Know How The Monty Hall Problem Impacts Your Split Decisions

If someone told you about a prize behind one of three doors, you would have about a 33 per cent chance of guessing the right one. Before you get the answer, you're told that the door you didn't pick was a loser. What are your chances now?

Photo by Marc Falardeau.

You might think you now have the same chance, but what you should really do is change your choice. In fact, according to one detailed and helpful rundown of what's known as the Monty Hall Problem:

So by changing doors you double the probability of winning the prize from 1/3 to 2/3.

If you don't switch, the probability remains 1/3.

If you are not convinced yet, play the game with a colleague as the host and try 30 times with and 30 times without switching. Then count the number of successes in each condition.

The principal applications of the Monty Hall conundrum -- that new information does in fact change what seems like simple odds -- are in A/B testing, scientific results and other maths-geek uses. But whenever you're asked to make a choice at picking the right item, and you can figure out somehow that choices you didn't pick are losers, keep Let's Make a Deal and Mr Hall in mind.

The famous Monty Hall problem and what it has to do with A/B testing [The Visual Website Optimizer] And Behind Door No. 1, a Fatal Flaw [New York Times]


Comments

    Am I missing something? Isn't this obvious? Or are you saying the prize could move prior to the 2nd guess?

      3 doors: so each door has a probability of 1/3 of being the prize. Let's ignore the game show for a minute: you pick a door, you open that door, you have a 1/3 chance that it will be the prize, as do the other doors.

      OK, so back to the show. You choose a door. So, the chances are: 1/3 that your door is a winner, and 2/3 chance that your door isn't the one. So if we group the doors, into 'yourdoor' and 'notyourdoor'. youdoor = 1/3, notyourdoor = 2/3. If you got to choose a group of doors, and open all the doors in the group to find the prize, you'd choose the group with more doors, yeah, so you'd get a higher chance of getting the right one. (1/3 + 1/3) versus (1/3). This is effectively what they do when they open the door without the prize behind it- you now just know which door of 'notyourdoor' to open!

      Not a guaranteed win, but 2/3 is bigger than 1/3.

      It's really confusing why it doesn't suddenly change to 1/2 1/2 when you eliminate one door, but it does work out in practice, they did this one on Mythbusters once.

        The other important thing is that humans have poor intuitive understanding of probability. Given this choice, without previous knowledge of the problem, the majority of people will choose not to change their choice, reasoning that they have a 50/50 chance of being right when in fact they're more likely to be wrong. They actually become more confident in their original decision once one of the incorrect answers is eliminated.

    Mythbusters did this ages ago. Don't know what episode, but I'm pretty sure that they confirmed it, or it was at least plausible.

      Yeah they did that one last year in Season 9? I think it was.

      Marilyn vos Savant put this forth in 1990 to a lot of controversy. She is one of the most brilliant minds in the world so Mythbusters just proved what every one in the mathematics community already knew.

    Interestingly, if the 'host' chooses one of the doors you didn't pick at random, and it's revealed that it wasn't the winner, there's no longer any advantage in switching.

      That's incorrect, watch the mythbusters ep as they do a good job of explaining it... with pictures.

      wrong Kieren the whole problem relies on the host picking one that wasn't the winner.

        Actually Bob and Aztech, you are both wrong. Kieran said "if the host RANDOMLY manage to choose a non-winner". In this case the dependence that makes the Mony Hall problem counter intuitive is lost, and the advantage is gone.

        Of course, this also means we are no longer talking about the Monty Hall problem.

    It's easier to imagine if you have 1000 doors. If you pick one door and then 998 doors that don't hide the prize are opened, does the prize lay behind your original pick (1/1000) or the single other remaining door (999/1000)?

      Of course, it relies on the host never opening the prize door.. Which is an odd condition, but nevertheless.

        The point of the game show is that the host knows where the prize is! :)

    Also nicely explained in "21" with Kevin Spacey.
    Link: http://www.youtube.com/watch?v=cXqDIFUB7YU

    I will never get my head around this thing - I just don't get why it's not 1/2 chance.

    What if two people are playing, and there are the three doors; A, B, and C. Person A chooses door A, and Person B chooses door B. The host reveals that the prize isn't behind door C. So, theoretically, they should swap picks to increase their chances? How does that work?

      Furthermore, what if you flip a coin after the initial wrong answer has been removed?

      Now that it's 50-50, surely a coin flip gives you a 50-50 chance of getting the correct door? What if the door the coin 'chooses' is the same one that you initially chose? Does that mean the coin just picked a 1/3 chance? How does that work?

        Don't worry- this did my head in. I'm a visual person, so I got it after drawing it, which I shall attempt to do for you here with spectacularly bad ascii art.

        [o][o][o] our three doors. Pick one

        yours: [o] not yours: [o] [o] now, you understand that your door has a 1 in 3 chance of having the prize, right? So each of the doors in group 'not yours' has a 1/3 chance, there is a 2/3 chance it will be in this group, not in the door you chose.

        So, the options: (P being prize, X meaning empty)
        yours: [P] NY: [X] [X]
        yours: [X] NY: [P] [X]
        yours: [X] NY: [X] [P]

        Ignore the host opening the door for a sec: what if Monty said: "I really like you chump, and I want you to win, so if you want, you can have BOTH of these other doors, or you can stick with your one door." - phrased like that, you'd want the 2 chances, right? This is effectively what happens when he shows you an empty door that isn't yours.

        Confused as to how your two player version works. Also, flipping coin is not relevant- yes, it's a 50/50 which one will land up, but the coin has no influence over the size of the groups. If you rely on the coin, you have a 1/2 chance of picking the one door group or the two door group.

        A: [X] B: [X] C: [P] - in this case, Monty can't show them the empty door. C HAS to be empty for this to work, so if the prize can't be behind C, then it's 50/50 whether the prize is behind A or B. In the problem, Monty always shows an empty door, which he can do, because you've only chosen one door.
        A: [P] B: [X] C: [X]
        A: [X] B: [P] C: [X]

        Let's say A and B are a couple, choose a door each, then all the doors are opened at once: they have a 2/3 chance of winning, whether they switch with each other or not.

      Think about this instead: after the host shows you a losing door, if you switch the only possible way for you to lose is if your first choice was the winner (1/3).

      If you originally picked either of the losers (2/3) then switching definitely gives you the winner.

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