Monty Hall Dilemma

The Monty Hall Dilemma was discussed in the popular "Ask Marylin" question-and-answer column of the Parade magazine. Details can also be found in the "Power of Logical Thinking" by Marylin vos Savant, St. Martin's Press, 1996.

Marylin received the following question:

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?
Craig. F. Whitaker
Columbia, MD

Marylin's response caused an avalanche of correspondence, mostly from people who would not accept her solution. Several iterations of correspondence ensued. Eventually, she issued a call to Math teachers among her readers to organize experiments and send her the charts. Some readers with access to computers ran computer simulations. Years ago I concluded this paragraph with the sentence, At long last, the truth was established and accepted.

In fact truth proved to be more complex. With the accumulated experience, it is safe to assert that the major thrust of the controversy was not directed at Marylin's solution but at her interpretation of Craig's formulation. Indeed, there are two ways to interpret Monty's behavior as described by Craig Whitaker, "... and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat."

One interpretation stems from the disregard of the piece of information conveyed by the phrase "the host, who knows what's behind the doors." According to this interpretation, Craig's statement nowhere says that the host acted on his knowledge of what's behind the doors. He just opens the door behind which (apparently by pure chance) is found a goat. The only reason to mention his knowledge one may think is to avoid the need to mention his lack of surprise at the find. During the whole operation the host remained as nonchalant as he is supposed to be.

Marylin, on the other hand, chose another interpretation. She sensed that there is a reason that host's knowledge of the location of the items behind the doors has been mentioned explicitly. And although it was not followed by an equally explicit statement to the effect that, based on his knowledge, the host always opens the door to reveal a goat, this is what the problem was about. For more on the controversy and its history please check the Monty Hall Problem article at the wikipedia.

There are two simulations based on Marylin's interpretation. One is plain and another with a twist that speeds up the simulation progress and, perhaps adds insight on what's happening with Monty's Dilemma.

Solution #1

There is a 1/3 chance that you'll hit the prize door, and a 2/3 chance that you'll miss the prize. If you do not switch, 1/3 is your probability to get the prize. However, if you missed (and this with the probability of 2/3) then the prize is behind one of the remaining two doors. Furthermore, of these two, the host will open the empty one, leaving the prize door closed. Therefore, if you miss and then switch, you are certain to get the prize. Summing up, if you do not switch your chance of winning is 1/3 whereas if you do switch your chance of winning is 2/3.

Solution #2

After the host opened one door, two remained closed with equal probabilities of having the prize behind them. Therefore, regardless of whether you switch or not you have a 50-50 chance(i.e, with probabilities 1/2) to hit or miss the prize door.

Remark 1

The above simulation tool has the virtue of being quite suggestive - three quantities, viz.,

  • the number of hits
  • the number of wins with no switching
  • the number of losses with switching

are all equal. It's really better to see once...

Remark 2

S. K. Stein in his book Strength in Numbers makes use of the Monty Hall Dilemma to demonstrate a mathematician's approach to problem solving. First run 50 experiments. Next think of the results. (In the following he uses 35 mm film canisters to simulate doors in the stage performance.)

If, after thinking some more about the question, you still are not sure about the answer and are not ready to explain it, then do the following. (Keep in mind that just citing experimental data is not an explanation. The data may convince you that something is true, but they do not explain it.)

Get one more canister and perform a similar experiment, using four canisters instead of three. Put a wad of paper in one canister. After your friend chooses a canister, look in the remaining three and show the friend two empty canisters. The friend then faces a choice between the two other canisters. Carry out the same experiments as before. Think over the results you get. What do they suggest? Do you see a way to explain what happens?

Performing these experiments not only gives you some clues, it also slows you down from the common frenzy of everyday life, so you can focus on just one thing for a period of time.

If you still do not see how to explain what is going on, then use ten canisters. Put the wad in one of them. After your friend chooses a canister, look in the other nine. Show your friend eight empty canisters out of those nine and remove all eight. Again that leaves just two canisters. Conduct a similar experiment.

I am confident that you will solve this problem, so confident that I do not include the answer anywhere in the book, not even in fine print upside down hidden in the back matter. You mill probably, along the way, calculate the fraction of times that switching will pick the car and the fraction of times that not switching will pick the car. Using these fractions, you will be able to explain the brain teaser completely. Then you will have to admit that you can think mathematically. You just needed the opportunity.

Remark 3

Another solution can be obtained via the Principle of Proportionality.

Terry Pascal offered his variant of the solution. Also check Ashutosh Joshi's Description of an approach that helped him come to terms with Marylin's solution. Bruno Barros found a different approach. Yet another view has been presented by Peter Stikker.

One of the solutions that come naturally to me has been sent by Michael Gerard Wilson:

Hi Alex,

It was a while ago that I accepted the idea that switching doors was the correct play every time because it improves your chances of winning, but I had trouble convincing my friends that it was the correct answer. However, a friend of mine just came up with this explanation that I think should really make it obvious.

Let's say that you choose your door (out of 3, of course). Then, without showing what's behind any of the doors, Monty says you can stick with your first choice or you can have both of the two other doors. I think most everyone would then take the two doors collectively.

Kenneth Kaplan found that counting the losing probabilities helps arrive at the right conclusion:

I'm sure others have lent this insight, but I've always found it easier to consider the chance of losing when (1) you don't switch, and (2) you do.

You'll lose 2/3 of the time if you don't switch. If you do switch, then the prize is still available 2/3 of the time, and you'll make the wrong choice 1/2 the time, so the chance of losing when you switch is only (2/3)·(1/2) = 1/3. You've halved your chance of losing = doubled your chances of winning by switching.

A similar argument with a credit to Erich Neuwirth has been published in The College Mathematical Journal (vol 30, no 5, November 1999, p. 369. Erich put it this way:

I think I have found the shortest possible solution. Just imagine two players, the first one always staying with the selected door and the second one always switching. Then, in each game, exactly one of them wins. Since the winning probability for the strategy "don't switch" is clearly 1/3, the winning probability for the second one is 2/3 and therefore switching is the way to go.

Keith Devlin has come with a new twist to the problem:

  ... suppose you are playing a seven door version of the game. You choose three doors. Monty now opens three of the remaining doors to show you that there is no prize behind it. He then says, "Would you like to stick with the three doors you have chosen, or would you prefer to swap them for the one other door I have not opened?" What do you do? Do you stick with your three doors or do you make the 3 for 1 swap he is offering?

Marcus Bizony came up with a device to help overcome a psychological barrier many encounter pondering the problem:

Or consider this approach: when you nominate a door you split the objects into His and Mine. He has two and you have one, so he probably has the car ('probably' here meaning 'more likely than not'). Now he shows you a goat - you always knew that he had at least one, so you have learned nothing and therefore still think he probably has the car. The only difference now is that you know WHERE the car is if he has it. So you argue: he probably has the car, and he certainly would put it THERE..... so it probably is THERE. Clearly you must change your choice.

Multi-Stage Monty Hall Dilemma

"In the three-door Monty Hall Dilemma, there are two stages to the decision, the initial pick followed by the decision to stick with it or switch to the only other remaining alternative after the host has shown an incorrect door. An intriguing extension of the basic Monty Hall Dilemma has been provided by M. Bhaskara Rao of the Department of Statistics at the North Dakota University. He analyzed what happens when the dilemma is expanded beyond the two stages. The number of stages can be as many as the number of doors minus one.

"Suppose there are four doors, one of which is a winner. The host says:

"You point to one of the doors, and then I will open one of the other non-winners. Then you decide whether to stick with your original pick or switch to one of the remaining doors. Then I will open another (other than the current pick) non-winner. You will then make your final decision by sticking with the door picked on the previous decision or by switching to the only other remaining door.

"Now there are three stages, and the four different strategies can be summarized as follows:

Stage123Probability of winning

"People who accept the correctness of the 2/3 solution in the basic Monty Hall Dilemma might assume that one does best by switching in both Stage 2 and Stage 3. However, as shown here, the counter-intuitive solution to the three-stage Monty Hall Dilemma is to stick in Stage 2 and to switch in Stage 3. These remarkable probabilities were published by Rao in the American Statistician. The underlying principle is that in a multi-stage Monty Hall Dilemma, one should stick with one's initial hunch until the very last chance and then switch."

Three Shell Game

Martin Gardner in his Aha! Gotcha describes the following variant:

Operator: Step right up, folks. See if you can guess which shell the pea is under. Double your money if you win.

After playing the game a while, Mr. Mark decided he couldn't win more than once out of three.

Operator: Don't leave, Mac. I'll give you a break. Pick any shell. I'll turn over an empty one. Then the pea has to be under one of the other two, so your chances of winning go way up.

Poor Mr. Mark went broke fast. He did not realize that turning an empty shell had no effect on his chances. Do you see why?


The problem is actually the same but looked at from a different perspective. Since Mr. Mark has made his choice no Operator's action can change his chances. So, to me at least, the Shell Game makes it pretty obvious that unless you switch in the Monty Hall Dilemma (i.e. if you play the Shell Game), you chances remain 1 to 3. However, if you switch, you select one door out of two.


  1. A. K. Dewdney, 200% of Nothing, John Wiley & Sons, Inc., 1993
  2. M. Gardner, aha! Gotcha. Paradoxes to puzzle and delight, Freeman & Co, NY, 1982
  3. J. dePillis, 777 Mathematical Conversation Starters, MAA, 2002, pp. 143-146
  4. J. Havil, Impossible?, Princeton University Press, 2008
  5. M. vos Savant, The Power of Logical Thinking, St. Martin's Press, NY 1996
  6. S. K. Stein, Strength in Numbers, John Wiley & Sons, 1996

Monty Hall Problem

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