A probability distribution lists every value and its chance. One number summarizes where it is centered: the expected value, the long-run average of the variable over many repetitions.
Key idea
The expected value (mean) of a discrete random variable is $$E(X) = \mu = \sum x \cdot P(x),$$ each value weighted by its probability. It is the average you would see in the long run, and it does not have to be a possible value of $X$.
Why it works
Imagine repeating the process $N$ times, with $N$ large. The value $x$ shows up about $N \cdot P(x)$ times, so the total of all results is about $\sum x \cdot N P(x)$. Divide by $N$ to get the average: $\sum x P(x)$. It is the same as a mean computed from a frequency table, with probabilities in place of relative frequencies.
Worked example: One die
Find the expected value of one roll of a fair die.
- Each value from 1 to 6 has probability $\frac{1}{6}$.
- $$E(X) = \frac{1}{6}(1 + 2 + 3 + 4 + 5 + 6) = \frac{21}{6} = 3.5$$
- No roll ever shows 3.5. It is the long-run average of many rolls.
Answer: $3.5$
Worked example: From a table
For the number of pets $X$ with $P(0) = 0.30$, $P(1) = 0.25$, $P(2) = 0.25$, $P(3) = 0.15$, $P(4) = 0.05$, find $E(X)$.
- $$E(X) = 0(0.30) + 1(0.25) + 2(0.25) + 3(0.15) + 4(0.05)$$
- $$= 0 + 0.25 + 0.50 + 0.45 + 0.20 = 1.4$$
Answer: $1.4$ pets per household, on average.
Worked example: A game
A game costs \$2. You win \$10 with probability $0.1$, \$3 with probability $0.2$, and nothing otherwise. What is the expected net gain per game?
- Expected winnings: $$10(0.1) + 3(0.2) + 0(0.7) = 1 + 0.6 = 1.6$$ dollars.
- Net: $1.6 - 2 = -0.4$.
- On average you lose 40 cents per game; over 100 games, about \$40.
Answer: $-\$0.40$ per game.
Common mistake
Thinking the expected value must be a value the variable can take, or the most likely value. The expected number of pets is 1.4, though no household has 1.4 pets. Check: E(X) is an average, a balance point, not a prediction for one trial.