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05 · Independence

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Probability, lesson 5

Independence

When knowing one thing tells you nothing, and the traps around it.

Some news changes nothing: learn that the red die shows 3, and a total of 7 still has chance 1/6. This lesson is about events like that, and why they let chances multiply. You need P(A | B) and the rule P(A and B) = P(A) × P(B | A).

By the end, you can:

  1. Test whether two events are independent.
  2. Multiply the chances of independent events, and find "at least one".
  3. Spot the gambler's fallacy, and the case where a streak does tell you something.

Part 1. News that changes nothing

Test whether two events are independent.

Learn that at least one die shows a 6, and a total of 7 moves from 1/6 to 2/11. Learn that the red die shows 3, and it stays at 1/6. Some news matters and some doesn't. Here's how to tell which, for any two events.

Example 1.1

An even red die and a total of 7

Roll a red die and a blue die. Does learning that the red die is even change the chance of a total of 7?

  1. The chance of a 7 on its own6/36 = 1/6

    The six squares that run corner to corner.

  2. Keep the squares the news allows3 × 6 = 18

    The red die shows 2, 4 or 6, with any blue face.

  3. Count the 7s among them3

    2 + 5, 4 + 3 and 6 + 1.

  4. The chance of a 7, given the news3/18 = 1/6

    The same as before. The news changed nothing.

AnswerNo. The chance stays 1/6.

234 567 345 67 845 678 956 78 9106 789 1011 78 91011 12 11223 3445 566 blue die red die Red die even 18 squares left Total of 7 3 of them 3 / 18 = 1/6 the same as 6 / 36
234 567 345 67 845 678 956 78 9106 789 1011 78 91011 12 11223 3445 566 blue die red die Red die even: 18 squares left 3 of them total 7: 3/18 = 1/6
Figure 1The news keeps the 18 squares with an even red die. 3 of them total 7: the same 1/6 share that the 7s have of all 36.

Why does it stay put? Whatever the red die shows, exactly one blue face makes 7. So every row holds the same share of 7s, and keeping some rows can't change it. A total of 8 is different: the even rows hold 3 of the 5 eights, so the same news lifts its chance from 5/36 to 3/18.

Independent events

A and B are independent when knowing one happened doesn't change the chance of the other: P(B | A) = P(B). Then the multiplication rule for chances loses its condition. For the dice, 1/2 × 1/6 = 1/12, and 3 of the 36 squares are both.

P(A and B) = P(A) × P(B)

The trap is mixing up independent with can't happen together. A double and a total of 7 can't both happen, since no double adds to 7. So learning you rolled a double drops the chance of a 7 from 1/6 to 0. Events that rule each other out are as far from independent as events get.

Your turn

Roll a red die and a blue die. Are "the red die is even" and "the total is even" independent?

Each has chance 1/2. Both happen when both dice are even: 3 × 3 = 9 squares, and 9/36 = 1/4 = 1/2 × 1/2. Given an even red die, the total is even only if the blue die is even, still a 1/2 chance. The total depends on the red die, but the chance of an even total doesn't.

Part 2. Chances that multiply

Multiply the chances of independent events, and find "at least one".

Keep going with Pro

You've done the first part. Pro opens the rest of this lesson, and every other one.

Still to come
  1. Chances that multiply
  2. The coin has no memory
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Next lessonExpected valueThe average outcome, and why it decides almost every bet. Start lesson 6

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