What Does E(X) = 0 Mean in the Context of a Fair Game?

In a fair game, $E(X)=0$ means that the player’s expected net gain per game is zero. Over many repetitions, the player is expected to neither gain nor lose overall.

This is part of IB Mathematics: Analysis and Approaches, SL 4.7: Discrete random variables. The conclusion assumes that $X$ represents the player’s net gain, including both winnings and the cost of playing.

The expected value of a discrete random variable is

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

Each possible net gain, $x$, is multiplied by its probability, and the products are added. Consider a game that costs $2$ points to play. A player has probability $\frac{1}{4}$ of receiving $8$ points and probability of receiving nothing.

Therefore,

$$E(X)=6\left(\frac{1}{4}\right)+(-2)\left(\frac{3}{4}\right)=0.$$

The game is fair because the expected gain is zero. This does not mean that every player receives zero in each game. Individual outcomes are still either a gain of $6$ points or a loss of $2$ points; zero describes the long-run average.

More formally, expected value is a probability-weighted mean, not a guaranteed outcome. After $n$ independent plays, the expected total net gain is $nE(X)$. Here, this remains zero, although the actual total can vary because of randomness. A positive expected value favours the player; a negative expected value favours the organizer.

For a fair-fee question, let the entry fee be $c$, write each net outcome as prize minus $c$, and solve $E(X)=0$.

A common misconception is to use the prize value rather than the net gain. If there is a cost to play, subtract it from every possible prize before calculating $E(X)$.

In an IB exam response, define $X$, show the full expected-value calculation, and interpret the result in context. If asked to determine a fair entry fee, let the fee be an unknown and solve $E(X)=0$. This method is applicable to both AA SL Paper 1 and Paper 2, depending on the complexity of the probabilities.