6 responses to “A fun (and potentially infuriating) mental exercise”

  1. Chris

    By getting an opportunity to choose another door you’re being given 33% back on your odds, making your odds 66% of choosing the right door. Staying with the door leaves you only with a 33% chance. There was a great example of this in the movie “21”.

  2. dpilling

    The easist way for me to think about it Chris is that there was a 2/3rd chance the prize was behind one of the two doors you did not pick. When the host opens one of those two doors, all of that 2/3rds probability is left with the remaining single door you did not choose originally. Therefore, if you switch doors, you have a 2/3rd chance of finding the prize.

    Did not see 21, but will check it out.

  3. AJ Choy

    The thing about probability is that only the curent state matters. Everything that led to the current state is history and cannot change the present.

    The current state is that there are two doors, each with a 0.5 probability.

    I realize this is an IT website, and that none of you are mathematicians, but still, you guys are smart and should know better.

  4. dpilling

    Actually AJ, what matters is the path taken to reach the current state, not just the current state, because the host’s opening one of the doors is not an indepedent, random event. The host’s opening of a door that does not contain the prize reveals new information that must be factored into the probabilities.

    For evidence, here are some links for you. The first is a peer reviewed paper on how human congnitive processes consistently get the answer to the Monty Hall dilemma wrong:


    The second is a simulation of the Monty Hall dilemma that my colleage ran over 5,000 observations. The version posted is static, but if you would like, I can send you the dynamic version so that you can run it yourself. It shows quite clearly that there is a 2 to 1 advantage to switching doors.


    Don’t fret, you are falling prey to a common cognitive error.

  5. Varuna

    Please allow me to step through the logic.

    The first assumption is that the probability of the prize existing behind doors 2 or 3, should be 2/3. Correct.

    Then, when door 2 is shown to be empty, you transferred the probability attached to door 2 across to door 3. Oops, not correct.

    The probability attached to door 2 (1/3) does NOT magically transfer across to just door 3 because all non-opened doors are equal therefore door 2’s probability transfers across to both door 1 and door 3. In effect, door 1 and door 3 get a probability top up together.

    If you still disagree, which is likely, perhaps you have the inclination to answer my extension to the puzzle.

    Assume two people were playing this game at the same time, independently and unknown to each other. Both players can co-exist without spoiling the puzzle as it is just a shift in perspective.

    There are still 3 doors. Player A chooses door 1. Player B chooses door 3. When it is shown that door 2 is empty, what happens under your model?

    For player A, door 3 now has a 2/3 chance.
    For player B, door 1 now has a 2/3 chance.

    Door 1 and door 3 cannot each have a 2/3 probability, so something is wrong somewhere. Can you explain?

    My solutions works for this scenario, and I’m waiting eagerly to here your explanation. I realize your time is limited as you have a family and a business to run (and so do I), but for some reason this puzzle has found traction with me.

    P.S. If you can crack this two player problem with your model, then what-if there was player C who selected door 2 to begin with. This is a special case because it has perfect symmetry between door 1 and door 3. When door 2 is shown to be empty, if player C was allowed to choose another door, there is no reason why door 3 should have a 2/3 probability.

    The key to solving this puzzle is to realize that there are many points of view that can co-exist, and to show that your explanation is water tight under each and every point of view.

  6. dpilling

    Thanks for the thought-provoking response. Actually, I do believe that the probability (1/3)shifts from door #2 to door #3 in the first sceanio you outlined. It is not however, magic. Again, I’d refer you to the wikipedia post on this, which is rather comprehensive and outlines exactly how the 1/3 probability from door #2 is shifted to door #3.

    As for your two player scenario, I’m no theoretical mahtemaetician, but I believe the issue with the scenario is that it excludes all scenarios (1/3rd of them in fact) where the prize is behind door #2. The host cannot open this door #2 as a result. Because the host cannot open this door, the analysis applies only to that very specific case. This changes hte problem from the original problem presented.

    One could, however, think of player A and player B as two separate observations, where the host chooses to open door #2; and yes, in each case, the probability of sucess from switching is 2/3rds.

    But don’t take my word for it. Sometimes the best way to understand the non-intuitive is to experience it. For this reason, I’ve posted a number of links and simulations on my blog (http://www.derekpilling.com). In some of them, you can play the game yourself. In others, you canh choose a strategy and simulate it over a series of observations. I recommend trying the strategy where you keep the original door you have chosen. You will find that you win 1/3 of the time. You will find that the strategy of swithing wins the prize 2/3 of the time. Should those not convince, you do a quick search on “Monte Hall dilemma”; you will find a wealth of information posted online.

    To loosely quote Buffet’s Partner Charles Munger, “You seem smart and I’m right, so eventually, you will agree with me. 🙂