Question

you pick a card at random from an ordinary deck of 52 cards. if the card is an ace, you get 9 points. if not, you lose 1 point.
what is the expected value? should you play the game?
write the equation for the expected value.
$e(v)=sum_{i = 1}^{n}a_{i}p(x_{i})$

Explanation:

Step1: Identify probabilities and values

The probability of picking an ace, $P(\text{Ace})=\frac{4}{52}$ and the value for an ace is $9$ points. The probability of not - picking an ace, $P(\text{Not Ace})=\frac{48}{52}$ and the value for non - ace is $- 1$ point.

Step2: Use the expected - value formula

The expected - value formula is $E(X)=\sum_{i}x_ip_i$. Here, $E(X)=\frac{4}{52}\times9+\frac{48}{52}\times(-1)$.
$E(X)=\frac{36}{52}-\frac{48}{52}=\frac{36 - 48}{52}=-\frac{12}{52}=-\frac{3}{13}$ points.
For a fair game, $E(X) = 0$. Let the value of an ace be $x$ for a fair game. Then the equation for expected value is $0=\frac{4}{52}x+\frac{48}{52}\times(-1)$.
Solve for $x$:
\[

$$\begin{align*} 0&=\frac{4}{52}x-\frac{48}{52}\\ \frac{48}{52}&=\frac{4}{52}x\\ x& = 12 \end{align*}$$

\]

Answer:

The expected value of the game is $-\frac{3}{13}$ points, so you should not play the game. For a fair game, the value of an ace should be $12$ points.