Question

In a card game (using a standard deck of cards), you pay $5 to draw a single card. If you draw an Ace, you win $20. If you draw a face card, you win $10. If you draw a 2, you win $5.

a) Fill in the probability distribution below:

X 5 10 20
P(X)


b) What is the expected value?
c) Is the game fair? Why or why not?

Answer 1

@jim_thompson5910

Answer 2

The probability of drawing an ace is \(\dfrac{4\text{ aces}}{52\text{ total cards}}\), or \(\dfrac{\binom41}{\binom{52}1}=\dfrac{4}{52}\).
The same probability for drawing a 2, since there are only four possible 2's.
There is a total of 12 face cards, which has a probability of \(\dfrac{\binom{12}1}{\binom{52}1}=\dfrac{12}{52}\).

Answer 3

The expected value of \(X\), assuming \(X\) denotes gain only (as opposed to net gain, which would be prize money minus what you pay), is
\[E(X)=\sum_{\text{all }x}x~p(x)=5\cdot\frac{1}{52}+10\cdot\frac{12}{52}+20\cdot\frac{1}{52}\]
If \(X\) is supposed to denote net gain, then the expected value is different:
\[E(X)=\sum_{\text{all }x}x~p(x)=(5-5)\cdot\frac{1}{52}+(10-5)\cdot\frac{12}{52}+(20-5)\cdot\frac{1}{52}\]

Answer 4

thank you! @SithsAndGiggles