May 28, 2014

Monty Hall problem and data

Tonight’s Mythbusters episode on Prime looked at the Monty Hall/Pick-a-Door problem, using experimental data as well as theory.

For those of you who haven’t been exposed to it, the idea is as follows:

There are three doors. Behind one is a prize. The contestant picks a door. The host then always opens one of the other doors, which he knows does not contain the prize. The contestant is given an opportunity to change their choice to the other unopened door. Should they take this choice?

The stipulation that the host always makes the offer and always opens an empty door is critical to the analysis. It was present in the original game-show problem and was explicit in Mythbusters.

A probabilistic analysis is straightforward. The chance that the prize is behind the originally-chosen door is 1/3.  It has to be somewhere. So the chance of it being behind the remaining door is 2/3.  You can do this more carefully by enumerating all possibilities, and you get the same answer.

The conclusion is surprising. Almost everyone, famously including both Marilyn vos Savant, and Paul Erdős, gets it wrong. Less impressively, so did I as an undergraduate, until I was convinced by writing a computer simulation (I didn’t need to run it; writing it was enough).  The compelling error is probably an example of the endowment effect.

All of the Mythbusters live subjects chose to keep their original choice,ruining the comparison.  The Mythbusters then ran a moderately large series of random choices where one person always switched and the other did not.  They got 38 wins out of 49 for switching and 11 for not switching. That’s a bit more extreme than you’d expect, but not unreasonably so. It gives a 95% confidence interval (analogous to the polling margin of error)  from 12% to 37%.

The Mythbusters are sometimes criticised for insufficient replication, but in this case 49 is plenty to distinguish the ‘obvious’ 50% success rate from the true 33%. It was a very nicely designed experiment.

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Thomas Lumley (@tslumley) is Professor of Biostatistics at the University of Auckland. His research interests include semiparametric models, survey sampling, statistical computing, foundations of statistics, and whatever methodological problems his medical collaborators come up with. He also blogs at Biased and Inefficient See all posts by Thomas Lumley »

Comments

  • avatar
    Wesley Burr

    I was under the impression that Marilyn vos Savant was actually the person who got this problem _right_. The Wikipedia article you linked certainly seems to agree with that. September 9, 1990, her response says that you should always switch (under the condition that the host always opens a door which is a losing option).

    2 months ago Reply

  • avatar

    Not quite sure how I managed it, but on being introduced to this is first year by a psych student, the answer seemed obvious to me.

    It may have been the way they explained it to me – I remember my first response being – does the host know where the prize is? I suspect I just completely ignored the first choice the person made, and just looked at what I saw as two possibilities:

    the doors the host was choosing between either had two consolation prizes, or had a consolation prize and the car. Given they knew which one was where, if one of the two they were choosing between had the car, their choice was forced and the other one had the car. If they know the answer, it’s not exactly random chance.

    This may not, in fact, answer, the problem, but my instinct, based on (I think) this reasoning told me intuitively to switch.

    2 months ago Reply

  • avatar
    David Hood

    My way of understanding it came from imagining the prize was always behind door 1. There were three possible outcomes for changing after door removal.
    1) you picked door 1 initially and the only choice is to change to a wrong door
    2) you picked door 2 initially and the only choice is to change to the correct door
    3) you picked door 3 initially and the only choice is to change to the correct door

    Since you don’t know which door the prize was behind, you do not know if you are currently at door 1, 2, or 3.

    2 months ago Reply

    • avatar
      Thomas Lumley

      That gives the right answer for the right reasons, but it might be hard to defend against a skeptic. The tricky part is that the actual location of the prize forces which door is ’1′ in your argument, but which of the remaining doors is ’2′ is arbitrary.

      2 months ago Reply

  • avatar

    “The stipulation that the host always makes the offer and always opens an empty door is critical to the analysis.”

    In your statement of the problem you state that the host knows he is going to open an empty door. However, that is irrelevant. What matters is whether the contestant knows this. It’s the contestant’s probabilities that determine the best decision for the contestant.

    2 months ago Reply

    • avatar
      Thomas Lumley

      Yes, I should have made that explicit. The host will behave as I describe, and this is known to everyone.

      Without that specification the problem is underdetermined, and outside the context of a game show it would be a very unnatural assumption. That’s one reason intuition doesn’t work well.

      2 months ago Reply

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