# 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 \[\begin{equation} P(B | A) = P(B). \tag{7.1} \end{equation}\]

**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:
\[\begin{align*}
P(A \textbf{ and } B) &= P(A) P(B | A) & \text{by the Multiplication Rule} \\
&= P(A) P(B) & \text{by assumption}.
\end{align*}\]

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

- 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?
- 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?
- 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)?