Expected Value Help!

I posted a question really similiar to this one earlier.
This one is worded a little different, and I don't know if it needs you to do anything differently.
I don't know why this is so confusing for me, even after I talked to the teacher about it.
If you could go step by step through this problem as well, that'd be great!

Thank you for your time!
It's greatly appreciated!

A card is drawn from a 52-card deck. If the card is a diamond, you win $10; otherwise you lose $4. What is the expected value of the game?

Hello, Sammy!

Okay, step-by-step . . .

A card is drawn from a 52-card deck.
If the card is a diamond, you win $10; otherwise you lose $4.
What is the expected value of the game?

Click to expand...

\(\displaystyle \begin{array}{ccccccc}P(\text{Diamond}) &=& \frac{13}{52} &=& \frac{1}{4} & \text{win \$10} \\ \\[-3mm] P(\text{Other}) &=& \frac{39}{52} &=& \frac{3}{4} & \text{ lose \$4}\end{array}\)


\(\displaystyle EV \;=\;\left(\tfrac{1}{4}\right)(+10) + \left(\tfrac{3}{4}\right)(-4) \;=\;-\tfrac{1}{2}\)


You can expect to lose an average of 50 cents per game.

It is multiplication for each
Find the number of possible wins (w) and number of possible losses (l)

(w)(10)+(l)(-4)=Expected value