** Module: Basic statistics**

- T00. Introduction
- T01. Basic concepts
- T02. The rules of probability
- T03. The game show puzzle
- T04. Expected values
- T05. Probability and utility
- T06. Cooperation
- T07. Summarizing data
- T08. Samples and biases
- T09. Sampling error
- T10. Hypothesis testing
- T11. Correlation
- T12. Simpson's paradox
- T13. The post hoc fallacy
- T14. Controlled trials
- T15. Bayesian confirmation

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There are two ways to slide easily through life: to believe everything or to doubt everything. Both ways save us from thinking.

- Alfred Korzybski

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- Analogical arguments

As a final test of your understanding of probability, try the following rather famous puzzle.

Imagine that you are a contestant on a television game show. You are shown three large doors. Behind one of the doors is a new car, and behind each of the other two is a goat. To win the car, you simply have to choose which door it is behind. When you choose a door, the host of the show opens one of the doors you have not chosen, and shows you that there is a goat behind it. You are then given a choice; you may stick with your original choice, or you may switch to the remaining closed door.

What should you do to maximize your chances of winning the car? Think about it for a while, and when you have decided, read the two arguments below and decide which is right.

**Argument 1. No need to switch**: Suppose you choose door number 1. The
probability that the car is behind door 1 is initially 1/3
(since there are three doors, and the car has an equal chance
of being behind each). Then suppose the host opens door
number 3 and shows you that there is a goat behind it. We
then need to calculate a conditional probability--the
probability that the car is behind door 1, *given* that
there is a goat behind door 3. Since there are only two doors
left, and there is an equal chance that the car is behind each
of them, this probability is 1/2. But similarly, the
probability that the car is behind door 2, given that there is
a goat behind door three, is also 1/2. So whether you stick
with door 1 or switch to door 2, your chance of winning is
1/2. So it really makes no difference whether you switch or
not.

**Argument 2. You should switch**: Suppose you choose door number 1. There are
three possibilities; either the car is behind door 1, or door
2, or door 3. Each of these possibilities has the same
probability (1/3). In each of the three cases, consider which
door the host will open. If the car is behind door 1, the
host could open either door 2 or door 3. In this case, if you
stick with your original choice you win the car, but if you
switch to the remaining door you lose. If the car is behind
door 2, the host will open door 3. In this case, if you stick
with your original choice you lose, but if you switch, you
win. Finally, if the car is behind door 3, the host will open
door 2. Again, if you stick with your original choice you
lose, but if you switch, you win. Remember that each of the
three possibilities has a probability of 1/3, and note that
they are mutually exclusive (the car is only behind one door).
If you switch, you will win in two cases out of three
(probability 2/3), but if you stick you will only win in one
case out of three (probability 1/3). So you should switch
doors, since it doubles your chance of winning.

Which argument do you think is right? Choose your answer:

If you're still not convinced about which answer is correct and why, don't worry - try it yourself. The simulation below keeps track of which strategy you use, and how frequently you win using each strategy, so you can see which strategy wins more often. Incidentally this game show puzzle is also called "The Monty Hall Problem", and it has got an interesting story behind.