Key takeaway: The expected value of a random outcome is the sum of each possible value times its probability, giving the long-run average result per trial.
Grade 12 – PrecalculusS-MD.A.2Probability and Counting
Definition of Expected Value
The expected value is what you would earn on average if a chance process were repeated many times . It need not be a value that can actually occur on a single trial. A fair die has an expected roll of 3.5, a number no roll can show.
To compute it, multiply every possible payoff by its probability and add the products. Likely outcomes pull the average toward themselves, and rare ones pull only slightly. Insurers, game designers and investors use expected values to decide whether a bet is fair: a negative net figure means the player loses money over time.
Formula and notation
For \(k\) outcomes \(x_i\), each with probability \(p_i\):
\[E(X)=\sum_{i=1}^{k} x_i\,p_i.\]
The probabilities must add to 1. The net figure equals the expected payout minus the cost to play.
Key facts
- A game is fair when its net expected value equals zero.
- The result describes the long run, not what happens in any single play.
- Adding a constant payout to every outcome raises the average by that constant.
- Check first that all the probabilities in your list add up to exactly 1.
Expected Value example
At a school carnival, a game costs \$3 to play. You win \$20 with probability \(\frac{1}{12}\), win \$5 with probability \(\frac{1}{4}\), and win nothing otherwise. What is the expected prize, and the expected net result per play?
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Probability of winning nothing: \(1-\frac{1}{12}-\frac{3}{12}=\frac{8}{12}\).
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Expected prize: \(20\cdot\frac{1}{12}+5\cdot\frac{3}{12}+0\cdot\frac{8}{12}=\frac{20+15}{12}=\frac{35}{12}\approx 2.92\).
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Net result: \(\frac{35}{12}-3=-\frac{1}{12}\approx -0.08\).
The expected prize is about \$2.92, so the game costs the player about \$0.08 per play.
Common mistake
Watch out
Hamza averages the three prizes, \(\frac{0+5+20}{3}\approx 8.33\), and decides the game is a bargain.
Why it is wrong
The prizes are not equally likely, so a plain average is wrong. Each prize must be weighted by its probability. The \$20 prize comes up only one time in twelve, which pulls the true average down to \$2.92.
Related terms and lessons
- Precalculus · Expected Value and DecisionsLessonWorksheetQuiz
- AP Statistics · Probability and Random VariablesLessonWorksheetQuiz
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