Skill: Interpret expected value in context

Expected value is a probability-weighted average. It tells you the long-run average result if the same situation is repeated many times.

For a discrete random variable $X$,

$$\mathbb{E}(X)=\sum xP(X=x).$$

This means: multiply each possible outcome by its probability, then add the products.

Example: A game costs $3 to play. The prize distribution is:

\vert Prize $x$ \vert Probability $P(X=x)$ \vert \vert --- \vert ---: \vert \vert 0$\vert$0.50$\vert \vert$$4$\vert$0.40$\vert \vert$$20$\vert$0.10$ \vert

Let $X$ represent the prize money won.

Step 1: Multiply each outcome by its probability.

$$0(0.50)=0$$ $$4(0.40)=1.60$$ $$20(0.10)=2.00$$

Step 2: Add the products.

$$\mathbb{E}(X)=0+1.60+2.00=\$3.60$$

The expected prize is 3.60$per game. This does **not** mean that a player can win exactly$$3.60 in one game; the possible prizes are only 0$,$$4$, and$$20$.

Over many plays, however, the average prize won per game would approach 3.60$. Since the game costs$$3$, the expected net result is

$$\$3.60-\$3.00=\$0.60.$$

So, in the long run, a player would gain an average of 0.60$ per game, even though individual games may result in a loss or a much larger win.