Monty Hall
I have always been amused and intrigued by responses to “The Monty Hall Problem”, especially when I talk about it to audiences with a high concentration of engineers and mathematicians. If you are familiar with it, but you’ve always struggled with an unsettled feeling of “this can’t be right”, read further and let me know if my explanation of the solution helps to alleviate the discomfort. If you are not familiar, I guarantee you will give your brain a workout by reading on.
First posed to statisticians in 1975, “The Monty Hall Problem” is well-known among academics because it still sparks debate. Many seem to think that disagreements about its solution stem from issues in the clarity of the problem, but I contend that it really stems from human flaws in the way that we process information.
I often discuss this problem in statistics and cognitive psychology courses for several reasons. It is a great exercise in probability calculation and it can be used to teach basic mathematical modeling (and its purpose). An added benefit, since almost all of my students were psychology majors, is that it also illustrates a flaw in human cognition as well as a pattern of problem solving. Even a knowledgeable statistician feels the need to run simulations to see the solution in action. Even then, fully grasping the mechanisms behind the answer often requires brute force cognition.
In general, human beings have a very difficult time wrapping their brains around concepts of probability. It is much like a visual illusion; we know that the lines are parallel/the circles are the same size/there is no motion, but we can’t make our brains process it in a way that represents that reality. It’s just not how our visual system works. I hypothesize that one of the reasons that probability is such a difficult field for most people is that it involves theory and models, which are distinct from observations and we must represent them differently in our minds to properly deal with them. Applications of probability often involve switching gears from the realm of models to data or vice versa and this is where I think most mathematicians get side-swiped in The Monty Hall Problem.
The Poser
In essence, here’s the problem:
You are a contestant on Let’s Make a Deal! and Monty loves your creative costume (a teddy bear carrying a human doll), so he calls on you to make a deal. Monty says, “There are three doors – Door #1, Door #2, and Door #3. Pick one and you get to keep whatever is behind it.”
You’ve seen the show (you weren’t just walking down Ventura Boulevard in a teddy bear costume for fun), so you know that it is highly likely that there is a coveted BRAND NEW CAR! behind one of those three doors. If you choose wrong, however, you might end up with an ostrich…
Everyone hopes for a car. Some get donkeys or other animals.
You choose Door #3.
Monty then says, “Let’s see what’s behind Door #1!” and the door opens to reveal one of the many consolation prizes (and product placements), a lifetime supply of Rice a la Roly.
Cool! You might get that car after all!
Well, the show was successful because the shell-game-huckster-style of Monty Hall rarely stopped there. In this case, he does what he often does, offers to let you switch from your first choice (Door #3) to the only remaining option, Door #2.
Should this woman switch?
Should you? Does it matter?
Not the Problem
Before I get into the solution, let me first deflect a common complaint from mathematicians. The most well-known version of the problem, from its Wikipedia entry:
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 No. 1 [but the door is not opened], and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
This version does not specifically state the name of the show or indicate the way that game shows of its era worked. If you have never seen the television show (i.e., you are younger than 35), or any game show of its kind, let me explain. Monty is in control of almost everything that happens. The only thing “contestants” can do is make choices when Monty offers them. As you will see, they had more control over their odds of winning than once thought, but Monty manipulates some of the build-up by choosing which items to reveal at different steps in the game.
Unfortunately, many probability theorists and mathematicians took issue with the lack of clarity in the problem (context is important sometimes). This provides a face-saving ‘other version’ for the geeks who get it wrong the first time. But whenever I hear comments like, “Okay, given this version, that Monty knows where the car is.” I usually think, “Of COURSE he knows where the car is! There is no other way to play the game!” and wish that people were more able to accept that they are just as human as everyone else.
The problem itself is written clearly, though: it specifically states that a door without a car behind it is revealed before you are given the option to switch. If the situation was a fully-randomized, double-blind game (like “Deal or No Deal”), then the option to switch would not even be on the table if the car is behind the revealed door. There would be no problem in that case. Therefore, the problem is a question of whether you should switch in a controlled setting – one in which the only participant who doesn’t know the location of the big prize is you.
The issue of knowledge is a factor in our processing of the problem, but it’s not what Monty knows that matters. It’s what you (the subject of the problem) know.
So, let’s put that complaint behind us and get back to the problem.
The Answer, and How to See it for Yourself
Hopefully, if the problem is new to you, you’ve spent some time trying to solve it instead of going with your first gut feeling, which was probably, “It doesn’t matter.”
It does. You should switch.
If you don’t believe me, try running some simulations. You’ll have to run a lot in order to get a large enough sample to be certain to see the trend, but here are a few ways to do it:
- Use your favorite program (MATLAB, R, etc.). There is a good database of pre-written simulators for this here. I am partial to Excel myself, even though it’s a bit more cumbersome. I just don’t remember enough code to use another program.
- Use a web-based simulator. Do it at least a hundred times, choosing to switch for half of the trials, and keep a tally your results.
- Use a die to simulate the outcome, assigning 1-3 to “Door #2” and 4-6 to “Door #3” (e.g., if you roll a 5, Door #3 is the one with the car). Roll at least a hundred times, choosing to switch for half of the trials (before rolling!). Keep a tally of the results.
What you will see is that switching will result in winning a car in approximately 2/3rds of the trials while staying will only provide a win in 1/3rd of them.
I know what you’re thinking. “But, there are only two doors left, so it should be 50/50!”
Why it is so Difficult to Accept
Human cognitive development is an interesting process. We learn to interpret information from the environment very quickly so that we can respond to that environment, but learning to reason hypothetically takes more time. Even adults with scientific training have a difficult time separating the concept of variables (each has a set of possible values) and data (values which are known).
In practice, hypothetical situations are often conditional (e.g., “If A, then B”). We tend to use information about what is to reason about what could be. We do this because it often works, but it is one of the ways in which our brains can lead us astray. For example, given the premise, “If I study, I will get a good grade on the exam”, what is the most sound conclusion when presented with a good grade? The most common response is, “I must have studied”, but that is not sound. In this premise, studying provides a guarantee for a good grade, but there is no statement that studying is the only way to get a good grade. It does not, for example, read, “If and only if I study…”
In the case of the Monty Hall Problem, the probability of winning is set before you pick a door. No matter which door you choose, the probability is 1/3rd. This is because there is a 1/3rd probability that the car is behind the door you chose given the information you had when you chose it. In reality, the car is behind one of the doors, so the probability it is behind Door #2 is 100% if it is there and 0% if it is not there. Probability is not a useful way to discuss what is or what happened; it is a tool for predicting what is likely to be true/happen.
The new information provided by revealing a loser changes the circumstances and this where we get trapped in our representations of models and data, possibilities and facts.
You had a 1/3rd chance of winning because there were three, equally-likely locations to choose from. It seems as if cutting the choice down to two should change the odds of winning to 1/2. It seems that way because we are focused on the probability that a given piece of information is true (e.g., that the car is behind Door #1) and not the probability that an event will occur (e.g., that we will win the car). The probability that we will win the car relies on the number of possible states of reality. This, in turn, initially relies on the number of locations for the car. When the situation changes, we try to adjust probabilities based on possible locations (which has changed) rather than on the number of possible states of reality (which has not).
Basically, when Monty makes the second offer, the offer is to switch from the door we have (#3) to the door we don’t have (#1 or #2). It does not matter that only one of those doors is left; there is still only a 1/3rd chance that our door has the car and a 2/3rd chance that the set of the other two contains the car.
Getting Un-Stuck
If you change the way you represent the problem from the beginning, the solution might seem more reasonable. Specifically, instead of thinking in terms of assigning probabilities to doors, think in terms of assigning probabilities to outcomes: winning versus losing.
Three possible states of reality, each with one winner and two losers
Let’s go step by step…
Monty asks you to pick a door from three choices. Behind one of those doors is a car. There are three possible locations and it must be in one of them, so there are three possible states of reality.
You choose to bet on Door #3; there is a 1/3rd chance that you will win the car.
There is a 2/3rd chance that you will not win the car.
This would be true no matter which door you chose.
Monty reveals that one of the remaining doors is a loser. At least one will be a loser since there is only one winner and you can choose only one. The car, however, does not move. Even though there are only two locations left, so there are still three possible states of reality. What’s changed is that we now know more about each of those possible states (there are fewer locations for the car to be):
So, if we model the problem in terms of the probability of winning with Door #3, the model itself does not change after the losing door is revealed. What changes is that we would no longer want to choose that door, so it is no longer among our options. This leaves us with only two options: keep the door we have or switch to the remaining door. The odds of winning/losing with Door #3 have not changed, but eliminating an option allows us to make a better choice – switch.
Barbara Drescher is a former educator and researcher, having taught research methods, statistics, and cognitive psychology at CSU Northridge for a decade. At ICBSEverywhere.com, Barbara evaluates claims and studies, discusses education, and promotes science and skepticism.
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I find it intriguing just how often the probabilities of the MHP continue to stump so many people. I’ve always found the correct answer of 2/3* to feel intuitive, but I certainly understand why it so often leads people astray.
(*= using the standard assumptions about the game for the sake of simplicity)
I do have a bit of an issue with some of what you wrote, though. Specifically: “Unfortunately, many probability theorists and mathematicians took issue with the lack of clarity in the problem (context is important sometimes). This provides a face-saving ‘other version’ for the geeks who get it wrong the first time. But whenever I hear comments like, ‘Okay, given this version, that Monty knows where the car is.’ I usually think, ‘Of COURSE he knows where the car is! There is no other way to play the game!’ and wish that people were more able to accept that they are just as human as everyone else. […] The issue of knowledge is a factor in our processing of the problem, but it’s not what Monty knows that matters. It’s what you (the subject of the problem) know.”
I find it troubling that you so eagerly dismiss real mathematical issues with the nature of the problem, panning them as a face-saving gesture. (Yes, all those poor, silly math nerds just didn’t understand what they were dealing with…) You mock one assumption about what concerns may be present in the wording of the initial problem, then proceed to ignore the main concerns that are *actually* raised. I guess it’s not a complete straw-man, since the state of Monty’s knowledge is a legitimate inquiry, but I rarely see someone look so proud of themselves after tilting at windmills.
The problem stems from your over-simplification. The standard method of explaining the probability relies on several assumptions, of which Monty’s knowledge of the car’s location is one. (Contrary to what you claim, it is absolutely possible to play with Monty lacking that knowledge, but from a practical standpoint it simply wouldn’t make sense to, so we tend to dismiss it. I don’t have a problem with making the assumption, but it is still nothing more than an assumption. Nothing in the way the problem was stated requires it.) The assumptions, taken together, comprise something crucial: the *rules of the game*. Those rules are unstated in the problem as written, and that is what the “probability theorists and mathematicians” take issue with.
For example: all we know so far is that, of doors {1,2,3} we chose an arbitrary door X from the set, then a door Y was revealed and was not the prize. We do not know whether the rules specified that a door Y (which is not the prize) MUST be opened under all circumstances after we chose X, only that it happened in this particular case. If the reveal of Y is not a standard, the 2/3 probability on switching does not hold in the instances where Y has been revealed. It could be that revealing a Y is more common when X is the correct door, and that when X is not correct we will occasionally have the host simply reveal our error and the game will be over. This is important because, well, that’s how the real Monty Hall ran the actual game! The show metered its chances of the contestant winning in order to control against excess loss.
In the end, the MHP is a great way of showing people that common intuitions about probability are often wrong, but you should be clear that you are explicitly working with a simplistic version that makes certain key assumptions. Instead, you hand-waved legitimate objections away, mocking those who raise them, and pretending that the problem is written clearly (and by extension that those who disagree are just wrong) when it is most assuredly NOT written clearly (the paragraph cut out by the ellipses in my quote above even begins with “The problem itself is written clearly, though”) and it is plainly incorrect to act as if it is.
Incidentally, I have a method for explaining the basis of the MHP to those who still seem to have trouble after a direct explanation. The key, I’ve found, is to exaggerate the probabilities without making the numbers too large.
Here’s what I do: Take a full deck of 52 standard playing cards. Shuffle them thoroughly, then hold them fanned out so you can see their values but the person you’re trying to convince cannot. Tell them the object is for them to get the Ace of Spades in order to win. Have them pick one card and place it face-down on the table without letting them see what it is. Then pick one card of your own and place it face down. Be sure that between your two cards, one of them was the Ace of Spades.
Then take the rest of the fanned-out deck and spread them out face-up and tell them to confirm that the Ace of Spades is not present in the remaining 50 that have been revealed. Ask them which is more likely: that they picked the Ace on their initial pick or that your card is the Ace. Ask them if they want to switch before turning over the cards to see if they’ve won. Ask them to explain their choice.
Stirling, Sarah and I were just talking about the MHP a couple weeks ago, it is really hard to understand (at least for me). Stirling (who is just beginning his math major) states that you are more likely to have chosen wrong the first time, so you always switch. That seemed to make more sense to me.
But I love the 52 card demo. I think that makes a lot of sense and you could even try it with less cards until they understand.
The one I like is the one I was told when I first heard of it: Consider a situation where there aren’t 3 doors but 100 doors. After you have made your selection Monty Hall reveals 98 incorrect choices leaving you with only two doors. Do you switch?
Clearly you do, as the chance of your original door being correct was 1/100. The odds were in favour of you being wrong and therefore are now in favour of being right if you switch.
Honestly, for me personally, that doesn’t make a difference. Without the explanation above, it’s the same end-result: it appears to me, if I don’t grasp the above, that the odds are 50/50 at that point, and makes switching irrelevant. Barbara’s visual was the first thing that made it easy for me to truly grasp it. Simply increasing the number and then ending up at two doesn’t change the part that seems to confuse the average person about this.
The clearest way I’ve seen to explain the outcome:
After you choose a door, Monty says “I could open a door with a zonk behind it, because there’s one car and two zonks behind the doors, and offer you the chance to switch. But I’m lazy, and the zonks are worthless, so I’ll just give you the chance to trade your one door for BOTH of the remaining two doors.”
Now it’s easier to see that switching wins.
After reading about this problem in Marilyn vos Savant’s column years ago, I wrote a simple BASIC program to test it and was delighted to see the 2/3rds advantage of switching doors proven empirically. I told one of my brothers about it and he refused to believe it. He wrote his own BASIC program and “proved” that the odds were 50/50. I asked him to send me the program and discovered that he had included a line of code that forced the answer to come out his way … not intentionally, but as a result of his steadfast, but subconscious bias. It wasn’t until I pointed out his trick to him that he finally accepted Marilyn’s math.
And once again we see the benefits of the peer review process!
The real issue with people properly understanding the problem is it’s rarely explained properly. I’ve heard people use the example of 1000 doors instead of three to drive the point home, but then they still don’t emphasize the REAL issue — which is that it’s not random! Most people that try to model it probably assume the placement of the car is random, and that the door Monty chooses to open is also random — but neither one is! The only thing that’s really random is the contestant’s initial choice.
It’s important to bear in mind Monty is actively trying to make the contestant lose (I assume). In other words if the person picks door number 1, and the car is actually behind door number 1, I suspect Monty is going to pull out all the stops to try to get you to switch from door 1, including body language, emphasis on certain words, and even revealing the fourth door/curtain they occasionally sprung on the contestant, just to increase the odds further against the contestant sticking with the door they chose.
There’s a problem with the usual setup of the problem though, because all the times I watched the show, I don’t remember there being a “super” prize and then two joke prizes — there was usually a “great” prize, a “joke/useless” prize, and a “nice” prize as a consolation. If the great prize was a car worth $2000 in 1970s dollars, there’d usually be a prize of a new washer and dryer, or a trip somewhere, worth maybe $300-400. I doubt Monty really tried to steer people to the joke prize, I imagine he tried to steer them to the medium, consolation prize — they leave happy, the sponsor is happy, Monty doesn’t go to bed with a guilty conscience, etc. At least if I was the producer of the show that’s what I’d try to do.
Personally I like Ryan’s use of the cards, but I think 52 is more than you need to make the point. I think 10 would be sufficient. Even only one or two additional ones change the equation enough to make the solution more apparent.
Now here’s a question to ponder: do folks think Monty, and the show’s producers and sponsors understood about the 1/3-2/3 different in probability? Or did they go through the run of the show blissfully unaware of the math?
I understand the math in this problem and find it fascinating. Yet, I believe there is a fallacy in the calculation that hasn’t been addressed. The fallacy is readily apparent if we change the game just a wee bit.
Let’s say there are two contestants in the game instead of one. Each contestant may pick any door he or she prefers. If they pick the same door and stick with their choices, each contestant would win whatever prize is behind that door. However, one contestant chooses door # 2, and the other chooses door #3. Monty then reveals that the prize behind door #1 is a zonk. Now he gives each constestant the option of switching doors.
According to the theory that Barbara lays out, each contestant has a two-thirds chance of winning the big prize by swaaping doors. How can that be?
I maiintain that this conundrum exists even when only one contestant is playing the game. If the contestant had chosen door #2 initially, he or she would have greater odds by switching to door #3. But, if that contestant had chosen door #3 initially, he or she would be more likely to win by switching to door #2.
There is no fallacy in the argument. The issue with having 2 contestants is that the unpicked door will contain the car 33.3% of the time, which Monty can’t open or if he does both contestants instantly lose.
Of the remaining 66.6% of the time there is no benefit in switching,each contestant is likely to win 50% of the time, or overall each has a 33.3% of winning.
The increased likelihood of winning by switching (in the regular game) , is because Monty can NEVER open the car door. If the game is changed such that Monty doesn’t know where the car is and accidently opens a goat door, then there is no benefit in switching.