What Does E(X) = 0 Mean for a “Fair Game”?

A game is fair when $E(X)=0$, where $X$ represents the player’s net gain. Over many repetitions, the player is expected to neither gain nor lose overall.

The Reasoning

For a discrete random variable, the expected value is

$$E(X)=\sum xP(X=x).$$

Each possible net gain $x$ is multiplied by its probability, and these products are added. Expected value is a long-run average, not a prediction of the result from one game.

Fairness is based on the player’s perspective. If $X$ were defined as the organizer’s net gain, the signs would reverse, although zero would still represent fairness. Therefore, define the variable before calculating.

Suppose a game costs 2 points to play. The player has a probability of $0.4$ of receiving 5 points and a probability of $0.6$ of receiving nothing.

Therefore,

$$E(X)=3(0.4)+(-2)(0.6)=1.2-1.2=0.$$

The game is fair because the expected net gain is zero. Although the player loses more often than they win, the larger winning amount exactly balances the more frequent losses in the long run.

The calculation also shows why the most likely outcome does not determine fairness. Here, losing 2 points is more likely, but its probability-weighted effect is exactly offset by the less likely gain of 3 points.

A common misconception is that a fair game must give equal probabilities of winning and losing. This is false: fairness depends on the probability-weighted net gains, not equal outcome probabilities.

Exam Technique

This is SL 4.7: Discrete random variables. In an IB Math AI response, define $X$ as net gain, show the full expected-value calculation, and conclude in context: “Since $E(X)=0$, the game is fair.” Do not use prize money alone without subtracting the cost of playing.