Expected Value

[sammy1245]

A 5-card hand is dealt from a standard 52 card deck. If the hand contains at least one king, you win $10; otherwise you lose $1, what is the expected value of the game.

I'm not sure how to even start this one out.
Any help is greatly appreciated!

 

[soroban]

Hello, Sammy!

I must assume you know the Expected Value formula.
Then all we need are the probabilities.

A 5-card hand is dealt from a standard 52 card deck.
If the hand contains at least one King, you win $10; otherwise you lose $1.
What is the expected value of the game?

$\displaystyle \text{There are: }\:{52\choose5} \,=\,2,598,\!960\text{ possible hands.}$

$\displaystyle \text{To get }no\text{ Kings, we must get 5 cards from the 48 Other cards,}$
. . $\displaystyle \text{There are: }\,{48\choose5} \,=\,1,712,304\text{ ways.}$

\(\displaystyle \text{Hence: }\(\text{no Kings}) \;\;=\;\;\frac{1,712,304}{2,598,960} \;\;=\;\;\frac{35,673}{54,145} \quad \hdots\text{ lose \$1}\)

\(\displaystyle \text{and: }\(\text{at least one King}) \;\;=\;\; 1-\frac{35,673}{54,145} \;\;=\;\;\frac{18,472}{54,145} \quad\hdots \text{ win \$10}\)


$\displaystyle \text{Therefore: }\:EV \;\;=\;\;\left(\frac{18,472}{54,145}\right)(+10) + \left(\frac{35,673}{54,145}\right)(-1) \;\;=\;\;\frac{149,047}{54,145}\;\;=\;\;2.752738...$


$\displaystyle \text{You can expect to }win\text{ an average of \$2.75 per game.}$