Expected Value Help!

[sammy1245]

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?

 

[soroban]

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?

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


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


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

 

[papermate2004]

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

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