# Lesson 7 Independence

## Motivating Example

Many gamblers believe that after a string of losses, they are “due” for a win. Consider a gambler who repeatedly bets on reds in roulette, an event with probability $$18 / 38 \approx 0.474$$. The ball has not landed in a red pocket on any of the last 4 spins of the wheel. Does this make it more likely that he will win on the next spin?

This is really a conditional probability question in disguise. We want to know the probability that the ball will land in a red pocket on the 5th spin, given that it did not land in a red pocket on any of the first 4 spins: $P(\text{red on 5th spin}\ |\ \text{not red on first 4 spins}).$

## Theory

To verify that the two probabilities are exactly the same, we do the calculation:

\begin{align*} P(\text{red on 5th spin}\ |\ \text{not red on first 4 spins}) &= \frac{P(\text{not red on first 4 spins} \textbf{ and } \text{red on 5th spin}) }{P(\text{not red on first 4 spins})} \\ &= \frac{ 20^4 \cdot 18 \big/ 38^5 }{20^4 \big/ 38^4} \\ &= \frac{18}{38}. \end{align*}

We see that the conditional probability is $$18 / 38$$, which is the same as the probability that the 5th spin is red if we did not know the outcome of the first 4 spins. In mathematical notation, \begin{align*} P(\text{red on 5th spin}\ |\ \text{not red on first 4 spins}) &= P(\text{red on 5th spin}) \end{align*} When conditioning on one event (e.g., “not red on first 4 spins”) does not change the probability of another event (e.g., “red on 5th spin”), we say that the two events are independent. In this case, whether the gambler wins on the 5th spin is independent of the fact that he has lost each of the last 4 spins. The folk belief that one is “due” for a win after a series of losses is plain wrong and is known as the gambler’s fallacy.

Definition 7.1 (Independence) Two events $$A$$ and $$B$$ are said to be independent if the $\begin{equation} P(B | A) = P(B). \tag{7.1} \end{equation}$
The next result follows by applying the Multiplication Rule (6.1) to the definition of independence.
Theorem 7.1 Two events $$A$$ and $$B$$ are independent if and only if their probabilities multiply: $\begin{equation} P(A \textbf{ and } B) = P(A) P(B). \tag{7.2} \end{equation}$

Proof. First, we assume (7.1) and show that (7.2) holds: \begin{align*} P(A \textbf{ and } B) &= P(A) P(B | A) & \text{by the Multiplication Rule} \\ &= P(A) P(B) & \text{by assumption}. \end{align*}

Conversely, we assume (7.2) and show that (7.1) holds: \begin{align*} P(B | A) &= \frac{P(A \textbf{ and } B)}{P(A)} & \text{by the definition of conditional probability} \\ &= \frac{P(A) P(B)}{P(A)} & \text{by assumption} \\ &= P(B). \end{align*}

Here is an example that combines several concepts from the past few lessons.

Example 7.1 You and a fellow castaway are stranded on a desert island, playing dice for the last banana. Two dice will be rolled. If the biggest number is 1, 2, 3, or 4, then Player 1 wins. If the biggest number is 5 or 6, then Player 2 wins. Would you rather be Player 1 or Player 2?
Solution. Player 2 has a $$20/36 = 0.5556$$ chance of winning. Here’s why: \begin{align*} P(\text{biggest number is 5, 6}) &= 1 - P(\text{biggest number is 1, 2, 3, 4}) \\ &= 1 - P(\text{1st die is 1, 2, 3, 4} \textbf{ and } \text{2nd die is 1, 2, 3, 4}) \\ &= 1 - \frac{4}{6} \cdot \frac{4}{6}\ \ \text{(by independence)} \\ &= 1 - \frac{16}{36} \\ &= \frac{20}{36} \end{align*}

## Examples

1. One card is dealt off the top of a well-shuffled deck of cards. Is the event that the card is a heart independent of the event that the card is an ace?
2. Two cards are dealt off the top of a well-shuffled deck of cards. Is the event that the first card is a heart independent of the event that the second card is a heart?
3. In the dice game Yahtzee, five dice are rolled. The outcomes of the five dice are independent. What is the probability of rolling a “Yahtzee” (i.e., when all five dice show the same number)?