9.2 Expected value

When outcomes differ, an ordinary average gives each outcome equal weight. Expected value weights each outcome by how likely it is. This lets you compare games, costs, and decisions that may have different outcomes each time.

This is intermediate probability. Work through OpenStax's expected-value examples after learning discrete distributions. The calculation extends the weighted averages and finite sums from Module 1.4: probability supplies each outcome's weight.

A basic idea: weight each possible result

If $X$ takes values $x_1,\ldots,x_m$ with probabilities $p_1,\ldots,p_m$, its expectation is

$$E[X]=\sum_{i=1}^{m}x_i p_i.$$

The probabilities must add to one. Expected value has the same units as $X$: rupees for a payoff, minutes for travel time. It describes a long-run average under repeated comparable trials; it need not be a value that can occur in one trial.

A quick tour of properties

Adding a fixed amount $c$ to every outcome adds $c$ to the expectation: $E[X+c]=E[X]+c$. Scaling every outcome by $a$ gives $E[aX]=aE[X]$. For a cost $C$ and reward $R$, define net payoff $N=R-C$ before calculating. Keeping gross and net values separate avoids a common error.

Application 1: a paid game

A game costs ₹$2$. With probability $0.25$ it pays ₹$8$; otherwise it pays nothing. What is the expected net payoff?

1. A win nets $8-2=6$ rupees; a loss nets $-2$ rupees.
2. $E[N]=0.25(6)+0.75(-2)=1.5-1.5=0$ rupees.
3. Equivalently, expected reward is $0.25(8)=2$ rupees, then subtract the ₹$2$ fee.

A zero expectation does not mean every play pays zero.

Application 2: travel time

A route takes $20$ minutes on a clear day with probability $0.7$ and $50$ minutes in congestion with probability $0.3$. Its expected travel time is

$$E[T]=20(0.7)+50(0.3)=29\text{ minutes}.$$

No actual trip lasts 29 minutes in this two-outcome model. The number is useful for comparing this route with alternatives over many trips.