Question:
I draw a card from a standard 52-card deck.  If I draw an Ace, I win 1 dollar.  If I draw a 2 through 10, I win a number of dollars equal to the value of the card.  If I draw a face card (Jack, Queen, or King), I win 20 dollars.  If I draw a $\clubsuit$, my winnings are doubled, and if I draw a $\spadesuit$, my winnings are tripled.  (For example, if I draw the $8\clubsuit$, then I win 16 dollars.)  What would be a fair price to pay to play the game?  Express your answer as a dollar value rounded to the nearest cent.

Your answer should be a number with two digits after the decimal point, like $21.43$.

Answer:
Let $E_1$ be the expected winnings if a $\heartsuit$ or $\diamondsuit$ is drawn.  Since the probability that any particular rank is drawn is the same for any rank, the expected value is simply the average all the winnings for each rank, so \[ E_1 = \frac{1}{13}(\$1+\$2+\cdots+\$10+(3\times\$20)) = \$\frac{115}{13}. \]Let $E_2$ be the expected winnings if a $\clubsuit$ is drawn and $E_3$ the expected winnings if a $\spadesuit$ is drawn.  Since drawing a $\clubsuit$ doubles the winnings and drawing a $\spadesuit$ triples the winnings, $E_2 = 2E_1$ and $E_3 = 3E_1$.  Since there is an equal chance drawing each of the suits, we can average their expected winnings to find the overall expected winnings.  Therefore the expected winnings are \[ E = \frac{1}{4}(E_1 + E_1 + E_2 + E_3) = \frac{1}{4}(7E_1) = {\$\frac{805}{52}}, \]or about $\boxed{\$15.48}$, which is the fair price to pay to play the game.