A game pays $10 if you roll a 6 on a fair die, and $0 otherwise. What is the expected payout?
A$0
B$1.67
C$5.00
D$10.00
E(X) = 10 × (1/6) + 0 × (5/6) = 10/6 ≈ $1.67. This is the long-run average payout per game — not the $10 you could win or the $0 you usually get. Expected value weights each outcome by its probability.
Question 2 True / False
If the expected value of a random variable is 3.5, you should expect the outcome 3.5 to occur frequently in repeated trials.
TTrue
FFalse
Answer: False
Expected value is a long-run average, not a prediction for individual trials. For a fair six-sided die, E(X) = 3.5, but 3.5 never actually occurs — it is the average you converge to as you roll many times. This is the most common misconception about expected value.
Question 3 Short Answer
Why is expected value described as a 'weighted average' rather than a simple average of possible outcomes?
Think about your answer, then reveal below.
Model answer: Because each possible outcome is weighted by its probability. Outcomes that are more likely contribute more to the expected value than rare outcomes, even if the rare outcomes have large magnitudes.
A simple average treats all outcomes equally. Expected value uses Σ x · P(X = x), so an outcome worth 100 with probability 0.01 contributes only 1, while an outcome worth 5 with probability 0.5 contributes 2.5. The weighting by probability is what makes expected value meaningful as a predictive quantity.