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.
7 thoughts on “A Brief Monty Hall Problem Digression”
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.
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.
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.
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
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.
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.
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.
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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