The problem
There was (and apparently still is! Wayne Brady runs it now!) a game show called Let's Make a Deal. It had a game in which the host--Monty Hall, after whom this problem is named--would show a contestant three doors. (I'm going to assume it's a female contestant here for the sake of clarity, to distinguish her from the distinguished Monty.) Behind one door was a car; behind each of the other two doors was a goat. The contestant would choose a door, and the host would open one of the other doors to reveal a goat. At this point, the contestant was given the choice to keep her original choice or to switch to the remaining unopened door. If the door she decided on in the end had the car behind it, the contestant won the car. (I don't know if they were allowed to keep the goat if they chose it. Probably they got to substitute another prize.) The question is: is there any advantage to switching doors?
| What do you think? Would you switch doors, or would you keep the same one, or does it not matter? |
Math background
All you really need to know is how to add and multiply. Easy enough, right? You could teach this to elementary school kids. (If anyone wants me to come to their classroom and teach it to their kids, I will do it!)
The paradox
There are three answers that people commonly come up with.
Answer 1: There can be no advantage. The door was originally chosen randomly, and the car is behind a random door. There are two doors after a goat is revealed, so the contestant has a 50% chance of having chosen the correct door.
Answer 2: Switching doors will give you a 2/3 chance of winning. There was a 1/3 chance of choosing the car in the first place, and then a goat was revealed, but that doesn't change the odds of finding the car behind the chosen door. If the contestant switches, there is a 2/3 chance of finding the car behind the other door.
Answer 3: Every door has something random behind it. No matter which door is choosen in the end, there is only a 1/3 chance of winning.
| Which answer of these three do you think is best? Can you figure out why the others are wrong? |
A more precise statement of the problem
- Two goats and a car are randomly placed behind 3 doors.
- The contestant picks any one door.
- If the contestant's door has the car behind it, the host picks either of the other doors. If the contestant's door has a goat behind it, the host picks the remaining door with a goat behind it.
- The contestant is given the option of keeping her door or switching her choice to the remaining unopened door.
- The contestant wins if her final choice of door has the car behind it.
What are the odds of winning if the contestant chooses to keep the same door? What if she switches?
The assumptions here are that all placements of the car are random, all episodes of the show are shown on TV (no bias in selection), and that the host always reveals a door with a goat behind it in step 3 (this is a bias, however). This is critical to the problem. If you assume that some episodes are never seen on TV, then you have an incomplete (and very possibly biased) view of the probabilities. In particular, if you exclude any scenario in which a car might be revealed, you get a different answer than if you don't.
| Take a minute and think about it again. Do you agree with your earlier idea? |
The solution
The more precise statement of the problem should clue you in to what's going on, especially step 3. If the host always had to choose to reveal one of the other doors at random, then each door would indeed be equally likely to have the car behind it. But that's not the case. However, just for the sake of comparison, let's look at each of the possibilities for this scenario. I've drawn these out in a tree form, because it's way easier to understand this way.
In the first column, I've shown the possibilities for which door the contestant first chooses. There is a 2/3 chance of choosing a goat, and 1/3 of choosing the car. In the second branch, I've shown the chances of the host picking either the car or a goat to reveal. Note that this doesn't happen in the game show; I'm just showing it for the purpose of comparison. In the final branch, I've shown what happens if the contestant switches doors or not. If the she wins the car, that's labelled win. If she gets a goat, it's lose.
For each winning possibility, I've multiplied all the probabilities from the left side of the tree to the win, and put that number at the end. For each loss, I put down 0 as the result. To get the total odds of winning by switching, add up each of the switch branches. To get the odds of winning by not switching, add up those branches.
In this case, there is a 1/3 chance of winning by switching, and a 1/3 chance of winning by not switching. There is also a 1/3 chance of the host opening the door with the car, and in that case you're not going to win no matter what you do. So there's no advantage to switching or not. Note that this diagram covers two of the commonly given answers above. If someone claims that there's a 50% chance of winning by switching, odds are she has calculated that the odds of winning and losing are the same and thinks that that means those odds are 50%. She has neglected the chance that it is impossible to win. If someone claims that the chance of winning by is 1/3 no matter what, then she has also chosen to use this diagram, and realizes that there's a chance of not winning.
However, I'd argue that that's not what the problem really is. On the TV show, the host always opened a door with a goat. It would make sense that if the contestant chose a goat in the first place, they always opened the other door with the goat. Otherwise, they'd have to throw out a third of their footage! (However, since we don't know for sure that this is the case, I won't argue too vigorously against it.) This distinction will become important when I discuss the Two Coins Paradox. In fact, it is a fundamental feature of experimental design: you have to know what your experiment is going to be before you do the experiment. In this case, the experiment might be a test to see if the "always switching" strategy works. To figure that out, you must know exactly how the test was done, and in particular, whether or not any possibilities are being discarded.
This diagram shows the correct solution. It is almost the same as the last diagram, except that the option of the host revealing the car is removed. You can see clearly that the odds of winning when you switch are 2/3, while the odds of winning if you don't switch are only 1/3. That is because no matter what your initial choice was, there is always a chance of winning. The bottom branch of the first diagram didn't have that choice.
| Was your first idea right? Did you change your mind to the right answer after the problem was reworded? |
Conclusion
Experimental design is important! You need to clearly define what you're trying to measure, or you can't do statistics on it!
If you clearly define the Monty Hall problem as "no matter what the contestant chooses, the host always picks a door with a goat," then the contestant doubles her chances of winning by switching doors. If you think, "the host opened a door at random, but they just didn't show the episodes where the host revealed the door with the car behind it," then there is no advantage to switching, and the contestant has a 1/3 chance of winning the car no matter what.
This is related to a statistical problem called Berkson's Paradox, and the "selection bias" that leads to the wrong answer is more formally called an ascertainment bias. The difference between this problem and Berkson's is that there is supposed to be an ascertainment bias here.
Sources:
1. Too much TV as a kid.
2. Image of Monty Hall from http://www.letsmakeadeal.com/images/mh-1975.jpg. I'm calling it fair use, for educational purposes.
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