you pick door A (A is 1 through 10) and Monty knows which door the prize is behind. now suppose Monty has opened 8 doors, all with duds, but skipped over door 5 (which is not door A). so the only doors not opened are door A and door 5. what is the probability that the prize is behind the door that you originally picked (door A) and what is the probability that the prize is behind some other door? but since all of the other remaining doors have been shown to be duds (except for door 5), that is the same as asking what the probability is that the prize is behind door 5.

  10. PalmerEldritch says:

    Another condition that needs to be stated is that if Monty has a choice of doors to open (i.e your initial pick is the prize) he picks a door at random, otherwise it ends up at 50/50.

  11. JJ says:

    To be honest, I’ve never really got the confusion, or why complex math models need to be introduced. I think I have a pretty simple way to explain it:

    2 out of 3 times you will originally pick a door with a goat. Because the host ALWAYS opens a door with a goat, that means that 2 out of 3 times you will be in a situation where you have picked a goat, and the other goat door stands open… meaning 2 out of 3 times switching will give you the car. Simple enough.

  12. Doomsday says:

    Still not getting it I suppose.
    Since sticking with your choice is as much a new choice as switching, both choices have a 1 in 2 chance of being correct, no ?

  13. PalmerEldritch says:

    No, both choices do not have a 1 in 2 chance of being correct. The previous comments explain why.

Leave a Reply

Your email address will not be published. Required fields are marked *

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>