# The Monty Hall Problem – Part 1

This post is off topic, but there are math and logic buffs out there who might enjoy a discussion of the Monty Hall problem, and I’m hoping to get some feedback on some thoughts I have about a standard solution to the problem.

Fig Leaf Justification
Atheists and Naturalists are a minority group. Most people in the USA believe in God. So those who are doubters and skeptics are generally clear on the idea that the majority can be in the wrong, and also that smart people can believe stupid things. The Monty Hall problem appears to provide some support for these general beliefs. Most people give the ‘wrong’ answer to the Monty Hall problem, and defend the ‘wrong’ answer, including many, many very bright people.

Here is the classical statement of the Monty Hall problem:

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, 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?

The ‘correct’ answer is that one should switch, and that doing so means having a two-in-three chance of winning the car, while sticking to the initial choice (door 1 in this example) will only give a one-in-three chance of winning. Most people believe there is no difference between sticking or switching; both choices are thought equally likely to win the car.

I don’t want to challenge the correctness of the ‘correct’ answer, but I do have doubts about the reasoning that is offered to justify this answer. It seems to me that two important concepts in probability are not being taken into account, or not explicitly dealt with in a standard justification of the answer. I will mention the two concepts first, before getting into the standard justification.

One basic principle of probability, is that the probability of an event can change as new information becomes available. There might only be a 50% chance of rain tomorrow, given the information available today. But come tomorrow morning, if dark clouds are gathering, and if the barometric pressure is dropping, etc., the chance of rain might bump up to 80%. And as soon as raindrops have been falling for a minute or so, the chance becomes 100%, and the probability becomes 1.0 (certain).

A second concept of probability is a bit more difficult (for me) to explain. There is probably standard terminology for this distinction, but I’m not an expert on probability, so I don’t know the technical terms to use. Anyway, there is a difference between statistical probability or probability in the long run, and probability in a specific case, or all-things-considered probability.

If one out of three chemistry students at my daughter’s high school likes to eat at Taco Bell, then the probability that a randomly selected chemistry student from that school likes to eat at Taco Bell is 1/3. But suppose a chemistry student is selected at random, and I know the student who was selected, and I also know that this student hates to eat at Taco Bell. What is the probability that this randomly selected student likest to eat at Taco Bell? One could say that the probability is 1/3, based on the statistical frequency of Taco Bell lovers in that population, but since I know that this particular student hates to eat at Taco Bell, I also know that the probability that he likes to eat at Taco Bell is 0 (i.e. the statement that he likes to eat at Taco Bell is certainly false). Nevertheless, it remains true that one out of three chemistry students at that school like to eat at Taco Bell.

A standard solution or justification of the ‘correct’ answer to the Monty Hall problem goes something like this:

1. The door initially selected by the contestant either has the car behind it or it does not.
2. If the car is not behind the door initially selected by the contestant, then it is behind one or other of the two remaining doors.
Thus,
3. The probability that one or other of the remaining two doors has the car is equal to 1.0 minus the probability that the car is behind the initially selected door.
4. The probability that the initially selected door has the car behind it is 1/3.
Thus,
5. The probability that one or other of the remaining two doors has the car is 2/3.
6. If the contestant sticks with the door that was initially selected, then the probability of the car being behind the finally selected door is equal to the probability of having initially selected the door with the car.
Thus,
7. If the contestant sticks with the door that was initially selected, then the probability of the car being behind the finally selected door is 1/3.

8. The probability that the remaining non-selected door (after Monty Hall reveals a goat behind one of the two non-selected doors) has the car is equal to 1.0 minus the probability of the car being behind the initially selected door.
Thus,
9. The probability that the remaining non-selected door (after Monty Hall reveals a goat behind one of the two non-selected doors) has the car is 2/3.
Thus,
10. If the contestant switches from the initially selected door to the remaining non-selected door (after Monty Hall reveals a goat behind one of the two non-selected doors), then the probability of the car being behind the finally selected door is 2/3.
Thus,
11. The probability that the finally selected door will have the car is greater if the contestant switches to the remaining door rather than sticking with the initially selected door. (from 7 and 10).