A Brief Monty Hall Problem Digression

At lunch at the Spoonbill Bowl on Saturday, I was privileged to volunteer with a group of students, faculty, and researchers. It was a long day. Lunch was provided, and I got to sit down with a colleague and a couple of faculty members from USF St. Petersburg. One of them posed a brain-teaser question. I followed up with broaching the Monty Hall problem.

Just to make sure everyone is on the same page, I’ll briefly describe the Monty Hall problem. In the television game show, “Let’s Make a Deal”, host Monty Hall would offer a contestant an opportunity to win a major prize, let’s say a new automobile. The stage would show three doors (“Door #1”, “Door #2”, and “Door #3”). The major prize is behind one of the doors. Behind the other two are booby prizes, let’s say that they are goats. The contestant is allowed to pick a door. Rather than simply opening the contestant’s pick, Monty would have a door the contestant did not pick opened to reveal a goat. Then, Monty would offer the contestant a choice: she could stay with her original pick, or she could switch to the other door that had not been opened.

The Monty Hall problem poses the question of strategy: Is it better to always stay with the original choice, to always switch to the other remaining door, or does it not matter one way or the other? This question was posed many years ago in a column hosted by Marilyn Vos Savant and gave her months of correspondence as people argued with her advice to always switch. Marilyn was right, of course. The problem and just how counterintuitive the result is has proven a popular topic since then, and Jason Rosenhouse even has authored a book about it.

Back to my luncheon discussion. Me bringing up the Monty Hall problem led to about twenty minutes of trying to explain to one of my lunch companions why always switching was the right choice. It was a microcosm of the entire history of the public history of the problem, and I found it frustrating that I wasn’t able to more clearly and simply put it so that my companion could be convinced of the correctness of the answer. What finally made sense to my companion was that if one enumerated all the permutations, staying won in one-third of them, and switching won in two-thirds of them.

So I decided that I would make up a set of business cards to make future discussions of the Monty Hall problem go faster. Here is my graphic:

While I can’t include all the text that I would like on something the size of a business card, I can use this to quickly demonstrate why switching is actually the better strategy. The card shows all nine possible ways that the game can be played. It also shows that in only three of those does staying with the initial pick work out to a win for the contestant. In the other six ways the game works out, the contestant only wins if they switched.

I think I’ll put a version on a T-shirt.

Update: During lunch today, I tested out my card as a tool on a Monty Hall Problem-naive colleague. Her initial hunch was that staying with the initial choice was the strategy to pursue. I said that I would try to convince her that switching was the correct strategy and produced a card. I pointed out that every possible way the game could go was represented, and in only the top row did staying work out to a win. Within two minutes, I had convinced her of the correctness of the switching strategy. So that’s one data point.

Also, I’ve updated the graphic here. I’ve changed the color scheme. Diane pointed out that it would be hard for color-blind people to distinguish differences in the original. I’ve also added door numbers to make it clearer that each block of three rectangles represents one set of doors. And I added drop shadows to the doors just because I think it looks better that way.

Wesley R. Elsberry

Falconer. Interdisciplinary researcher: biology and computer science. Data scientist in real estate and econometrics. Blogger. Speaker. Photographer. Husband. Christian. Activist.

7 thoughts on “A Brief Monty Hall Problem Digression

  • 2012/03/05 at 12:07 am

    Good idea

    I usually try to argue by suggesting the case with a 1000 doors

    you pick one and Monty opens 998, will you switch?

    This seems more intuitive for most people than the 3 door instance.

  • 2012/03/05 at 6:41 am

    I did try that exact case, and I still was not convincing my companion that with two doors left there was anything other than even odds. At least with the full enumeration, it fits on the card and is relatively easy to survey for correctness.

  • 2012/03/05 at 12:27 pm

    I like to point out that the odds are not even because Monte knows where the car is and will never open that door. This is what makes the 1000 doors argument compelling, and is also what in effect moves the chance of the door that he opens to the remaining door.

  • 2012/03/05 at 11:01 pm

    Ha! Looking back in my files, I found a column I wrote for the Fort Worth Star-Telegram’s StarText online service back in 1991 about the Monty Hall Problem. I wrote the Pascal column, but for some reason decided to go after this in C.


    —-=( PASCAL )=—-

    A column about structured programming, among other things.

    Hosted by Wesley R. Elsberry
    (Subscriber contributions for guest columns welcome)

    No. 24 (~6 Kbytes)

    Topic: Behind the door on the left… (in C)


    The off-topic topic of choice at work today concerned the tempest raging
    around the Parade magazine column “Ask Marilyn.” Apparently, Marilyn has
    stirred up a large controversy (large by Marilyn’s standards, anyway) by
    giving a answer to the “door problem.”

    What is the “door problem”? Well, if you wandered into the show, “Let’s
    Make a Deal,” and managed to do well, at some point Monty Hall would show
    you three doors. “Behind one of these doors,” Monty would say to you
    as you smiled at millions of folks at home from inside your banana suit,
    “there is a brand new car worth $34,995. Behind the other two doors, there
    are a couple of cranky goats. You get to keep whatever is behind the door
    you pick. So pick a door…” You pick a door. “That could be a very nice
    door,” says Monty. “But let’s see what’s behind the door on the left!” At
    this point, the door on the left would be opened to reveal a goat
    attempting to eat the skimpy clothing worn by one of the pretty girls hired
    by Hollywood to handle cranky goats. Having started with three doors, now
    you realize that there are only two doors left unrevealed: the one you
    picked, and one that Monty now will make a deal on.

    “Now you see one of the goats. Are you sure you want the door you’ve
    picked? I’ll let you choose this one over here if you want…”

    The question is: Is it to your advantage to switch over to the door you
    did not originally select?

    Marilyn’s answer, which is (unfortunately) correct, is that it is to your
    best advantage, in a probabilistic way, to switch over.

    Apparently, thousands of otherwise mathematically astute people wrote in to
    tell her she was out of her gourd. The people I work with, who are all
    engineers and pretty bright, were almost evenly divided between doubt and
    credence of this answer. “But when it gets down to two doors, the odds are
    50/50!” was the typical protest.

    In order to short-circuit the longish debates on probability that can arise
    in this situation, I have come up with a short program that will handle the
    general case, and perform a simulation using Monte Carlo (rather than Monty
    Hall) techniques.

    So, generically, you start with some number of doors. You pick one.
    Eventually, you will be given a choice between sticking with your original
    choice or switching to the one remaining unknown door. The prize resides
    behind one or the other. Which strategy should you choose? Sticking or

    /* monte.c 2-22-91

    Monte Carlo Door Problem Simulation
    Copyright 1991 by Wesley R. Elsberry

    This program will do a Monte Carlo simulation of the Monty Hall
    door problem: A good prize lurks behind one of several doors. You
    make an initial choice. The host thens reveals all the unchosen
    non-winning doors. Question: As a strategy, is it in your best interest
    to switch to the door that you didn’t choose originally whose contents
    are still unknown?

    Example invocation: monte 3 1500 543

    which would simulate having three doors at the start, then running
    the scenario 1500 times, and using a pseudo random number generator
    seed of 543.

    This program was written in Turbo C++ 1.0, in ANSI C mode.


    #define NUMDOORS 3
    #define TRIES 1000
    #define SEED 42

    void main(argc,argv)
    int argc;
    char *argv[];
    long numdoors, seed, ii, jj, kk, ll;
    long tries;
    long s_stay, s_go;
    double dstay, dgo;

    printf(“\nMonte Carlo Door Problem Simulation\n”);
    printf(“Copyright 1991 by Wesley R. Elsberry\n”);
    printf(” Usage: monte [#doors tries seed]\n\n”);

    /* get arguments if they are available */
    if (argc == 4) {
    if (numdoors < 3) { numdoors = NUMDOORS; } sscanf(argv[2],"%D",&tries); if (tries < 1) { tries = TRIES; } sscanf(argv[3],"%D",&seed); } else { numdoors = NUMDOORS; tries = TRIES; seed = SEED; } /* seed pseudo random number generator */ srand(seed); /* initialize */ s_stay = 0; s_go = 0; /* simulate for the number of tries */ for (ii = 0; ii < tries; ii++) { /* get the number of the door that hides the prize */ kk = rand() % numdoors; /* Note: this is a horrid kludge to limit the size of the number, and probably wreaks havoc upon the distribution characteristics of the PRNG */ /* pick a door at random */ ll = rand() % numdoors; /* strategy success evaluation */ if (ll == kk) { /* you were a winner all along */ s_stay++; } else { /* you win if you switched */ s_go++; } } /* report the results */ dstay = ((double) s_stay) / tries; dgo = ((double) s_go) / tries; printf("Results using %ld doors over %ld tries using a seed of %ld\n", numdoors,tries,seed); printf("Probability of success when keeping the first choice: %f\n", dstay); printf("Probability of success when moving to remaining choice: %f\n\n", dgo); } *=* --={}=-- ** Copyright 1991 ** ** Wesley R. Elsberry ** ** MC 207547 ** *** Note: The user must determine suitability and applicability of any code or constructs used. Neither Startext nor this author assume any liability for any use or application of ideas or code published in this column. While all code will be checked for accuracy and function, user must verify correct functioning for his/her own purposes. "Turbo Pascal" is a trademark of Borland International. "QuickPascal" is a trademark of MicroSoft Corporation. ***

  • 2012/03/16 at 9:53 am

    If I understand your diagram correctly, the first column illustrates the possibilities if the contestant picks Door #1. In the first row, she stays with #1, and wins, and in the lower two, she stays with #1 and looses. But that doesn’t seem quite right. Let us say that Monty opens #3 to reveal a goat, in that case the bottom line disappears because she can no longer pick #3and you’re left with a 50-50 choice between #1 and #2.

    I realize you’re say that if you picked #1 you would have a 2/3 chance of one of the other doors being correct, and that the 2/3 chance remains in the unknown door you didn’t pick after the reveal. But that doesn’t make any sense to me. Suppose Monty showed you the three doors and said: I will reveal before you choose that #3 is a goat and you must now choose between #1 and #2. In that case it has to be a 50-50 chance, and how can it be different because you had made an initial guess?

    But I suppose there’s a reason I became a Classicist rather than a mathematician.

  • 2012/03/16 at 10:59 pm

    I use the Monty Hall problem as an illustration of how the human brain finds correct assessment of probabilities tough going. That said, your take is not quite it. The column on the left illustrates all three games that are possible when the contestant chooses door number 1. If the contestant guesses right (top game), Monty can reveal either door number 2 or door number 3. If the prize is behind one of the other doors, that corresponds to one or the other of the two other games, so there’s no “disappearing” of the items here. Given an initial choice of door number 1, there’s only three possible games to be played, and the contestant loses two of them if she stays with her initial choice. That’s what the column is supposed to illustrate.

    If someone wandered into the studio and was asked to pick between the two remaining doors without knowing which one had been selected by the contestant, then they have a 0.5 probability of choosing correctly. If they are told which one the contestant picked, though, they can do better by choosing the other door.

  • 2012/03/17 at 8:27 am

    I should note that in the case of a latecomer choosing between two doors, the overall 0.5 probability of success does not imply an unbiased distribution of the prize. One can derive the overall probability as

    p = (0.5 * 0.3333) + (0.5 * 0.6666) = 0.5 * (0.3333 + 0.6666) = 0.5

    and that reflects the underlying situation where there is an unknown (to the late door picker) bias in the prize distribution.

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