Expected value and fair price

Idea: One number for an uncertain payoff

Multiply each payoff by its probability and add them up.

• It is the average per play over a long run
• It need not be a result that can happen

Formula: Expected value

$E[X]=\sum x\times P(x)$

Sum over every outcome: the payoff times its probability.

Example: Win 10 or lose 2

A bet wins $10$ with probability $\frac{1}{4}$ and otherwise loses $2$. What is its expected value?

1. $\frac{1}{4}\times 10-\frac{3}{4}\times 2=1$
2. In $400$ plays: $100\times 10-300\times 2=400$, so $\frac{400}{400}=1$

Idea: The mean of a die

A fair die pays its face: $\frac{1+2+3+4+5+6}{6}=\frac{7}{2}$.

• $n$ sides from $1$: the mean is $\frac{n+1}{2}$
• $8$ sides: $\frac{8+1}{2}=\frac{9}{2}$
• $2$ dice: $2\times \frac{7}{2}=7$

Example: A game with a fee

A die pays $18$ on a $6$, $6$ on a $5$, nothing else. Playing costs $5$. What is the game worth?

1. Payout: $\frac{1}{6}\times 18+\frac{1}{6}\times 6+\frac{4}{6}\times 0=4$
2. Net at the fee: $4-5=-1$, so pass
3. Fair price: the expected payout, $4$

Common mistake: Gross versus net

In net payoffs the fee comes off every face, including those that pay nothing.

Instead Write $-5$ there, not $0$: $\frac{1}{6}\times 13+\frac{1}{6}\times 1+\frac{4}{6}\times (-5)=-1$.

Idea: A probability is an expected value

A ticket paying $1$ if an event happens is worth its probability.

• A total of $7$ on $2$ dice: $\frac{6}{36}=\frac{1}{6}$
• At least one $6$: $1-{\left(\frac{5}{6}\right)}^2=\frac{11}{36}$

Quick check: Even faces pay

A fair die pays its face value when it is even and nothing otherwise. What is the fair price to play?

Idea: Say it out loud

Name the outcomes, their probabilities, and whether amounts are before or after the fee.

• End with a decision: "worth $4$, so at $5$ I pass"
• The fair price answers the follow-up