2.7 Monty Hall Problem
One of the most popular probability problems is called the Monty Hall problem, and is based on the television game show “Let's Make a Deal.” It can take on many forms, but a common form is as follows
Example 1 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 change your choice to door No. 2?" Is it to your advantage or disadvantage to switch your choice, or does it matter whether you switch your choice or not?
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The result is that it is always better to switch , where the probability of getting the car moves up from 1/3 to 2/3 by switching!
Because this problem is particularly unintuitive, we will break it up into smaller pieces. The critical aspect of this is that a change in our assignment of probability to an event must be somehow tied to a change in our information about that event . In order to understand the problem, we must then understand where the extra information is coming from.
Note
Most people will state that, because we are left with 2 choices, it must be 50-50. However there is added information in the system which moves us from knowing nothing about the two choices (i.e. 50-50 chance) to knowing a little bit more about the two choices (i.e. not 50-50 chance).
We will step up to the full problem listed, but for now we explore some simpler versions of the problem.
Two Doors with Information
Example 2 Imagine we have a game with two doors: Behind one door is a car; behind the other is a goat. You pick a door, say No. 1 (but the door is not opened), and the host, who knows what's behind the doors, says that there is a 90% chance that the car is behind door No. 2. Is it to your advantage to switch your choice?
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Initially there is a two-door choice, with no information about either choice, so we assign equal probabilities to the choices: P ( car behind No. 1 ) = P ( car behind No. 2 ) = 0.5
What if the host were a little less direct? Perhaps something like
Example 3 The host, who knows what's behind the doors, points to a door, choosing the correct door 90% of the time and the incorrect one 10%. You pick a door, say No. 1, and the host points to door No. 2. Is it to your advantage to switch your choice?
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This amounts to an identical situation as the previous one - the host is giving you correct information 90% of the time, and we are in a much better position switching.
Three Doors with Information
We return to the three-door case with a slight variation
Example 4 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, says that another door, say No. 3, has a 0% chance of having a car, and that the remaining door (that you haven't chosen - i.e door No. 2) has a 66% of having the car. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
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In this case, switching to door No. 3 would be ridiculous - we know the car isn't there, because the (honest) host knows that it is not there. The host also has told us that there is a 66% chance of the car behind door No. 2, and thus we have P ( car behind No. 1 ) = 0.34 P ( car behind No. 2 ) = 0.66
It isn't the number of choices that is important, it is the information we have about those choices. When you have no information, we assign equal probabilities. When we have information, we can assign non-equal probabilities.
Three Doors Down To Two
Back to our original problem, we have
Example 5 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 change your choice to door No. 2?" Is it to your advantage or disadvantage to switch your choice, or does it matter whether you switch your choice or not?
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The key part is that, no matter what happens,
the host never opens your door the host always opens a door with a goat Given that your first choice, with three equal probability choices (i.e. you have no information about any of the choices), we expect to be correct only about 33% of the time. If we happened to get lucky with our first choice, then the host has a pick of two doors with goats and has some freedom. If we happened to get unlucky with our first choice (and there is a goat behind it), then the host has no freedom at all , because there is only one remaining door with a goat. So, about 66% of the time the host is forced to reveal some of his information, because the door he leaves closed (other than your door) must have the car. Thus, 66% of the time the host is telling you where the car is , just a little indirectly.
Note
Another way to look at this is to imagine a game with 1000 doors, car behind only one, and the host has to open up 998 doors (not yours and not the prize - if the prize is different than yours). Once you pick, say door number 1, and the host opens every door except door 576, and gives you the opportunity to switch is it a good choice? Of course! Ones intuition realizes that my initial 1/1000 chance of getting it right (and thus have the other door have a goat) is swamped by the 999/1000 chance of getting it wrong, and the host being forced to open every door without the prize.
Formally, we need to involve model comparison, so we postpone this particular analysis until Section 5.5.
Adapted from Statistical Inference for Everyone, by Brian Blais (Bryant University), licensed under CC BY-SA 4.0 (dual-licensed under the GNU FDL 1.2 or later; this adaptation uses the CC BY-SA grant). Changes were made; this adaptation is distributed under the same license. License: CC-BY-SA-4.0 .
